Answer:
0.08
Explanation:
The alpha particle suffers a head-on collision with the gold nucleus, so it retraces it path after the collision.
Let us take the masses of the particles in atomic mass units.
The initial momentum and kinetic energy of the gold nucleus is 0(since it is stationary). So, applying conservation of momentum and energy, we get the following two equations:
[tex]m_{1}u_{1}=m_{1}v_{1}+m_{2}v_{2}[/tex] ..........(1)
[tex]\frac{1}{2}m_{1}u_{2}^{2}=\frac{1}{2} m_{1} v_{1}^{2} +\frac{1}{2} m_{2} v_{2}^{2}[/tex] ..........(2)
where,
[tex]m_{1}[/tex] = mass of the alpha particle = 4 units
[tex]m_{2}[/tex] = mass of the gold nucleus = 197 units
[tex]u_{1}[/tex] = initial velocity of the alpha particle
[tex]v_{1}[/tex] = final velocity of the alpha particle
[tex]v_{2}[/tex] = final velocity of the gold nucleus
Now, we shall substitute the value of [tex]v_{2}[/tex] from equation (1) in equation (2). After some simplifications, we get,
[tex]u_{1}^{2}=v_{1}^{2}+\frac{m_{1}}{m_{2}} (u_{1}^{2}+v_{1}^{2}-2u_{1}v_{1})[/tex]
Dividing both sides by [tex]u_1^2[/tex] and substituting [tex]x=\frac{v_1}{u_1}[/tex] and [tex]k=\frac{m_1}{m_2}[/tex] , we get,
[tex]1=x^2+k(1+x^2-2x)\\[/tex]
or, [tex]x^2(k+1)-2kx+(k-1)=0[/tex]
Here, [tex]k=\frac{m_1}{m_2}=\frac{4}{197}=0.02[/tex]
Therefore, [tex]x=\frac{2(0.02)\pm\sqrt{(2\times0.02)^2-(4\times1.02\times-0.98)} }{2\times1.02}[/tex]
or, [tex]x = 1, -0.96[/tex]
Our required solution is -0.96 because the final velocity([tex]v_1[/tex]) of the alpha particle will be a little less the initial velocity([tex]u_1[/tex]). The negative sign comes as the alpha particle reverses it's direction after colliding with the gold nucleus.
Fractional change in kinetic energy is given by,
[tex]\delta E=\frac{\frac{1}{2} m_1u_1^2-\frac{1}{2}m_1v_1^2 }{\frac{1}{2}m_1u_1^2 }=1-x^2=0.078\approx0.08[/tex]
The alpha particle can lose a significant amount of its kinetic energy in a head-on elastic collision with a gold nucleus due to the gold nucleus's much larger mass. The original kinetic energy of the alpha particle is converted to potential energy before being transferred mostly to the gold nucleus. Specific loss would depend upon the alpha particle's original kinetic energy.
Explanation:The question pertains to the concept of elastic collisions, specifically between an alpha particle and a gold nucleus. In an elastic collision, both momentum and kinetic energy are conserved. However, while total energy is conserved, individual kinetic energies of colliding particles may change. Since the gold nucleus, which was initially at rest, is significantly more massive (197 units) than the alpha particle (4.0 units), the alpha particle can lose a significant amount of its kinetic energy in a head-on collision.
To calculate the fractional loss of kinetic energy for the alpha particle in this instance, we would use the principle of conservation of kinetic energy and momentum. The kinetic energy of an alpha particle before the collision is transformed into both kinetic and potential energy during the collision as it approaches the gold nucleus until its original energy is converted to potential energy.
Upon collision, a good proportion of this energy is transferred to the gold atom, given its much larger mass. However, necessary calculations would require specific knowledge of the kinetic energy of the alpha particle before the collision, which may vary depending upon the specific nuclear decay process involved.
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A 1 in diameter solid round bar has a groove 0.1 in deep with a 0.1 in radius machined into it. The bar is made of AISI 1040 CD steel and is subjected to purely reversed torque of 1800 lbf∙in. Determine the maximum shear stress taking the effect of the groove into account.
Maximum shear stress in a round steel bar with a groove, subjected to reversed torque, is found by computing the nominal shear stress and multiplying it by the stress concentration factor of the groove. The stress concentration factor must be known to complete the calculation.
Explanation:To determine the maximum shear stress in an AISI 1040 CD steel round bar which is subjected to purely reversed torque of 1800 lbf∙in, and has a groove machined into it, the effect of the shape modification by the groove needs to be considered. This groove effect is described using stress concentration factor (Kt), which shows the increase in maximum stress over the nominal stress because of the change in geometry.
The method involves determining the nominal shear stress (τnom) which equals the torque (T) divided by the polar moment of inertia (J) given as J = (π * (d/2)^4)/2 for a round rod. Then, multiply τnom by the stress concentration factor of the groove to find the maximum shear stress τmax = Kt * τnom.
However, I see the Kt value for the particular groove shape and size is not provided. This value is usually looked up in standard tables or calculated using specific formulas/fixtures based on the groove's size and shape. Once Kt is known, you can compute the maximum shear stress precisely.
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A baseball of mass m1 = 0.45 kg is thrown at a concrete block m2 = 7.25 kg. The block has a coefficient of static friction of μs = 0.74 between it and the floor. The ball is in contact with the block for t = 0.185 s while it collides elastically.
a. Write an expression for the minimum velocity the ball must have, vmin, to make the block move.
b. What is the velocity in m/s?
Answer:
a. [tex]v = \frac{mu_sm_2g\Delta t}{m_1}[/tex]
b. 21.64 m/s
Explanation:
Let g = 9.81m/s2
a. The weight of the block is product of its mass and gravitational acceleration
[tex]W = m_2g = 7.25*9.81 = 71.1225N[/tex]
which is also the normal force acting on the block from the floor so it stays balanced.
N = 71.225N
The static friction of the block is product of its normal force from the floor and the friction coefficient
[tex]F_s = \mu_sN = \mu_sW = mu_sm_2g[/tex]
For the block to move, the force generated by the impact must be at least equal to the static friction.
[tex]F = F_s = mu_sm_2g[/tex]
The impulse is product of this force and time duration of impact.
[tex]I = F\Delta t = mu_sm_2g\Delta t[/tex]
As impulse is generated by change in momentum of the ball, which is product of its mass and velocity v
[tex]I = \Delta p = m_1\Delta v[/tex]
[tex]mu_sm_2g\Delta t = m_1 v[/tex]
[tex]v = \frac{mu_sm_2g\Delta t}{m_1}[/tex]
b. [tex]v = \frac{mu_sm_2g\Delta t}{m_1} = \frac{0.74*7.25*9.81*0.185}{0.45} = 21.64 m/s[/tex]
From Newton's second law, the minimum velocity the ball must have, to make the block move is 21 m/s
COLLISION
There are two types of collision
Elastic collisionInelastic collisionIn elastic collision, both momentum and energy are conserved.
