Answer:
Q = 3.363 x 10⁻² C
Explanation:
given,
Voltage, V= 169 V
Capacitance of the capacitor, C = 199 μF
Charge in the capacitor = ?
We know,
Q = CV
Q = 169 x 199 x 10⁻⁶
Q = 3.363 x 10⁻² C
Hence, the Charge stored in the capacitor is equal to Q = 3.363 x 10⁻² C
During the battle of Bunker Hill, Colonel William Prescott ordered the American Army to bombard the British Army camped near Boston. The projectiles had an initial velocity of 41 m/s at 38° above the horizon and an initial position that was 35 m higher than where they hit the ground. How far did the projectiles move horizontally before they hit the ground? Ignore air resistance.
Answer:
[tex]s=203.149\ m[/tex]
Explanation:
Given:
initial velocity of projectile, [tex]u=41\ m.s^{-1}[/tex]angle of projection above horizontal, [tex]\theta=38^{\circ}[/tex]height of the initial projection point above the ground, [tex]y=35\ m[/tex]
Vertical component of the velocity:
[tex]u_y=u.\sin\theta[/tex]
[tex]u_y=41\times \sin38[/tex]
[tex]u_y=25.242\m.s^{-1}[/tex]
The time taken in course of going up:
(at top the final velocity will be zero)
[tex]v_y=u_y-g.t[/tex]
[tex]0=25.242-9.8\times t[/tex]
[tex]t=2.576\ s[/tex]
In course of going up the maximum height reached form the initial point:
(at top height the final velocity is zero. )
using eq. of motion,
[tex]v_y^2=u_y^2-2\times g.h[/tex]
where:
[tex]v_y=[/tex] final vertical velocity while going up.=0
[tex]h=[/tex] maximum height
[tex]0^2=25.242^2-2\times 9.8\times h[/tex]
[tex]h=32.5081\ m[/tex]
Now the total height to be descended:
[tex]h'=h+y[/tex]
[tex]h'=32.5081+35[/tex]
[tex]h'=67.5081\ m[/tex]
Now the time taken to fall the gross height in course of falling from the top:
[tex]h'=v_y.t'+\frac{1}{2} g.t'^2[/tex]
[tex]67.5081=0+4.9\times t'^2[/tex]
[tex]t'=3.7118\ s[/tex]
Now the total time the projectile spends in the air:
[tex]t_t=t+t'[/tex]
[tex]t_t=2.576+3.7118[/tex]
[tex]t_t=6.2878\ s[/tex]
Now the horizontal component of the initial velocity:
(it remains constant throughout the motion)
[tex]u_x=u.\cos\theta[/tex]
[tex]u_x=41\times \cos38[/tex]
[tex]u_x=32.3084\ m.s^{-1}[/tex]
Therefore the horizontal distance covered in the total time;
[tex]s=u_x\times t_t[/tex]
[tex]s=32.3084\times 6.2878[/tex]
[tex]s=203.149\ m[/tex]
Answer:
Explanation:
initial velocity, u = 41 m/s
angle, θ = 38 °
height, h = 35 m
Let the time is t.
Use second equation of motion in vertical direction
h = ut + 1/2 gt²
- 35 = 41 Sin 38 t - 0.5 x 9.8 x t²
4.9t² - 25.2 t - 35 = 0
[tex]t = \frac{25.2 \pm \sqrt{25.2^{2}+4\times 4.9\times 35}}{2\times 4.9}[/tex]
t = 6.3 second
Horizontal distance traveled in time t is
d = uCos 38 x t
d = 41 x Cos 38 x 6.3
d = 203.54 m
A cube that is 20 nanometer on an edge contains 399,500 silicon atoms, and each silicon atom has 14 electrons and 14 protons. In the silicon we replace 4 silicon atoms with phosphorus atoms (15 electrons and 15 protons/atom), and we replace 7 silicon atoms with boron atoms (5 electrons and 5 protons/atom). How many "holes" are available to carry current at 300K? Holes look like positive mobile carriers. Three significant digits and fixed point notation.
Answer:
Total 3 holes are available for conduction of current at 300K.
