A steel piano wire is 0.7 m long and has a mass of 5 g. It is stretched with a tension of 500 N. What is the speed of transverse waves on the wire? To reduce the wave speed by a factor of 2 without changing the tension, what mass of copper wire would have to be wrapped around the wire?

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

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]

The frequency of the loop is 58 Hz and the amplitude of the emf induced in the loop is 7.73 mV and this can be determined by using the given data.

Given :

A stiff wire bent into a semicircle of radius 'a' is rotated with a frequency f in a uniform magnetic field.

According to the data, the angular speed is 58 rev/sec that is, 364.4 rad/sec, the magnetic field is 15 mT that is, 15 [tex]\times[/tex] [tex]10^{-3}[/tex] T, and the radius 'a' is 3 cm that is, 3  [tex]\times[/tex] [tex]10^{-2}[/tex] m.

a) The frequency is 58 rev/sec that is 58 Hz.

b) The amplitude of the emf induced in the loop can be calculated as:

[tex]\rm \epsilon = \dfrac{\omega B \pi a^2}{2}[/tex]

Now, substitute the values of the known terms in the above formula.

[tex]\epsilon = \dfrac{364.4\times 15\times10^{-3}\times \pi \times (3\times 10^{-2})^2}{2}[/tex]

Further, simplify the above expression.

[tex]\rm \epsilon = 0.007728\;V[/tex]

[tex]\rm \epsilon = 7.73\;mV[/tex]

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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

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.

Answer:

[tex]P=1139.16384mmHg[/tex]

Explanation:

Given data

[tex]R=0.08206(\frac{L.Atm}{mol.K} )\\Density=2.697g/L\\Temperature=298K\\f.wt=44(g/mol)\\[/tex]

To find

Pressure

Solution

From Ideal gas law we know that

[tex]PV=nRT\\P=(nR\frac{T}{V} )=(R(\frac{mass}{f.wt} )(\frac{T}{V} ))\\P=R(\frac{mass}{volume}) (\frac{T}{f.wt} )=R(Density)(\frac{T}{f.wt} )[/tex]

Substitute the given values to find pressure

So

[tex]P=(0.08206\frac{L.Atm}{mol.K} )(2.697g/L)(298K)(44g/mol)^{-1}\\ P=1.4989Atm\\[/tex]

Convert Atm to mmHg

Multiply the pressure values by 760

So

[tex]P=1139.16384mmHg[/tex]

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;

The 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

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!!!

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.

Answer:

R =  16.67 m

Explanation:

Given:

- Initial Temperature T_i = 20 C

- Thickness of both strips t = 0.001 m

- Final Temperature T_f = 30 C

- Length of the strip L = 0.1 m

- coefficient of linear expansion for brass a_b = 19*10^-6

- coefficient of linear expansion for steel a_s = 13*10^-6

Find:

What is the radius of curvature of this arc R?

Solution:

- The radius of curvature R in relation to dT and a_b, a_s:

                               R = t / dT*(a_b - a_s)

                               R = 0.001 / (30-20)*(19-13)*10^-6

                               R =  16.67 m

Answer:

[tex]g \approx 7.4 m/s^2[/tex]

Explanation:

Assuming the following data:

Heigth (m): 0 , 1, 2, 3, 4, 5

Time (s): 0.00, 0.54,0.73, 0.91, 1.01, 1.17

We know from kinematics that the height is given by the following expression:

[tex] h_f = h_i + v_i t + \frac{1}{2} g t^2[/tex]

Assuming for this case that the initial velocity is [tex]v_i[/tex] we can find a polynomial with degree 2 in order to have an estimation for the height with the time.

We can use excel for this and we can see the polynomial adjusted for the data given.