Given that a baseball of mass m1 = 0.45 kg is thrown at a concrete block m2 = 7.25 kg. The block has a coefficient of static friction of μs = 0.74 between it and the floor. The ball is in contact with the block for t = 0.185 s while it collides elastically.
For the concreate block to move, the force applied must be greater than the friction between the block and the floor.
The frictional force = μN
where N = mg
Friction = 7.25 x 9.8 x 0.72
Friction = 51.156
Let assume that the force applied will be equal to the friction. From Newton's second law,
F = Change in momentum / time taken.
That is
F = [tex]m_{1}[/tex]V / t
since the ball is starting from rest, the initial velocity u = 0
a. The expression for the minimum velocity the ball must have, to make the block move will be
F = [tex]m_{1}[/tex]V / t
Make V the subject of formula
V = Ft / [tex]m_{1}[/tex]
substitute F into the formula
V = μ[tex]m_{2}[/tex]g t / [tex]m_{1}[/tex]
b. The velocity in m/s will be calculated by substituting all the parameters into the formula above.
V = (51.156 x 0.185) / 0.45
V = 9.46386 / 0.45
V = 21 m/s
Therefore, the minimum velocity the ball must have, to make the block move is 21 m/s
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A flat sheet with an area of 3.8 m 2 is placed in a uniform electric field of magnitude 10 N/C. The electric flux through the sheet is 6.0 Nm 2 /C . What is the angle (in degrees) between the electric field and sheet's normal vector?
Answer:
The angle between the electric field and sheet's normal vector is 80.96 degrees.
Explanation:
Given that,
Area of the flat sheet, [tex]A=3.8\ m^2[/tex]
Electric field, E = 10 N/C
Electric flux of the sheet, [tex]\phi=6\ Nm^2/C[/tex]
The electric flux is through the sheet is given by the dot product of electric field and the area vector. It is given by :
[tex]\phi=E{\cdot} A[/tex]
or
[tex]\phi=EA\ cos\theta[/tex]
[tex]\theta[/tex] is the angle between electric field and sheet's normal vector
So,
[tex]cos\theta=\dfrac{\phi}{EA}[/tex]
[tex]cos\theta=\dfrac{6}{10\times 3.8}[/tex]
[tex]\theta=cos^{-1}(0.157)[/tex]
[tex]\theta=80.96^{\circ}[/tex]
So, the angle between the electric field and sheet's normal vector is 80.96 degrees. Hence, this is the required solution.
Final answer:
The angle between the electric field and the flat sheet's normal vector, given the electric flux of 6.0 Nm²/C and field magnitude of 10 N/C, is approximately 81.2 degrees.
Explanation:
The question involves calculating the angle between an electric field and a flat sheet's normal vector, given the electric flux and the field magnitude. The formula for electric flux (Φ) is given by Φ = E * A * cos(θ), where E is the electric field strength, A is the area through which the field lines pass, and θ is the angle between the field and the normal to the surface. In this case, we have the electric flux (Φ = 6.0 Nm²/C), the electric field (E = 10 N/C), and the area (A = 3.8 m²). To find the angle θ, we rearrange the equation to solve for the cosine of the angle: cos(θ) = Φ / (E * A).
Substituting the given values, we get cos(θ) = 6.0 / (10 * 3.8), which simplifies to cos(θ) = 0.1579. Taking the arccosine of both sides, we find θ ≈ arccos(0.1579). By calculating this, we find that θ ≈ 81.2°.
Thus, the angle between the electric field and the sheet's normal vector is approximately 81.2 degrees.
A cart is initially moving at 0.5 m/s along a track. The cart comes to rest after traveling 1 m. The experiment is repeated on the same track, but now the cart is initially moving at 1 m/s. How far does the cart travel before coming to rest?
A. 1 m B. 2 m C. 3 m D. 4 m E. 8 m
Answer:
D. 4 m
Explanation:
According to the work-energy theorem, the work done by the force acting on a body modifies its kinetic energy.
[tex]W=\Delta K\\W=K_f-K_i\\W=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}[/tex]
The car comes to rest after a given distance, so [tex]v_f=0[/tex]. Recall the definition of work [tex]W=Fdcos\theta[/tex], here F is the constant force acting on the car, d is its traveled distance and [tex]\theta[/tex] is the angle between the force and the displacement, since friction force acts opposite to the direction of motion [tex]\theta=180^\circ[/tex] :
[tex]-Fd=-\frac{mv_i^2}{2}\\d=\frac{mv_i^2}{2F}[/tex]
We have:
[tex]v_i'=1\frac{m}{s}\\v_i=0.5\frac{m}{s}\\v_i'=2vi[/tex]
Thus:
[tex]d'=\frac{2m(v_i')^2}{2F}\\d'=\frac{2m(2v_i)^2}{2F}\\d'=4\frac{2mv_i^2}{2F}\\d'=4d\\d'=4(1m)\\d'=4m[/tex]
A 0.032-kg bullet is fired vertically at 230 m/s into a 0.15-kg baseball that is initially at rest. How high does the combined bullet and baseball rise after the collision, assuming the bullet embeds itself in the ball?
Answer:
[tex]heigth=83.44m[/tex]
Explanation:
Given data
Baseball mass m₁=0.15 kg
initial speed v₁=0
Bullet mass m₂=0.032 kg
final speed v₂=230 m/s
To find
height h=?
Solution
From conservation of momentum we know that
[tex]m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v\\ (0.032kg)(230m/s)+(0.15kg)(0m/s)=(0.15kg+0.032kg)v\\7.36+0=0.182v\\v=7.36/0.182\\v=40.44m/s[/tex]
Now from the conservation of mechanical energy
[tex]P.E=K.E\\mgh=(1/2)mv^{2}\\ gh=(1/2)v^{2}\\ (9.8m/s^{2} )h=(1/2)(40.44m/s)^{2}\\ (9.8m/s^{2} )h=(817.7m^{2} /s^{2} )\\h=(817.7m^{2} /s^{2} )/9.8m/s^{2}\\ heigth=83.44m[/tex]
Final answer:
To determine the rise after a bullet embeds itself in a baseball, use conservation of momentum to calculate the final velocity of the combined mass, then use conservation of energy to find the maximum height the system achieves.
Explanation:
The student is asking about the vertical rise of a combined system of a bullet and a baseball after a collision in which the bullet embeds itself into the baseball. To determine the height to which the combined mass rises, we first need to apply the principle of conservation of momentum to find the velocity of the combined mass right after the collision. Then we use the conservation of energy to find the maximum height the combined mass reaches.
The conservation of momentum before and after the collision gives us:
Initial momentum = mass of bullet × velocity of the bullet
Final momentum = (mass of bullet + mass of baseball) × final velocity
Since initial and final momentum are equal (assuming no external net forces), we can set them equal to each other to solve for the final velocity.