Explanation:
In order to develop a semiconductor, two type of impurities can be added as given below:
N-type Impurities: Pentavalent impurities e.g. Phosphorous, Arsenic are added to have an additional electron in the structure. Thus a pentavalent impurity creates 1 additional electron.P-type Impurities: Trivalent impurities e.g. Boron, Aluminium are added to have a positive "hole" in the structure. Thus a trivalent impurity creates 1 hole.Now for estimation of extra electrons in the impured structure is as
[tex]N_{electrons-free}=n_{pentavalent \, atoms}\\N_{electrons-free}=4\\[/tex]
Now for estimation of "holes" in the impured structure is as
[tex]N_{holes}=n_{trivalent \, atoms}\\N_{holes}=7\\[/tex]
Now when the free electrons and "holes" are available in the structure ,the "holes" will be filled by the free electrons therefore
[tex]N_{holes-net}=N_{holes}-N_{electrons-free}\\N_{holes-net}=7-4\\N_{holes-net}=3[/tex]
So total 3 "holes" are available for conduction of current at 300K.
According to the Guinness Book of World Records (1999) the highest rotary speed ever attained was 2010 m/s (4500 mph) The rotating rod was 15.3 cm long. Assume that the speed quoted is that of the end of the rod. a. What is the centripetal acceleration of the end of the rod? (2.64 x 107 m/s2) b. If you were to attach a 1.0 g object to the end of the rod, what force would be needed to hold it on the rod?
Answer:
a. 2.645 * 10^7 m/s^2
b. 2.645 * 10^4 N
Explanation:
Parameters given:
Velocity of rod = 2010m/s
Length of rod = 15.3cm = 0.153m
Mass of object placed at the end of the rod = 1g = 0.001kg
a. Centripetal acceleration is given as:
a = (v*v)/r
Where v = velocity
r = radius of curvature.
The radius of curvature in this case is equal to the length of the rod, since the rod makes the circular path of the motion.
Hence, centripetal acceleration at the end of the rod:
a = (2010*2010)/(0.153)
a = 26432156.86 m/s^2 = 2.64 * 10^7 m/s
b. The force needed to hold the object at the end of the rod is equal to the centripetal force at the end of the rod. Centripetal force is given as:
F = ma = (m*v*v)/r
Where a = centripetal acceleration
F = 0.001 * 2.64 * 10^7
F = 2.64 * 10^4N
The Centripetal acceleration at the end of rod is [tex]\bold {2.64 x 10^7\ m/s}[/tex] and Centripetal force required to hold the object is [tex]\bold { 2.64x 10^4\ N}[/tex].
Given here:
Velocity of rod = 2010 m/s
Length of rod = 15.3cm = 0.153 m
Mass of object placed at the end of the rod = 1g = 0.001 kg
(A). Centripetal acceleration can be calculated by
[tex]\bold {a =\dfrac { v^2}{r}}[/tex]
Where
v = velocity
r = radius of curvature.
Put the values,
[tex]\bold {a = \dfrac {(2010^2)}{(0.153)}}\\\\\bold {a = 26432156.86\ m/s^2 } \\\\\bold {a = 2.64 x 10^7\ m/s}[/tex]
(B). Centripetal force can be calculated by
[tex]\bold {F = ma = \dfrac {(m v^2)}{r}}[/tex]
Where,
a = centripetal acceleration
Thus the force
[tex]\bold {F = 0.001 \times 2.64x10^7}\\\\\bold {F = 2.64x 10^4\ N}[/tex]
Therefore, the Centripetal acceleration at the end of rod is [tex]\bold {2.64 x 10^7\ m/s}[/tex] and Centripetal force required to hold the object is [tex]\bold { 2.64x 10^4\ N}[/tex].
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A small, solid sphere of mass 0.9 kg and radius 47 cm rolls without slipping along the track consisting of slope and loop-the-loop with radius 4.75 m at the end of the slope. It starts from rest near the top of the track at a height h, where h is large compared to 47 cm. If the g = 9.8 m/s^2 and I(solid sphere) = 2/5 mr^2, what is the minimum value of h such that the sphere completes the loop?
The minimum height that the sphere should start from to complete a loop-the-loop is 23.75 meters, as calculated through the conservation of energy and dynamics principles.
Explanation:In this physics problem, the minimum height (h) that the solid sphere needs to start from to ensure it completes the loop-the-loop involves applying principles of conservation of energy and dynamics. Initially, the sphere has potential energy equal to mgh, and no kinetic energy as it starts from rest. As it descends, it gains kinetic energy and loses potential energy.