As we can see the best polynomial of degree 2 is given by:

[tex] h(x)= -0.0178 +0.066 x+ 3.6801 x^2[/tex]

For our case x = time and we can rewrite the expression like this

[tex] h(t)= -0.0178 +0.066 t+ 3.6801 t^2[/tex]

And if we are interested on the gravity we want on special the last term of this equation, we can set equal the following terms:

[tex] 3.6801 t^2 = \frac{1}{2} g t^2[/tex]

And solving for g we got:

[tex] 2*3.6801 = g= 7.36 \frac{m}{s^2}[/tex]

So then a good approximation for the gravity of the planet rounded to 2 significant figures is 7.4 m/s^2

To determine the free-fall acceleration on Planet X, you can use the formula: acceleration = 2 * height / fall time^2. Calculating the acceleration for each data point, and the uncertainty can be determined by finding the range of values.

a.

To determine the free-fall acceleration on Planet X, we can use the formula: acceleration = 2 * height / fall time^2. We can calculate the acceleration using the given data:

For height 1m, fall time 0.54s, the acceleration is 7.485 m/s^2

For height 2m, fall time 0.72s, the acceleration is 8.660 m/s^2

For height 3m, fall time 0.91s, the acceleration is 9.337 m/s^2

For height 4m, fall time 1.01s, the acceleration is 9.704 m/s^2

For height 5m, fall time 1.17s, the acceleration is 9.335 m/s^2

b.

To determine the uncertainty in the free-fall acceleration, we can calculate the range of values by subtracting the smallest acceleration from the largest acceleration. The smallest acceleration is 7.485 m/s^2 and the largest acceleration is 9.704 m/s^2. Therefore, the uncertainty in the free-fall acceleration is 2.219 m/s^2.

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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

Answer:

Explanation:

Given

Voltage of battery [tex]V=45\ V[/tex]

electric field(E) is given by  

[tex]E=|-\frac{dV}{dx}|[/tex]

i.e. Change in Potential over a distance x

(a)When Plate are separated by 3 cm apart

[tex]E=|\frac{45}{3\times 10^{-2}}|[/tex]

[tex]E=1500\ V/m[/tex]

(b)When Plates are 5 cm apart

[tex]E=|-\frac{45}{5\times 10^{-2}}|[/tex]

[tex]E=900\ V/m[/tex]

Force on electron is given by

[tex]F=charge\times Electric\ Field[/tex]

[tex]F=q\times E[/tex]

For case a

[tex]F=1.6\times 10^{-19}\times 1500[/tex]

[tex]F=24\times 10^{-17}\ N[/tex]

(b)[tex]F=1.6\times 10^{-19}\times 900=14.4\times 10^{-17}\ N[/tex]

Answer

Given,

Periscope uses 45-45-90 prisms with total internal reflection adjacent to 45°.

refractive index of water, n_a = 1.33

refractive index of glass, n_g = 1.52

When the light enters the water, water will act as a lens and when we see the object from the periscope the object shown is farther than the usual distance.

Answer:

The question is incomplete,below is the complete question

"Calculate the number of atoms contained in a cylinder (1μm radius and 1μm deep)of (a) magnesium (b) lead."

Answer:

a. 1.35*10^{11} atoms

b. 1.03*10^{11} atoms

Explanation:

First, we determine the volume of the magnesium in the cylinder container

using the volume of a cylinder

[tex]V=\pi r^{2}h\\ r=10^{-6}m\\ h=10^{-6}m\\V=\pi *10^{-6*2}*10^{-6}\\V=\pi *10^{-18}\\V=3.14*10^{-18}m^{3}\\[/tex]

a. Next we determine the mass of the magnesium ,

using the density=mass/volume

since density of a magnesium

[tex]the density of magnesium =1.738*10^{3}kg/m^{3} \\mass=density * volume \\mass=1.738*10^{3}*3.14*10^{-18}\\mass=5.46*10^{-15}kg\\ \\mass=5.46*10^{-12}g\\[/tex]

Finally to calculate the number of atoms,

we determine the number of moles

mole=mass/molarmass

[tex]mole=5.46*10^{-12}/ 24.305\\mole=0.225*10^{-12}mol\\[/tex]