Next, to find the maximum height, we use the conservation of energy, where the initial kinetic energy of the combined mass is converted to gravitational potential energy at the peak of its rise. This gives us:
Kinetic energy = (1/2) × (mass of bullet + mass of baseball) × (final velocity)2
Potential energy at height = (mass of bullet + mass of baseball) × g (acceleration due to gravity) × height
By equating the kinetic energy right after the collision to the potential energy at the maximum height, we can solve for the height.
A raw egg can be dropped from a third-fl oor window and land on a foam-rubber pad on the ground without breaking. If a 75.0-g egg is dropped from a window located 32.0 m above the ground and a foam-rubber pad that is 15.0 cm thick stops the egg in 9.20 ms, (a) by how much is the pad compressed?(b) What is the average force exertedon the egg after it strikes the pad?
Answer:
N 204.13
Explanation:
Using equation of motion
v² = u² + 2as
u = 0 is the egg was dropped from rest.
v = 2 × 9.8 × 32 = √627.2 = 25.04 m/s
when the egg hit the foam-rubber, the acceleration can be calculated with
a = change in velocity / change in time = - 25.04 / 0.0092 = -2721.74 m/s²
a) how much it is compressed
v² = u² + 2as
- u² = 2 (-2717.4) s
- 627.2 / -5443.48 = s
s = 0.1152 m = 11.52 cm
b) average force exerted on the egg = mΔv / Δt = 25.04 × 0.075 / 0.0092 = 204.13
Write this large number in scientific notation. Determine the values of Vm) and (n) when the following mass of the Earth is written in scientific notation: 5,970,000,000,000,000,000,000,000 (rm kgl) Enter I(ml) and (nl), separated by commas.
Answer : The answer is, 5.97, 24
Explanation :
Scientific notation : It is the representation of expressing the numbers that are too big or too small and are represented in the decimal form with one digit before the decimal point times 10 raise to the power.
For example :
5000 is written as [tex]5.0\times 10^3[/tex]
889.9 is written as [tex]8.899\times 10^{-2}[/tex]
In this examples, 5000 and 889.9 are written in the standard notation and [tex]5.0\times 10^3[/tex] and [tex]8.899\times 10^{-2}[/tex] are written in the scientific notation.
If the decimal is shifting to right side, the power of 10 is negative and if the decimal is shifting to left side, the power of 10 is positive.
As we are given the 5,970,000,000,000,000,000,000,000 in standard notation.
Now converting this into scientific notation, we get:
[tex]\Rightarrow 5,970,000,000,000,000,000,000,000=5.97\times 10^{24}[/tex]
As, the decimal point is shifting to left side, thus the power of 10 is positive.
Hence, the answer is, [tex]5.97\times 10^{24}[/tex]
Now the answer is comparing to [tex]m.\times 10^n[/tex]
So, m = 5.97 and n = 24
Thus, the answer is, 5.97, 24
Two 1.0 cm * 2.0 cm rectangular electrodes are 1.0 mm apart. What charge must be placed on each electrode to create a uniform electric field of strength 2.0 * 106 N/C? How many electrons must be moved from one electrode to the other to accomplish this?
Answer:
The number of electrons that must be moved from one electrode to the other to accomplish this is 1.4 X 10⁹ electrons.
Explanation:
Step 1: calculate the charge on each electrode
Given;
Electric field strength = 2.0 X 10⁶ N/C
The distance between the electrode = 1mm = 1 X 10⁻³ m
Electric field strength (E) = Force (F)/Charge (q)
[tex]E =\frac{Kq}{r^2}[/tex]
where;
E is the electric field strength = 2.0 X 10⁶ N/C
K is coulomb's constant = 8.99 X 10⁹ Nm²/C²
r is the distance between the electrodes = 1 X 10⁻³ m
q is the charge in each electrode = ?
[tex]q = \frac{Er^2}{K} = \frac{(2X10^6)(1X10^{-3})^2}{8.99 X10 ^9}[/tex] = 0.2225 X 10⁻⁹ C
The charge on each electrode is 0.2225 X 10⁻⁹ C
Step 2: calculate the number of electrons to be moved from one electrode to the other.
1 electron contains 1.602 X 10⁻¹⁹ C
So, 0.2225 X 10⁻⁹ C will contain how many electrons ?
= (0.2225 X 10⁻⁹)/(1.602 X 10⁻¹⁹)
= 1.4 X 10⁹ electrons
Therefore, the number of electrons that must be moved from one electrode to the other to accomplish this is 1.4 X 10⁹ electrons.
Final answer:
To create a uniform electric field of 2.0 x 10^6 N/C between two electrodes 1.0 mm apart, a charge of 3.54 x 10^-6 C is required on each electrode. Approximately 2.21 x 10^13 electrons must be moved from one electrode to the other to achieve this.
Explanation:
To solve the immediate question, we first need to calculate the charge required to create a uniform electric field of strength 2.0 x 106 N/C between two rectangular electrodes that are 1.0 mm apart. The relationship between the electric field (E), the charge (Q), the permittivity of free space (ε0), and the distance (d) between the plates is given by the equation E = Q / (ε0 × A), where A is the area of one of the plates. Given the plates have dimensions of 1.0 cm * 2.0 cm, we firstly convert these into meters, giving an area of 0.0002 m2. The permittivity of free space (ε0) is 8.85 x 10-12 C2/N·m2.
Substituting the given values into the formula, we get Q = E × ε0 × A = 2.0 x 106 N/C × 8.85 x 10-12 C2/N·m2 × 0.0002 m2 = 3.54 x 10-6 C. Therefore, this is the charge required on each electrode.
To find the number of electrons that must be moved, we use the relationship between charge and number of electrons, which is given by Q = n × e, where n is the number of electrons and e is the charge of an electron (1.602 x 10-19 Coulombs). So, n = Q / e = 3.54 x 10-6 C / 1.602 x 10-19 C = 2.21 x 1013 electrons. Hence, approximately 2.21 x 1013 electrons need to be moved from one electrode to the other to create the desired electric field.
During the 440, a runner changes his speed as he comes out of the curve onto the home stretch from 18 ft/sec to 38 ft/sec over a 3 second time period. What was his average acceleration over that 3 second period?
Answer:
[tex]6.67ft/s^2[/tex]
Explanation:
We are given that
Initial velocity=u=18ft/s
Final velocity,v=38ft/s
Time=t=3 s
We have to find the average acceleration over that 3 s period.