For the ball to successfully complete the loop, the force at the top must be equivalent to the weight of the sphere plus the force necessary to maintain circular motion. This can be written as: mg + mv²/r = 5mg. From here, we can derive the equation for v²: v² = 4gr.
Since the kinetic energy at the top of the loop is (1/2)mv² and the potential energy is 2mgr, by equating total energy at the top of the loop (potential plus kinetic) to the initial potential energy (mgh), we obtain: mgh = (1/2)m(4gr) + 2mgr.
From this equation, we can solve for h and find that h = 5r = 5*4.75m = 23.75m. This is the minimum height the sphere must start from to complete the loop.
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A large box of mass M is moving on a horizontal surface at speed v0. A small box of mass m sits on top of the large box. The coefficients of static and kinetic friction between the two boxes are μs and μk, respectively.
Part A: Find an expression for the shortest distance dminin which the large box can stop without the small box slipping.
Part B: A pickup truck with a steel bed is carrying a steel file cabinet. If the truck's speed is 10 m/s , what is the shortest distance in which it can stop without the file cabinet sliding? Assume that μs=0.80.
Answer:
Part A:
[tex]d_{min} = \frac{v_0^2}{2g\mu_s}[/tex]
Part B:
[tex]d_{min} = 6.37~m[/tex]
Explanation:
Part A:
We should determine the free-body diagram of the small box.
For the first box, the only force exerted to the box is the static friction force in the direction of the motion.
(The direction of the static friction is always confusing to the students. The wrong idea is that the static friction is in the opposite direction with the motion. However, if you look at the Newton's Second Law, it states that the net force acting on an object is equal to the mass times acceleration. And in this case acceleration of the total system is equal to that of the small box, since it sits on the larger box.)
We can use the equations of kinematics to find the minimum distance to stop without the small box slipping.
[tex]v^2 = v_0^2 + 2a(\Delta x)\\0 = v_0^2 + 2ad_{min}[/tex]
The acceleration can be found by Newton's Second Law:
[tex]F = ma\\mg\mu_s = m(-a)\\a = -g\mu_s[/tex]
The negative sign comes from the fact that in order for the boxes to stop they have to apply a negative acceleration.
Now, we can combine the two equations to find the distance x:
[tex]0 = v_0^2 + 2ad_{min} = v_0^2 + 2(-g\mu_s)d_{min}\\d_{min} = \frac{v_0^2}{2g\mu_s}[/tex]
Part B:
We can apply the above formula to the truck and file cabinet.
[tex]d_{min} = \frac{v_0^2}{2g\mu_s} = \frac{10^2}{2(9.8)(0.80)} = 6.37~m[/tex]
The expression for the shortest distance which the large box can stop without the small box slipping is [tex]d_{min} = \frac{v_0^2 }{2\mu_s g}[/tex]
The shortest distance in which the pickup can stop without the file cabinet sliding is 6.38 m.
The given parameters;
mass of the bigger box, = Mspeed of the bigger box, = v0mass of the small box, = mcoefficient of static friction, = μscoefficient of kinetic friction, = μkThe expression for the shortest distance which the large box can stop without the small box slipping is calculated as follows;
Apply work-energy theorem;
[tex]\mu_s F \times d_{min} = \frac{1}{2} Mv_0^2\\\\\mu_s (Mg)d_{min} = \frac{1}{2} Mv_0^2\\\\\mu_s gd_{min} = \frac{v_0^2}{2} \\\\d_{min} = \frac{v_0^2}{2\mu_s g}\\\\[/tex]
At the given speed and coefficient of static friction, the shortest distance in which the pickup can stop without the file cabinet sliding is calculated as;
[tex]d_{min} = \frac{v^2}{2\mu_s g} = \frac{10^2 }{2\times 0.8 \times 9.8} = 6.38 \ m[/tex]
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An object is thrown straight up into the air and feels no air resistance. How can the object have an acceleration when it has stopped moving at its highest point?