Hence the number of atoms is

number of atoms=mole*Avogadro's constant

[tex]number of atoms = 0.225*10^{-12}*6.02*10^{23}\\number of atoms =1.35*10^{11} atoms[/tex]

b. for he lead, we determine the mass of the lead  ,

using the density=mass/volume

since density of a magnesium

[tex]the density of lead =11.34*10^{3}kg/m^{3} \\mass=density * volume \\mass=11.34*10^{3}*3.14*10^{-18}\\mass=35.60*10^{-15}kg\\ \\mass=35.60*10^{-12}g\\[/tex]

Finally to calculate the number of atoms,

we determine the number of moles

mole=mass/molarmass

[tex]mole=35.60*10^{-12}/ 207.2\\mole=0.1718*10^{-12}mol\\[/tex]

Hence the number of atoms is

number of atoms=mole*Avogadro's constant

[tex]number of atoms = 0.1718*10^{-12}*6.02*10^{23}\\number of atoms =1.03*10^{11} atoms[/tex]

Answer:

The maximum height of the package is 140 m above the ground. Jim Bond will not catch the package.

Explanation:

Hi there!

The equation of height and velocity of the package are the following:

h = h0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

h = height of the package at time t.

h0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.81 m/s² because we consider the upward direction as positive).

v = velocity of the package at a time t.

First, let´s find the time it takes the package to reach the maximum height. For this, we will use the equation of velocity because we know that at the maximum height, the velocity of the package is zero. So, we have to find the time at which v = 0:

v = v0 + g · t

0 = 50.5 m/s - 9.8 m/s² · t

Solving for t:

-50.5 m/s / -9.81 m/s² = t

t = 5.15 s

Now, let´s find the height that the package reaches in that time using the equation of height. Let´s place the origin of the frame of reference on the ground so that the initial position of the package is 10 m above the ground:

h = h0 + v0 · t + 1/2 · g · t²

h = 10 m + 50.5 m/s · 5.15 s - 1/2 · 9.81 m/s² · (5.15 s)²

h = 140 m

The maximum height of the package is 140 m above the ground. Jim Bond will not catch the package.

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

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:

Therefore, 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.

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.

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.

The question is incomplete but an analysis of the situation using  various useful physics concepts can still be made

Answer:

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].

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.

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?

An average adult of weigh 70 kg has approximately [tex]7.0 \times 10^{27}[/tex].

Explanation:

Estimation of Atoms in Human Body

We generally evaluate the amount of elements, liquid, or solid bones in a human body but, we can relate these counts to the atomic level as well. Talking about the approximate amount of Hydrogen, Oxygen and Carbon that we carry in our body, they all make 99% of it neglecting 1% of trace elements.

On the basis of this calculation, we can say that a human body of weight 70 kg has around [tex]7.0 \times 10^{27}[/tex] atoms that are differentiated as 2/3 of hydrogen atoms, 1/4 of Oxygen and only 1/3 of Carbon atoms in the total count.

Answer:

1.15 rad/s²

Explanation:

given,

angular speed of turntable = 45 rpm

                       =[tex]45\times \dfrac{2\pi}{60}[/tex]

                       =[tex]4.71\ rad/s[/tex]

time, t = 4.10 s

initial angular speed = 0 rad/s

angular acceleration.

[tex]\alpha = \dfrac{\omega_f-\omega_0}{t}[/tex]

[tex]\alpha = \dfrac{4.71-0}{4.10}[/tex]

[tex]\alpha = 1.15\ rad/s^2[/tex]

Hence, the angular acceleration of the turntable is 1.15 rad/s²

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".