We know that
Average acceleration,a=[tex]\frac{v-u}{t}{t}[/tex]
Using the formula
Average acceleration,a=[tex]\frac{38-18}{3}ft/s^2[/tex]
Average acceleration,a=[tex]\frac{20}{3}ft/s^2[/tex]
Average acceleration,a=[tex]6.67ft/s^2[/tex]
Hence, the average acceleration=[tex]6.67ft/s^2[/tex]
A ball is launched vertically with an initial speed of y˙0= 50 m/s, and its acceleration is governed by y¨=-g-cDy˙2, where the air drag coefficient cD is given by cD= 0.001 m-1. What is the maximum height that the ball reaches? Compare this to the maximum height achieved when air drag is neglected.
Answer:
Explanation:
Given
acceleration is given by
[tex]a=-g-c_Dv^2[/tex]
where [tex]\ddot{y}=a[/tex]
[tex]\dot{y}=v[/tex]
Also acceleration is given by
[tex]a=v\frac{\mathrm{d} v}{\mathrm{d} s}[/tex]
[tex]ds=\frac{v}{a}dv[/tex]
[tex]\int ds=\int \frac{v}{-g-0.001v^2}dv[/tex]
[tex]\Rightarrow Let -g-0.001v^2=t[/tex]
[tex]-0.001\times 2vdv=dt[/tex]
[tex]vdv=-\frac{dt}{0.002}[/tex]
[tex]at\ v_0=50\ m/s,\ t=-g-0.001(50)^2[/tex]
[tex]t=-g-2.5[/tex]
at [tex]v=0,\ t=-g[/tex]
[tex]\int_{0}^{s}ds=\int_{-g}^{-g-2.5}\frac{-dt}{0.002t}[/tex]
[tex]\int_{0}^{s}ds=\int^{-g}_{-g-2.5}\frac{dt}{0.002t}[/tex]
[tex]s=\frac{1}{0.002}lnt|_{-g}^{-g-2.5}[/tex]
[tex]s=\frac{1}{0.002}\ln (\frac{g+2.5}{g})[/tex]
[tex]s=113.608\ m[/tex]
when air drag is neglected maximum height reached is
[tex]h=\frac{v_0^2}{2g}[/tex]
[tex]h=\frac{50^2}{2\times 9.8}[/tex]
[tex]h=127.55\ m[/tex]
Peg P is driven by the forked link OA along the path described by r = eu, where r is in meters. When u = p4 rad, the link has an angular velocity and angular acceleration of u # = 2 rad>s and u $ = 4 rad>s2. Determine the radial and transverse components of the peg’s acceleration at this instant.
Answer:
The transverse component of acceleration is 26.32 [tex]m/s^2[/tex] where as radial the component of acceleration is 8.77 [tex]m/s^2[/tex]
Explanation:
As per the given data
u=π/4 rad
ω=u'=2 rad/s
α=u''=4 rad/s
[tex]r=e^u[/tex]
So the transverse component of acceleration are given as
[tex]a_{\theta}=(ru''+2r'u')\\[/tex]
Here
[tex]r=e^u\\r=e^{\pi/4}\\r=2.1932 m[/tex]
[tex]r'=e^u.u'\\r'=2.1932 \times 2\\r'=4.3864 m[/tex]
So
[tex]a_{\theta}=(ru''+2r'u')\\a_{\theta}=(2.1932\times 4+2\times 4.3864 \times 2)\\a_{\theta}=26.32 m/s\\[/tex]
The transverse component of acceleration is 26.32 [tex]m/s^2[/tex]
The radial component is given as
[tex]a_r=r''-r\theta'^2[/tex]
Here
[tex]r''=e^u.u'^2+e^u u''\\r''=2.1932 \times (2)^2+2.1932\times 4\\r''=17.5456 m[/tex]
So
[tex]a_r=r''-ru'^2\\a_r=17.5456-2.1932\times (2)^2\\a_r=8.7728 m/s^2[/tex]
The radial component of acceleration is 8.77 [tex]m/s^2[/tex]
The radial and transverse components of the peg's acceleration at the given instant are both 4eu m/s^2.
Explanation:In circular motion, there are two components of acceleration: the radial component (also called centripetal acceleration), directed towards the center of the circle, and the transverse component (also called tangential or azimuthal acceleration), which is perpendicular to the radial component and in the direction of increasing angle.
Given that r = eu, and u = p4 rad, u# = 2 rad/s (angular velocity), u$ = 4 rad/s^2 (angular acceleration), we can calculate these components as follows:
The radial component of acceleration (Ar) can be computed using the formula Ar = r(u#)^2. Substituting the given values, we get Ar = (eu)*(2)^2 = 4eu m/s^2. The transverse component of acceleration (At) can be computed using the formula At = r*u$. Substituting the given values, we get At = eu*4 = 4eu m/s^2.
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Calculate the number of vacancies per cubic meter in iron at 850 °C. The energy for vacancy formation is 1.08 eV/atom. Furthermore, the density and atomic weight for Fe are 7.65 g/cm3 and 55.85 g/mol, respectively.
Answer:
The number of vacancies per cubic meter is 1.18 X 10²⁴ m⁻³
Explanation:
[tex]N_v = N*e[^{-\frac{Q_v}{KT}}] = \frac{N_A*\rho _F_e}{A_F_e}e[^-\frac{Q_v}{KT}}][/tex]
where;
N[tex]_A[/tex] is the number of atoms in iron = 6.022 X 10²³ atoms/mol
ρFe is the density of iron = 7.65 g/cm3
AFe is the atomic weight of iron = 55.85 g/mol
Qv is the energy vacancy formation = 1.08 eV/atom
K is Boltzmann constant = 8.62 X 10⁻⁶ k⁻¹
T is the temperature = 850 °C = 1123 k
Substituting these values in the above equation, gives
[tex]N_v = \frac{6.022 X 10^{23}*7.65}{55.85}e[^-\frac{1.08}{8.62 X10^{-5}*1123}}]\\\\N_v = 8.2486X10^{22}*e^{(-11.1567)}\\\\N_v = 8.2486X10^{22}*1.4279 X 10^{-5}\\\\N_v = 1.18 X 10^{18}cm^{-3} = 1.18 X 10^{24}m^{-3}[/tex]
Therefore, the number of vacancies per cubic meter is 1.18 X 10²⁴ m⁻³
The number of vacancies will be "1.18 × 10²⁴ m⁻³".
Vacancy formationAccording to the question,
Number of atoms, [tex]N_A[/tex] = 6.022 × 10²³ atoms/mol
Iron's density, ρFe = 7.65 g/cm³
Iron's atomic weight, AFe = 55.85 g/mol
Energy vacancy formation, Qv = 1.08 eV/atom
Boltzmann constant, K = 8.62 × 10⁻⁶ k⁻¹
Temperature, T = 850°C or, 1123 K
We know the formula,
→ [tex]N_v[/tex] = N × e [[tex]-\frac{Qv}{KT}[/tex]]
= [tex]\frac{N_A\times \rho Fe}{AFe}[/tex] e [[tex]-\frac{Qv}{KT}[/tex]]
By substituting the above values, we get
= [tex]\frac{6.022\times 10^{23}\times 7.65}{55.85}[/tex] e [[tex]- \frac{1.08}{8.62\times 10^{-5}\times 1123}[/tex]]
= 8.2486 × 10²² × [tex]e^{(-11.1567)}[/tex]
= 8.2486 × 10²² × 1.4279 × 10⁻⁵
= 1.18 × 10¹⁸ cm⁻³ or,
= 1.18 × 10²⁴ m⁻³
Thus the answer above is correct.