Answer:
Explanation:
All objects on Earth are subjected to a constant gravitational acceleration g = 9.8m/s2, wherever they are on the surface of Earth and whatever their speed is. So if an object is being thrown to its highest point and stopped moving at that instant, that means the velocity at that instant is 0, not the acceleration. The acceleration is still g = 9.8m/s2
CP Bang! A student sits atop a platform a distance h above the ground. He throws a large firecracker horizontally with a speed. However, a wind blowing parallel to the ground gives the firecracker a constant horizontal acceleration with magnitude a. As a result, the firecracker reaches the ground directly below the student. Determine the height h in terms of v, a, and g. Ignore the effect of air resistance on the vertical motion.
Answer:
h = v₀ g / a
Explanation:
We can solve this problem using the kinematic equations. As they indicate that the air does not influence the vertical movement, we can find the time it takes for the body to reach the floor
y = [tex]v_{oy}[/tex] t - ½ g t²
The vertical start speed is zero
t² = 2t / g
The horizontal document has an acceleration, with direction opposite to the speed therefore it is negative, the expression is
x = v₀ₓ t - ½ a t²
Indicates that it reaches the same exit point x = 0
v₀ₓ t = ½ a t2
v₀ₓ = ½ a (2h / g)
v₀ₓ = v₀
h = v₀ g / a
Calculate the osmotic pressure at 36.6 degrees C of a solution made by dissolving 9.18 g of glucose in 34.2 mL of solution. Enter your answer using 2 decimal places!!!!
Answer:
38.35 bar
Explanation:
We are given that
Temperature=T=36.6 degree Celsius=36.6+273=309.6 K
Given mass of glucose=9.18 g
Molar mass of glucose([tex]C_6H_{12}O_6=6(12)+12(1)+6(16)[/tex]=180 g
Mass of c=12 g,mass of hydrogen=1 g, mass of O=16 g
Volume of solution=34.2 mL
Molarity of solution=[tex]\frac{given\;mass}{molar\;mass\times volume}\times 1000[/tex]
Where volume (in mL)
Molarity of solution=[tex]\frac{9.18}{180\times 34.2}\times 1000=1.49 M[/tex]
We know that
Osmotic pressure=[tex]\pi=MRT[/tex]
Where M=Molarity of solution
R=Constant=0.08314 Lbar/mol k
T=Temperature in kelvin
Using the formula
[tex]\pi=1.49\times 0.08314\times 309.6=38.35 bar[/tex]
Hence, the osmotic pressure=38.35 bar
A runner wants to run 11.8 km. Her running pace is 7.4 mi/hr. How many minutes must she run? Express your answer using two significant figures.
Answer:
She must run 59 min to run 11.8 km.
Explanation:
Hi there!
First let's convert mi/h into km/min:
7.4 mi/h · (1.61 km /1 mi) · (1 h / 60 min) = 0.20 km/min (notice how the units mi and h cancel).
The runner runs at 0.20 km/ min, i.e., every minute she travels 0.20 km.
If 0.20 km are traveled in 1 min, then 11.8 km will be traveled in:
11.8 km / 0.20 km/min = 59 min
She must run 59 min to run 11.8 km.
A car travels in the + x-direction on a straight and level road. For the first 4.00 s of its motion, the average velocity of the car is vav-x = 6.25 m/s. How far does the car travel in 4.00 s?
Answer:
25 m
Explanation:
The relationship between Velocity, distance and time is given as
S = v/t........................... Equation 1
Where S = average velocity of the car, d = distance covered by the car, t = time
Making d the subject of the equation,
d = vt.................... Equation 2
Given: v = 6.25 m/s, t = 4.00 s.
Substitute into equation 2,
d = 6.25(4)
d = 25 m.
Hence, the distance traveled by the car = 25 m
The distance traveled by the car will be "25 m".
The given values are:
Speed,
v = 6.25 m/s,Time,
t = 4.00 s
As we know the formula,
→ [tex]Distance = Speed\times Time[/tex]
or,
→ [tex]d = v\times t[/tex]
By substituting the values, we get
[tex]= 6.25\times 4[/tex]
[tex]=25 \ m[/tex]
Thus the above answer is right.
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A proton moving at 3.0 × 10^4 m/s is projected at an angle of 30° above a horizontal plane. If an electric field of 400 N/C is acting down, how long does it take the proton to return to the horizontal plane?