According 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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Answer:

So after stretching new resistance will be 0.1823 ohm

Explanation:

We have given initially length of the wire [tex]l_1=150m[/tex]

Radius of the wire [tex]r_1=0.32mm=0.32\times 10^{-3}m[/tex]

Resistance of the wire initially [tex]R_1=90ohm[/tex]

We know that resistance is equal to [tex]R=\frac{\rho l}{A}[/tex] ,here [tex]\rho[/tex] is resistivity, l is length and A is area

From the relation we can say that [tex]\frac{R_1}{R_2}=\frac{l_1}{l_2}\times \frac{A_2}{A_1}[/tex]

Now length of wire become 6.75 m

Volume will be constant

So [tex]A_1l_1=A_2l_2[/tex]

So [tex]\pi \times (0.32)^2\times150=\pi \times r_2^2\times 6.75[/tex]

[tex]r_2=1.508mm[/tex]

So [tex]\frac{90}{R_2}=\frac{150}{6.75}\times \frac{1.508^2}{0.32^2}[/tex]

[tex]R_2=0.1823ohm[/tex]

With the new length and cross-sectional area, we determine the new resistance to be approximately 1822.5 Ohms.

To determine the resistance of the wire after it is stretched, follow these steps:

Calculate the initial volume of the wire using the initial length and cross-sectional area.

The initial length (L1) = 1.50 m

The radius (r1) = 0.32 mm = 0.00032 m

Initial cross-sectional area (A1) = πr1² = π (0.00032 m)² = 3.216 × 10⁻⁷ m²

Initial volume (V) = A1 × L1 = 3.216 × 10⁻⁷ m² × 1.50 m = 4.824 × 10⁻⁷ m³

Since volume remains constant, calculate the new radius after stretching.

The new length (L2) = 6.75 m

Initial volume (V) = New volume (V)

V = A2 × L2; thus, A2 = V / L2 = 4.824 × 10⁻⁷ m³ / 6.75 m = 7.148 × 10⁻⁸ m²

New radius (r2) = √(A2 / π) = √(7.148 × 10⁻⁸ m² / π) ≈ 0.000151 m = 0.151 mm

Calculate the new resistance using the resistivity formula.

Resistance (R) = ρ × L / A

Assuming resistivity (ρ) is the same, R1 / R2 = (L1 / A1) / (L2 / A2)

New resistance (R2) = R1 × (L2 / L1)² = 90 Ω × (6.75 m / 1.50 m)²

R2 = 90 Ω × (4.5)² = 90 Ω × 20.25 ≈ 1822.5 Ω

Therefore, the resistance of the wire after stretching is approximately 1822.5 Ohms.

Answer:

C)T

Explanation:

The period of a mass-spring system is:

[tex]T=2\pi\sqrt\frac{m}{k}[/tex]

As can be seen, the period of this simple harmonic motion, does not depend at all on the gravitational acceleration (g), neither the mass nor the spring constant depends on this value.

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]

The electric field due to a charged ring E is given by

E = Qz/4πε₀[√(z² + R²)]³ where

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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Answer:

Explanation:

Given

velocity of diver [tex]u=-10\ m/s[/tex] i.e. downward motion

acceleration due to gravity [tex]a=g=-9.8\ m/s^2[/tex]

time [tex]t=1.16\ s[/tex]

using equation of motion

[tex]y=ut+\frac{1}{2}at^2[/tex]

[tex]y=(-10)\cdot 1.16-\frac{1}{2}(-9.8)(1.16)^2[/tex]

[tex]y=-11.6-6.593[/tex]

[tex]y=-18.19\ m[/tex]

I.e. in downward direction                    

The displacement of the diver during the last 1.16 seconds of her dive is -11.6 meters. The negative sign indicates a downward movement.

In physics, displacement is the overall change in position of an object. It is calculated by multiplying velocity and time. In this case, the velocity of the diver is -10.0 m/s (a negative sign indicating downward motion) and the time is 1.16 s. To find the displacement, multiply the velocity and the time: (-10.0 m/s) × (1.16 s) = -11.6 m. The minus sign still indicates downward direction, and it means that the diver moved 11.6 meters downward in the last 1.16 seconds of her dive.

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