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The inner planets formed:
a. by collisions and mergers of planetesimals.
b. in the outer solar system and then were deflected inward by interactions with Jupiter and Saturn.
c. when the Sun's heat destroyed all the smaller bodies in the inner solar system.
d. when a larger planet broke into pieces.
Answer:
a. by collisions and mergers of planetesimals.
Explanation:
Inner planets are planets within 1.5 AU distance from the sun. These are called terrestrial planets because they are somewhat similar to Earth, mainly made of rocks.
The main ingredient of these planets are solar nebula and interstellar dust condensation of which leads to formation of small rock particles. These particles come close to each other under in the influence of gravity and other forces. As the mass of the particles increase they form planetesimals, these planetesimals eventually merge to form planets.
How much horizontal force F must a sprinter of mass 52 kg exert on the starting blocks to produce this acceleration?
Answer:
The horizontal force is 780 N.
Explanation:
Given that,
Mass of sprinter = 52 kg
Suppose A world-class sprinters can accelerate out of the starting blocks with an acceleration that is nearly horizontal and has magnitude 15 m/s².
We need to calculate the horizontal force
Using formula of force
[tex]F = ma[/tex]
Where, m = mass of sprinter
a = acceleration
Put the value into the formula
[tex]F=52\times15[/tex]
[tex]F=780\ N[/tex]
Hence, The horizontal force is 780 N.
Does the KE of a car change more when it accelerates from 23 km/h to 33 km/h or when it accelerates from 33 km/h to 43 km/h?
a. From 23 km/h to 33 km/h
b. From 33 km/h to 43 km/h
c. More information is needed.
Answer:
b. From 33 km/h to 43 km/h
Explanation:
Lets take mass of the car = m
We know that The change kinetic energy KE is give as
[tex]KE=\dfrac{1}{2}m(v^2-u^2)[/tex]
When speed changes from 23 km/h to 33 km/h :
We know that 1 km/h= 0.27 m/s
[tex]KE=\dfrac{1}{2}m(v^2-u^2)[/tex]
[tex]KE=\dfrac{1}{2}\times m((0.27\times 33)^2-(0.27\times 23)^2)[/tex]
KE= 20.412m J
When speed changes from 33 km/h to 43 km/h :
We know that 1 km/h= 0.27 m/s
[tex]KE=\dfrac{1}{2}m(v^2-u^2)[/tex]
[tex]KE=\dfrac{1}{2}\times m((0.27\times 43)^2-(0.27\times 33)^2)[/tex]
KE= 27.702m J
Therefore we can say that when speed changes 33 km/h to 43 km/h ,the kinetic energy will changes more.
An object is moving along the x-axis. At t = 0 it has velocity v0x = 20.0 m/s. Starting at time t = 0 it has acceleration ax = - Ct, where C has units of m/s3. (a) What is the value of C if the object stops in 8.00 s after t = 0? (b) For the value of C calculated in part (a), how far does the object travel during the 8.00 s?
The value of C if the object stops in 8.00 s is 0.625 m/s³.
The distance traveled by the object before stopping in 8 seconds is 40 m.
The given parameters;
initial velocity, [tex]v_0[/tex] = 20.0 m/sinitial time of motion, t = 0acceleration of the object, a = -CtThe value of C is determined by using velocity equation as shown below;
[tex]\frac{dv}{dt} = -Ct\\\\dv = -Ctdt\\\\\int\limits^v_{v_0} \, dv = -\int\limits^t_{t_0} \, Ct \\\\v-v_0= -C[\frac{t^2}{2} ]^t_0\\\\v-v_0 = - \frac{1}{2} Ct^2\\\\0 - 20 = - \frac{1}{2}C(8)^2\\\\-20 = -32 C\\\\C = \frac{20}{32} = 0.625 \ m/s^3[/tex]
The acceleration of the object during 8 seconds is calculated as follows;
a = -Ct
a = -0.625(8)
a = -5 m/s²
The distance traveled by the object before stopping in 8 seconds is calculated as follows;
[tex]v^2 = u^2 + 2as\\\\0 = 20^2 + 2(-5)s\\\\0 = 400 - 10s\\\\10s = 400 \\\\s = \frac{400}{10} \\\\s = 40 \ m[/tex]
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A value of C that stops the object in 8 seconds is 1.25 m/s³. The object travels a total distance of 53.33 meters during this time.
An object is moving along the x-axis with an initial velocity of v₀x = 20.0 m/s and an acceleration of ax = -Ct where C is in m/s³. We'll solve for the value of C and the distance traveled in 8.00 s.
Part (a): Finding the value of C
To determine the value of C, consider the velocity function:
v(t) = v₀x + ∫ax dt = v₀x + ∫-Ct dt
Integrating the acceleration to get the velocity:
v(t) = 20.0 m/s - (C/2)t²
Given that the object stops at t = 8.00 s, set v(8.00) = 0:
0 = 20.0 m/s - (C/2)(8.00 s)²
Solving for C:
20.0 m/s = C × 32 s²
C = 40/32 = 1.25 m/s³
Part (b): Distance traveled in 8.00 s
The displacement function x(t) can be found by integrating the velocity function:
x(t) = ∫v(t) dt
x(t) = ∫[20.0 m/s - (C/2)t²] dt
x(t) = 20.0t - (C/6)t³
Using C = 1.25 m/s³ and t = 8.00 s:
x(8.00) = 20.0(8.00) - (1.25/6)(8.00)³
x(8.00) = 160.0 - (1.25/6)(512)
x(8.00) = 160.0 - 106.67
x(8.00) = 53.33 m
Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a distance "a" from the ring's center.
Answer:
E=[tex]\frac{KQ}{2\sqrt 2a^2}[/tex]
Explanation:
We are given that
Charge on ring= Q
Radius of ring=a
We have to find the magnitude of electric filed on the axis at distance a from the ring's center.