Answer:
The time it takes the proton to return to the horizontal plane is 7.83 X10⁻⁷ s
Explanation:
From Newton's second law, F = mg and also from coulomb's law F= Eq
Dividing both equations by mass;
F/m = Eq/m = mg/m, then
g = Eq/m --------equation 1
Again, in a projectile motion, the time of flight (T) is given as
T = (2usinθ/g) ---------equation 2
Substitute in the value of g into equation 2
[tex]T = \frac{2usin \theta}{\frac{Eq}{m}} =\frac{m* 2usin \theta}{Eq}[/tex]
Charge of proton = 1.6 X 10⁻¹⁹ C
Mass of proton = 1.67 X 10⁻²⁷ kg
E is given as 400 N/C, u = 3.0 × 10⁴ m/s and θ = 30°
Solving for T;
[tex]T = \frac{(1.67X10^{-27}* 2*3X10^4sin 30}{400*1.6X10^{-19}}[/tex]
T = 7.83 X10⁻⁷ s
Final answer:
The force exerted on a proton in a magnetic field can be calculated using the given formula with specified values.
Explanation:
To find the time it takes for the proton to return to the horizontal plane, we need to consider the motion in both the horizontal and vertical directions separately.
Given:
[tex]- Initial velocity (\(v_0\)) of the proton = \(3.0 \times 10^4\) m/s\\- Launch angle (\(\theta\)) = 30°\\- Electric field (\(E\)) = 400 N/C\\- We'll assume the gravitational acceleration (\(g\)) as \(9.8 \, \text{m/s}^2\) downward.[/tex]
Vertical Motion:
In the vertical direction, the proton undergoes uniformly accelerated motion under gravity.
Using the formula for vertical motion:
[tex]\[ v = u + at \] \\where: \\- \( v \) is the final velocity (which is 0 when the proton returns to the horizontal plane),\\- \( u \) is the initial vertical velocity,\\- \( a \) is the acceleration due to gravity,\\- \( t \) is the time taken.\\[/tex]
We can resolve the initial velocity into vertical and horizontal components:
[tex]\[ u_y = v_0 \sin(\theta) \]\[ u_y = 3.0 \times 10^4 \times \sin(30^\circ) \]\[ u_y \approx 3.0 \times 10^4 \times 0.5 \]\[ u_y = 1.5 \times 10^4 \, \text{m/s} \][/tex]
Now, let's find the time it takes for the vertical velocity to become zero:
[tex]\[ 0 = u_y - gt \]\[ t = \frac{u_y}{g} \]\[ t = \frac{1.5 \times 10^4}{9.8} \]\[ t \approx 1530.61 \, \text{s} \][/tex]
Horizontal Motion:
In the horizontal direction, there is no acceleration or deceleration acting on the proton. So, the time taken for horizontal motion is the same as the time taken for vertical motion.
The time it takes for the proton to return to the horizontal plane is approximately [tex]\( 1530.61 \) seconds.[/tex]
List three advantages of reflecting telescopes over refracting telescopes.
Answer:
Three advantages of Reflecting Telescope over refracting Telescope:
1. there is no chromatic aberration in the Reflecting Telescope being mirror as an objective, while Refracting Telescope can suffer chromatic aberration being using lenses.
2. image of Reflecting Telescope is brighter due to large, polished curved mirrors than Refracting Telescope
3. Reflecting Telescope is compact and portable in size than Refracting Telescope.
Final answer:
Reflecting telescopes have three advantages over refracting telescopes: they gather more light, do not suffer from chromatic aberration, and are easier and less expensive to manufacture.
Explanation:
There are three advantages of reflecting telescopes over refracting telescopes:
Light gathering: Reflecting telescopes can gather more light than refracting telescopes because the mirror in a reflecting telescope is larger in diameter than the lens in a refracting telescope. This allows reflecting telescopes to have a greater ability to capture faint objects in the sky.No chromatic aberration: Reflecting telescopes do not suffer from chromatic aberration, which is a distortion of colors caused by different wavelengths of light focusing at different points. This is because reflecting telescopes use mirrors instead of lenses to focus light, eliminating the issue of chromatic aberration.Ease and cost of manufacturing: Reflecting telescopes are easier and less expensive to manufacture than refracting telescopes. Only the front surface of a mirror needs to be accurately polished, whereas lenses in a refracting telescope require both sides to be polished to great accuracy. Reflecting telescopes also do not require high-quality glass throughout like refracting telescopes.3. In a physics lab, 0.500-kg cart (Cart A) moving rightward with a speed of 100 m/s collides with a 1.50-kg cart (Cart B) moving leftward with a speed of 20 m/s. The two carts stick together and move as a single object after the collision. Determine the post-collision speed of the two carts.