We know that the electric field at distance x from the center of ring of radius R is given by
[tex]E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}[/tex]
Substitute x=a and R=a
Then, we get
[tex]E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}[/tex]
[tex]E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}[/tex]
[tex]E=\frac{KQa}{2\sqrt 2a^3}[/tex]
[tex]E=\frac{KQ}{2\sqrt 2a^2}[/tex]
Where K=[tex]9\times 10^9 Nm^2/C^2[/tex]
Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center=[tex]\frac{KQ}{2\sqrt 2a^2}[/tex]
The magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a distance "a" from the ring's center is E = Q/[8√2πε₀a²]
Electric field due to a charged ring
The electric field due to a charged ring E is given by
E = Qz/4πε₀[√(z² + R²)]³ where
Q = total charge on ring, z = distance of point from axis of ring and R = radius of ring.Magnitude of electric field due to ring
Given that for this ring R = a and z = a, substituting these values into E, the magnitude of the electric field at a is given by
E = Qz/4πε₀[√(z² + R²)]³
E = Qa/4πε₀[√(a² + a²)]³
E = Qa/4πε₀[√(2a²)]³
E = Qa/4πε₀[2√2a³]
E = Q/[8πε₀√2a²]
E = Q/[8√2πε₀a²]
So, the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a distance "a" from the ring's center is E = Q/[8√2πε₀a²]
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Janet wants to find the spring constant of a given spring, so she hangs the spring vertically and attaches a 0.46 kg mass to the spring’s other end. The acceleration of gravity is 9.81 m/s 2 . If the spring stretches 3.7 cm from its equilibrium position, what is the spring constant?
Answer: The value of spring constant is 121.9 N/m
Explanation:
Force is defined as the mass multiplied by the acceleration of the object.
[tex]F=m\times g[/tex]
where,
F = force exerted on the object = ?
m = mass of the object = 0.46 kg
g = acceleration due to gravity = [tex]9.81m/s^2[/tex]
Putting values in above equation, we get:
[tex]F=0.46kg\times 9.81m/s^2=4.51N[/tex]
To calculate the spring constant, we use the equation:
[tex]F=k\times x[/tex]
where,
F = force exerted on the spring = 4.51 N
k = spring constant = ?
x = length of the spring = 3.7 cm = 0.037 m (Conversion factor: 1 m = 100 cm)
Putting values in above equation, we get:
[tex]4.51N=k\times 0.037m\\\\k=\frac{4.51N}{0.037m}=121.9N/m[/tex]
Hence, the value of spring constant is 121.9 N/m
A skydiver jumps out of an airplane. Her speed steadily increases until she deploys her parachute, at which point her speed quickly decreases. She subsequently falls to earth at a constant rate, stopping when she lands on the ground.
The question is incomplete but an analysis of the situation using various useful physics concepts can still be made
Answer:
When she immediately jumps out of the plane, the downward force(weight) is greater than any opposing forces upwards (such as air resistance). So the netforce is downwards and therefore the direction of acceleration is also downwards. The direction of acceleration is always in the direction of the netforce The person is not falling at the rate of free fall (9.8 m/s²) because that is for bodies falling in a vacuum and this person is not, air resistance is very much a factor hereUpon deployment of the parachute, upward forces (air resistance) increases matching the downward forces in size, causing the netforce to be zero. A zero netfroce means zero acceleration which is why the person stops accelerating and falls at a constant rateYou are on the west bank of a river that is flowing north with a speed of 1.2 m/s. Your swimming speed relative to the water is 1.5 m/s, and the river is 60 m wide. What is your path relative to the earth that allows you to cross the river in the shortest time? Explain your reasoning.
Answer:
head straight across the river (perpendicular to the bank).
Explanation:
To cross the river in the shortest time first your velocity should be relative to the earth has to have the largest possible component to the bank
suppose,
S be the swimmer
E be the earth
W be the water
[tex]u_{x/y}[/tex] be the velocity of X relative to Y
resultant velocity relative to E will be:
[tex]u_{S/E}=u_{S/W}+u_{W/E}[/tex]
[tex]u_{W/E}[/tex] is parallel to the bank so,
[tex]u_{S/E}[/tex] has its largest component perpendicular to the bank when [tex]u_{S/W}[/tex] is in that direction
so to cross the river in the shortest time you should straight across the current will then carry you downstream so your path relative to the earth is directed at angle downstream
Final answer:
To cross the river in the shortest time, you must swim perpendicularly to the current. In this case, swimming directly eastward with a speed of 1.5 m/s across a 60 m wide river flowing north at 1.2 m/s will take you across in 40 seconds without being carried downstream.
Explanation:
To cross the river in the shortest time, you must aim to minimize the time spent fighting the water current. The key is to swim in a direction such that your velocity relative to the water combines with the river's velocity to give a resultant path straight across the river. Since the river is flowing north with a speed of 1.2 m/s and your swimming speed relative to the water is 1.5 m/s, the shortest path across would be due east.
If you swim directly eastward, your swimming speed relative to the water ensures that you are moving across the river without being pushed downstream. Thus, the only velocity affecting your eastward crossing is your swimming speed, which is perpendicular to the current. Since the water's current is orthogonal to your motion, it does not affect the time it takes to cross. You'll cross the 60 m wide river in the shortest amount of time by moving at your maximum speed of 1.5 m/s directed perpendicularly to the current.
Considering a swimmer in the given scenario, here's an example to illustrate this concept with numbers:
Width of river: 60 mSpeed of swimmer relative to water: 1.5 m/sSpeed of river current: 1.2 m/sTime to cross = Width of river / Speed of swimmer relative to the water = 60 m / 1.5 m/s = 40 secondsTherefore, the time taken to cross the river is 40 seconds, and the path taken by the swimmer is perpendicular to the flow of the river, relative to the Earth.
You throw a baseball straight up in the air so that it rises to a maximum height much greater than your height. Is the magnitude of the ball’s acceleration greater while it is being thrown or after it leaves your hand? Explain.
The ball's acceleration is constant in magnitude and direction, from the instant it leaves your hand, until the instant it hits the ground, no matter what direction or speed you throw it.
It's the acceleration of gravity, on whatever planet you happen to be standing when you throw the ball.
Final answer:
The baseball's magnitude of acceleration is greater while being thrown than after it leaves the hand due to the additional force applied by the thrower, while in free fall, the ball is subject only to gravity. With air resistance, the ball takes longer to go up than to come back down.
Explanation:
The question pertains to the acceleration of a baseball when thrown straight up into the air. While being thrown, the ball experiences an acceleration greater than the acceleration due to gravity because of the force applied by the person's arm. After the ball leaves the hand, however, the only force acting on it is the force of gravity, which gives it a constant acceleration of approximately 9.81 m/s² downward, regardless of air resistance. In the absence of other forces, the magnitude of acceleration when the ball is in free fall is less than the acceleration imparted to the ball by the thrower's arm.
When air resistance is considered, it acts to slow down the ball as it rises and speeds up as it falls. Therefore, with air resistance, the time it takes for the ball to go up is greater than the time it takes to come back down, because air resistance removes kinetic energy from the ball on the way up, slowing it down more quickly than gravity alone would.