Answer:
The speed of the two carts after the collision is 10 m/s.
Explanation:
Hi there!
The momentum of the system Cart A - Cart B is conserved because there is no external force acting on the system at the instant of the collision. Then, the momentum of the system before the collision will be equal to the momentum of the system after the collision. The momentum of the system is calculated as the sum of momenta of cart A and cart B:
initial momentum = mA · vA1 + mB · vB1
final momentum = (mA + mB) · vAB2
Where:
mA = mass of cart A = 0.500 kg
vA1 = velocity of cart A before the collision = 100 m/s
mB = mass of cart B = 1.50 kg.
vB1 = velocity of cart B before the collision = - 20 m/s
vAB2 = velocity of the carts that move as a single object = unknown.
(notice that we have considered leftward as negative direction)
Since the momentum of system remains constant:
initial momentum = final momentum
mA · vA1 + mB · vB1 = (mA + mB) · vAB2
Solving for vAB2:
(mA · vA1 + mB · vB1) / (mA + mB) = vAB2
(0.500 kg · 100 m/s - 1.50 kg · 20 m/s) / (0.500 kg + 1.50 kg) = vAB2
vAB2 = 10 m/s
The speed of the two carts after the collision is 10 m/s.
The post-collision speed of the two carts is 10 m/s moving in the positive x-direction.
Explanation:In order to determine the post-collision speed of the two carts, we can use the principle of conservation of momentum. The total momentum before the collision is equal to the total momentum after the collision.
Before the collision, Cart A has a mass of 0.500 kg and a velocity of 100 m/s, while Cart B has a mass of 1.50 kg and a velocity of -20 m/s (negative because it is moving leftward). After the collision, the two carts stick together, so their masses add up to 2 kg.
Using the conservation of momentum equation: (Momentum before collision) = (Momentum after collision)
(0.500 kg × 100 m/s) + (1.50 kg × -20 m/s) = 2 kg × v
By solving this equation, we find that the post-collision speed of the two carts is 10 m/s moving in the positive x-direction.
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An airplane is dropping bales of hay to cattle stranded in a blizzard on the Great Plains. The pilot releases the bales at 150 m above the level ground when the plane is flying at 75 m/s in a direction 55° above the horizontal. How far in front of the cattle should the pilot release the hay so that the bales land at the point where the cattle are stranded?
Answer:
Δx=629.35 m
The pilot release the hay 629.35 m in front of the cattle so that the bales land at the point where the cattle are stranded.
Explanation:
Step 1:
Finding initial velocity components:
Initial velocity=v=75 m/s
α=55
[tex]v_{ox}=vcos\alpha\\v_{ox}=75cos55^o\\v_{ox}=43.018 m/s\\v_{oy}=vsin\alpha\\v_{oy}=75sin55^o\\v_{oy}=61.436 m/s[/tex]
Step 2:
[tex]y_o=150\ m[/tex]
Newton Second Equation:
[tex]y-y_o=v_{oy}t+\frac{1}{2}g t^2[/tex]
g=-9.8 m/s^2 (Downward direction)
[tex]v_{oy}=61.436\ m/s[/tex]
y=0 m
Above equation will become:
-150=(61.436)t-(4.90)t^2
Solving the above quadratic equation we will get:
t=-2.09 sec , t=14.63 sec
t= 14.63 sec
Step 3:
Finding the distance:
Using Again Newton equation of motion in x-direction:
[tex]x-x_o=v_{ox}t+\frac{1}{2}a_{x} t^2[/tex]
Since velocity is constant in x- direction, [tex]a_x[/tex] will be zero.
Above equation will be:
[tex]\Delta x=v_{ox}t[/tex]
Δx=(43.018)(14.63)
Δx=629.35 m
The pilot release the hay 629.35 m in front of the cattle so that the bales land at the point where the cattle are stranded.
The pilot should release the hay at a height of 629.35 m.