The visible spectrum of sunlight reflected from Saturn’s cold moon Titan would be expected to be (a) continuous; (b) an emission spectrum; (c) an absorption spectrum.
Titan is one of Saturn's largest satellites. The molecules on the Saturn's cold moon titan absorb the light from the Sun light because the atmosphere on the titan is at low temperature.Titan is made of thick layers of ice, hence it is relatively cold. If the sunlight reflects from saturns moon Titan, due to the prescence of cold atmosphere, abosrption spectrum arises. So the Spectrum formed by the reflected light from the titan is absorption spectrum
The correct option is C: Absorption spectrum
The visible spectrum of sunlight reflected from Titan, Saturn's moon, would be an absorption spectrum, as Titan's atmosphere absorbs some wavelengths of sunlight. The term 'absorption spectrum' refers to a spectrum produced when light passes through a cool, dilute gas.
Explanation:The visible spectrum of sunlight reflected from Saturn's moon Titan would be expected to be an absorption spectrum. This is because Titan's atmosphere and surface would absorb some wavelengths of sunlight and reflect the rest, producing an absorption spectrum. There are three types of spectrums: continuous, emission, and absorption. A continuous spectrum is one where all colors (wavelengths) are present without any gaps, which usually represents an ideal black body radiator. An emission spectrum is a spectrum of the electromagnetic radiation emitted by a source. The absorption spectrum, on the other hand, is a spectrum produced when light passes through a cool, dilute gas and atoms in the gas absorb at specific frequencies; since the re-emitted light is unlikely to be emitted in the same direction as the absorbed photon, this gives rise to dark lines (absence of light) in the spectrum.
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A piece of clay sits 0.10 m from the center of a potter’s wheel. If the potter spins the wheel at an angular speed of 15.5 rad/s, what is the magnitude of the centripetal acceleration of the piece of clay on the wheel?
Answer:
[tex]a=24.025\ m/s^2[/tex]
Explanation:
Given that
Distance from the center ,r= 0.1 m
The angular speed ,ω = 15.5 rad/s
We know that centripetal acceleration is given as
a=ω² r
a=Acceleration
r=Radius
ω=angular speed
a=ω² r
Now by putting the values in the above equation we get
[tex]a=15.5^2\times 0.1\ m/s^2[/tex]
[tex]a=24.025\ m/s^2[/tex]
Therefore the acceleration of the clay will be [tex]a=24.025\ m/s^2[/tex].
An object initially at rest experiences an acceleration of 1.90 m/s² for 6.60 s then travels at that constant velocity for another 8.50 s. What is the magnitude of the object's average velocity over the 15.1 s interval?
Answer:
The magnitude of the object's average velocity is 9.74 m/s (9.80 m/s without any intermediate rounding).
Explanation:
Hi there!
The average velocity is calculated as the displacement of the object divided by the time it takes the object to do that displacement.
The displacement is calculated as the distance between the final position of the object and the initial position. In this problem, the displacement is equal to the traveled distance because the object travels only in one direction:
a.v = Δx/t
Where:
a.v = average velocity.
Δx = displacement = final position - initial position
t = time
So, let's find the distance traveled while the object was accelerating. For that, we will use this equation:
x = x0 + v0 · t + 1/2 · a · t²
Where:
x = position of the object at time t.
x0 = initial position.
v0 = initial velocity.
t = time.
a = acceleration.
In this case, since the object is initially at rest, v0 = 0. If we place the origin of the frame of reference at the point where the object starts moving, then x0 = 0. So, the equation of the position of the object after a time t will be:
x = 1/2 · a · t²
x = 1/2 · 1.90 m/s² · (6.60 s)²
x = 41.4 m
The object traveled 41.4 m during the first 6.60 s.
Now, let's find the rest of the traveled distance.
When the velocity is constant, a = 0. Then, the equation of position will be:
x = x0 + v · t
Let's place now the origin of the frame of reference at the point where the object starts traveling at constant velocity so that x0 = 0:
x = v · t
The velocity reached by the object during the acceleration phase is calculated as follows:
v = v0 + a · t (v0 = 0 because the object started from rest)
v = 1.90 m/s² · 6.60 s
v = 12.5 m/s
Then, the distance traveled by the object at a constant velocity will be:
x = 12.5 m/s · 8.50 s
x = 106 m
The total traveled distance in 15.1 s is (106 m + 41.4 m) 147 m.
Then the displacement will be:
Δx = final position - initial position
Δx = 147 m - 0 = 147 m
and the average velocity will be:
a.v = Δx/t
a.v = 147 m / 15.1 s
a.v = 9.74 m/s
The magnitude of the object's average velocity is 9.74 m/s (9.80 m/s without any intermediate rounding).
A man stands on the roof of a building of height 14.0 m and throws a rock with a velocity of magnitude 26.0 m/s at an angle of 28.0 ∘ above the horizontal. You can ignore air resistance.
The question involves analyzing the projectile motion of a rock thrown from a building using physics principles like motion decomposition and energy conservation. It requires breaking down the initial velocity into horizontal and vertical components and applying kinematic equations to determine parameters such as max height, range, and flight time.
Explanation:This question involves the principles of projectile motion and energy conservation in physics. When the man throws the rock at an angle of 28 degrees above the horizontal with an initial velocity of 26.0 m/s from a height of 14.0 meters, we need to analyze the horizontal and vertical components of the motion separately to determine various aspects of the rock's trajectory, such as its range, maximum height, and time of flight. However, since the specific request is missing in this context, we'll focus on the general approach to solving such problems.
To solve problems involving objects thrown at an angle, we first decompose the initial velocity into its horizontal (vx = v*cos(θ)) and vertical (vy = v*sin(θ)) components, where v is the magnitude of the initial velocity and θ is the angle of projection. The horizontal motion is uniform, meaning the velocity remains constant, whereas the vertical motion is affected by gravity, leading to acceleration in the opposite direction of the initial vertical velocity component.
Energy conservation or kinematic equations can be used to find specific details like maximum height reached, time of flight, and range. For example, the formula s = ut + 0.5at² can be applied where s is displacement, u is initial velocity, a is acceleration (gravity in the case of vertical motion), and t is time. Remember, acceleration due to gravity (a) is -9.8 m/s², indicating it acts downwards. Ignoring air resistance simplifies calculations by omitting drag force considerations.
The maximum horizontal distance is approximately 61.7 meters. This is determined using the projectile motion equations for horizontal distance with initial velocity, angle, and height given.
To find the maximum horizontal distance the rock travels, we can analyze the projectile motion. The initial velocity of 26.0 m/s is broken down into horizontal and vertical components. The horizontal component is [tex]\( v_x = v \cdot \cos(\theta) \)[/tex], where \( v \) is the magnitude of the velocity (26.0 m/s) and \( \theta \) is the angle (28.0 degrees). The vertical component is [tex]\( v_y = v \cdot \sin(\theta) \).[/tex]
The vertical motion is affected by gravity, and the time it takes for the rock to hit the ground can be calculated using the equation [tex]\( h = v_y \cdot t - \frac{1}{2} g t^2 \),[/tex] where \( h \) is the initial height (14.0 m), \( g \) is the acceleration due to gravity (approximately 9.8 m/s\(^2\)), and \( t \) is the time of flight.
Using the quadratic formula to solve for \( t \), we find two solutions: one when the rock is at the initial height and one when it hits the ground. We use the positive solution for the time of flight to calculate the horizontal distance traveled using the equation [tex]\( d = v_x \cdot t \).[/tex]
Substituting the values, we find the maximum horizontal distance to be approximately 61.7 meters.
The question probably maybe: What is the maximum horizontal distance the rock travels before hitting the ground, given that a man stands on the roof of a building of height 14.0 m and throws a rock with a velocity of magnitude 26.0 m/s at an angle of 28.0 degrees above the horizontal, ignoring air resistance?
A car is accelerated from rest to 85 km/h in 10 s. Would the energy transferred to the car be different if it were accelerated to the same speed in 5 s?
The energy transferred to the car would be different if it were accelerated to the same speed in a shorter time period.
Explanation:The energy transferred to the car would indeed be different if it were accelerated to the same speed in 5 seconds instead of 10 seconds. This is because the rate of acceleration affects the amount of energy transferred. In the first scenario, the car would experience a lower rate of acceleration over a longer time period, resulting in a smaller energy transfer. In the second scenario, the car would experience a higher rate of acceleration over a shorter time period, resulting in a larger energy transfer.
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A 945- kg elevator is suspended by a cable of negligible mass. If the tension in the cable is 8.65 kN, what are the magnitude and direction of the elevator's acceleration?
Answer:
Acceleration= -0.6466 m/[tex]sec^{2}[/tex] Downward
Explanation:
Given: Mass M= 945 Kg, Tension T = 8.65 kN = 8650 N and g =9.8 m/[tex]sec^{2}[/tex]
Sol: Weight W = mg = 945 Kg 9.8 m/[tex]sec^{2}[/tex] = 9261 N
this show that T < W so the motion is downward so to find acceleration
mass × acceleration = T - W (putting values)
945 Kg × a = 8650 N - 9261 N
a= -0.6466 m/[tex]sec^{2}[/tex] (-ve sign shows the downward direction)
Answer:
- Magnitude of the acceleration = 0.65 m/s²
- The acceleration is directed upwards in the direction of the Tension in the suspending cables.
Explanation:
The force balance on the elevator consists of the Tension in the cable, acting upwards away from the elevator, the weight of the elevator, mg, acting downwards and the 'ma' force responsible for motion.
The direction of this 'ma' force depends on which side of the force balance is lesser or more.
That is, depending on the one that's higher,
ma = T - mg
OR
ma = mg - T
We need to find the higher force.
T = 8.65 KN = 8650 N
mg = 945 × 9.8 = 9261 N
mg > T, meaning the 'ma' force is on the upwards side of the tension, but motion of the elevator is definitely downwards.
Since mg > T,
ma = mg - T = 9261 - 8650 = 611 N
a = 611/m = 611/945 = 0.65 m/s²
Hope this helps!
Calculate the amount of heat (in kJ) required to convert 97.6 g of water to steam at 100° C. (The molar heat of vaporization of water is 40.79 kJ/mol.)
Answer:
221.17 kJ
Explanation: Note the heat of vaporization is in kJ/mol,then to determine the number of moles of water: divide the mass by 18. Then multiply the number of moles by the molar heat of vaporization of water.
N = 97.6 ÷ 18
Q=molar heat *moles
Q = (40.79) * (97.6 ÷ 18)
This is approximately 221.17 kJ
In a World Cup soccer match, Juan is running due north toward the goal with a speed of 8.00 m/s relative to the ground. A teammate passes the ball to him. The ball has a speed of 12.0 m/s and is moving in a direction 37.0o east of north, relative to the ground. What are the magnitude and direction of the ball’s velocity relative to Juan?
Answer:
19m/s
22.3 degrees
Explanation:
it is a case aof relative velocity.
the basic for relative velocity vector equation is :
V_b = V_j + V_(b/j)---------------1
V_b: ball velocity relative to ground
V_j : Jaun velocity relative to ground
V_(b/j): ball velocity relative to jaun
reference frame:
We take east and north as +ve x and +ve y
V_(b/j) = V_b - V_j
so for x-axis;
net x-component of V_(b/j) = 12 sin (37) + 0 = 7.22m/s
net y-component of V_(b/j) = 12 cos (37) + 8 = 17.6m/s
magnitude = ((7.22^2)+(17.6^2))^(0.5) = 19 m/s
*direction with respect to Jaun = angle between the vertical (North) and vector V_(b/j)
angle = arctan(7.22/17.6) = 22.3 degrees
A projectile thrown from a point P moves in such a way that its distance from P is always increasing. Find the maximum angle above the horizontal with which the projectile could have been thrown. Ignore air resistance.
Answer
70.52°
Explanation
The distance between projectile's position and it's starting point at any time is given by the relation
r² = x² + y²
where x = horizontal distance covered and y = vertical distance covered
According to projectile motion the horizontal displacement is given by
x = v(x)t = v cos(θ) t
Also the vertical component is given by
y = v(y) t - 0.5gt² = v sin(θ) t - 0.5gt²
Substituting the x and y values into the r-equation yields,
r² = (v cos(θ) t)² + (v sin(θ) t - 0.5gt²)²
r² = v²(cos²(θ))t² + v²(sin²(θ))t² – (vg sin(θ))t³+ 0.25 g²(t^4)
r² = v²t² (cos²(θ)+ sin²(θ)) – (vg sin(θ))t³ + 0.25 g²(t^4)
r² = v²t² – (vg sin(θ))t³ + 0.25 g²(t^4)
Differentiate r with respect to t
r(dr/dt) = 2v²t - 3vg sin(θ)t² + g²t³
At maximum angle the projectile could have been thrown above the horizontal, dr/dt = 0
2v²t - 3vg sin(θ)t² + g²t³ = 0
Divide through by t
2v² - 3vg sin(θ)t + g²t² = 0
g²t² - 3vg sin(θ)t + 2v² = 0
This can be solved using the general law for quadratic equations
(-b ± √(b² - 4ac))2a
a = g², b = -3vg sin(θ) c = 2v²
t = ((3vg sin(θ)) ± √(9v²g²sin²(θ) - 8g²v²))/2g²
This equation makes sense when the value under the square root is positive, that is, the square root exists.
9v²g²sin²(θ) - 8g²v² > 0
9sin²(θ) - 8 > 0
Meaning sin²(θ) = 8/9
Sin θ = (2√2)/3
θ = 70.52°
QED!!!