Given information,
Initial velocity = 75 m/s
Velocity For x-component,
[tex]\bold {V_0x = Vcos \alpha}\\\\\bold {V_0x = 75 cos 55^o}\\\\\bold {V_0x = 43. 018m/s}[/tex]
Velocity for Y-component
[tex]\bold {V_0y = Vsin \alpha}\\\\\bold {V_0y = 75 sin 55^o}\\\\\bold {V_0y = 61. 43m/s}[/tex]
Using Newton's second equation for y-axis,
[tex]\bold {y-y_0 = V_0t + \dfrac {1}{2} gt^2}[/tex]
Where,
g - gravitational acceleration
put the values in the equation,
[tex]\bold {-150=(61.436)t-(4.90)t^2}[/tex]
Solving this quadratic equation, we get 2 values
t = 14.29 s
To find the distance, use Newton's second equation,
[tex]\bold {x-x_0 = V_0t + \dfrac {1}{2} gt^2}[/tex]
Since acceleration is zero because the velocity is constant in x-axis hence .
So,
[tex]\bold {x-x_0 = V_0_xt }[/tex]
[tex]\bold {x- x_0=(43.018)(14.63)}\\\\\bold {x - x_0=629.35 m}[/tex]
Therefore, the pilot should release the hay at 629.35 m.
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(Schaum’s 18.25) A 55 g copper calorimeter (c=377 J/kg-K) contains 250 g of water (c=4190 J/kg-K) at 18o When a 75 g metal alloy at 100o C is dropped into the calorimeter, the final equilibrium temperature is 20.4o C. What is the specific heat of the alloy?
Answer:
1205.77 J/kg.K
Explanation:
Heat lost by alloy = heat gained by water + heat gained by the calorimeter
c₁m₁(t₂-t₃) = c₂m₂(t₃-t₁) + c₃m₃(t₃-t₁)................. Equation 1
Where c₁ = specific heat capacity of the alloy, m₁ = mass of the alloy, t₂ = initial temperature of the alloy, t₃ = equilibrium temperature, c₂ = specific heat capacity of water, m₂ = mass of water, t₁ = initial temperature of water and calorimter, c₃ = specific heat capacity of calorimter, m₃ = mass of calorimter.
Making c₁ the subject of the equation,
c₁ = c₂m₂(t₃-t₁) + c₃m₃(t₃-t₁)/m₁(t₂-t₃)........................ Equation 2
Given: c₂ = 4190 J/kgK, m₂ = 250 g = 0.25 kg, m₁ = 75 g = 0.075 kg, m₃ = 55 g = 0.055 kg, c₃ = 377 J/kg.K, t₁ = 18 °C, t₂ = 100 °C, t₃ = 24.4 °C.
Substitute into equation 2
c₁ = [0.25×4190×(24.4-18) + 0.055×377×(24.4-18)]/[0.075(100-24.4)]
c₁ = (6704+132.704)/5.67
c₁ = 6836.704/5.67
c₁ = 1205.77 J/kg.K
Thus the specific heat capacity of the alloy = 1205.77 J/kg.K
Final answer:
The student's question is about calculating the specific heat capacity of a metal alloy using the principles of calorimetry and the conservation of energy in a heat exchange process.
The student is asking about finding the specific heat capacity of a metal alloy using calorimetry. We know that when objects at different temperatures are combined, they will exchange heat energy until they reach thermal equilibrium. We can use the equation Q = mc ext{ extdegree}T (where Q is heat energy, m is mass, c is specific heat capacity, and ext{ extdegree}T is the change in temperature) to find the specific heat capacity. In this scenario, the heat lost by the metal alloy will equal the heat gained by the copper calorimeter and the water contained within it. By setting these two equations equal to each other and solving for the specific heat capacity of the alloy, we can find that value.
What are the first three overtones of a bassoon that has a fundamental frequency of 90.0 Hz? It is open at both ends. (The overtones of a real bassoon are more complex than this example, because its double reed makes it act more like a tube closed at one end.)
Answer:
[tex]f_{2}=180Hz,f_{3}=270Hz,f_{4}=360Hz\\[/tex]
Explanation:
Given data
Frequency f=90 Hz
To find
First three overtones of bassoon
Solution
The fundamental frequency of bassoon is found by substituting n=1 in below equation
f=v/λ=nv/2L
[tex]f_{1}=v/2L[/tex]
The first overtone of bassoon is found by substituting n=2
So
[tex]f_{2}=2v/2L\\f_{2}=2(v/2L)\\as \\f_{1}=v/2L\\So\\f_{2}=2f_{1}\\f_{2}=2(90Hz)\\f_{2}=180Hz[/tex]
The second overtone of bassoon is found by substituting n=3
So
[tex]f_{3}=3v/2L\\f_{3}=3(v/2L)\\as \\f_{1}=v/2L\\So\\f_{3}=3f_{1}\\f_{3}=3(90Hz)\\f_{3}=270Hz[/tex]
The third overtone of bassoon is found by substituting n=4
So
[tex]f_{4}=4v/2L\\f_{4}=4(v/2L)\\as \\f_{1}=v/2L\\So\\f_{4}=4f_{1}\\f_{4}=4(90Hz)\\f_{4}=360Hz[/tex]
The first three overtones of a bassoon with a fundamental frequency of 90.0 Hz are 180.0 Hz, 270.0 Hz and 360.0 Hz. The calculation is based on the behaviour of the bassoon as a tube open at both ends where overtones occur at integer multiples of the fundamental frequency.
Explanation:The question is asking for the first three overtones of a bassoon that has a fundamental frequency of 90.0 Hz. The bassoon is assumed to act like a tube that is open at both ends. For a tube open at both ends, the overtones, also called harmonics, occur at integer multiples of the fundamental frequency.
In this case, the fundamental frequency (first harmonic) is 90.0 Hz. The first overtone (which is the second harmonic) is then 2 * 90.0 Hz = 180.0 Hz. The second overtone (third harmonic) is 3 * 90.0 Hz = 270 Hz, and the third overtone (fourth harmonic) is 4 * 90.0 Hz = 360 Hz.
Thus, the first three overtones of a bassoon that has a fundamental frequency of 90.0 Hz, and is open at both ends, are 180.0 Hz, 270.0 Hz, and 360.0 Hz respectively.
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A spring with a spring-constant 2.2 N/cm is compressed 28 cm and released. The 5 kg mass skids down the frictional incline of height 31 cm and inclined at a 18 angle. The acceleration of gravity is 9.8 m/s^2. The path is frictionless except for a distance of 0.8 m along the incline which has a coefficient of friction of 0.3. k = 2.2 N/cm What is the final velocity vf of the mass? Answer in units of m/s
The final velocity of the mass is calculated using the principle of conservation of energy, which states that the potential energy in the spring is converted into kinetic energy and work done against friction as the mass moves down the incline. This total energy is then equal to the kinetic energy at the end of the path, from which the final velocity of the mass can be found.
Explanation:To calculate the final velocity of the mass after it has been released from the spring, we need to realize that the energy in the system is conserved. The potential energy stored in the spring when it is compressed is converted into kinetic energy and work done against friction as the mass skids down the incline. The spring energy is given by (1/2)kx^2, where 'k' is the spring constant and 'x' is the amount of compression. In this case, it is (1/2)*2.2 N/cm*(28 cm)^2. We convert all values to standard SI units before calculation for accurate results.
The work done against the friction in the path of 0.8 m is friction force times the distance, which = μmgcosθ*d, where 'μ' is the friction coefficient, 'm' the mass, 'g' gravity, 'θ' the angle of incline, and 'd' the distance with friction. Here, it is 0.3*5 kg*9.8 m/s^2*cos(18 degrees)*0.8 m.
The total energy is thus the sum of spring energy and work done against friction. As this energy is equal to the kinetic energy at the end of the path (1/2)*m*v^2, we can solve this to find the final velocity 'v'
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The final velocity of a 5 kg mass sliding down a frictional incline is calculated using energy conservation and work-energy principles, resulting in a final velocity of approximately 2.25 m/s.
Calculating the Final Velocity:
To determine the final velocity of a 5 kg mass after it slides down a frictional incline, we need to consider both energy conservation and work-energy principles.
Step-by-Step Solution
Spring Potential Energy: The spring is compressed by 28 cm, and the spring constant k is 2.2 N/cm. First, convert these units to meters and N/m: k = 220 N/m and compression x = 0.28 m. The potential energy stored in the spring is given by:
PE_spring = 0.5 x k x x².
PE_spring = 0.5 x 220 x (0.28)² = 8.624 J.
Gravitational Potential Energy: As the mass slides down the incline of height h = 0.31 m, the gravitational potential energy is converted to kinetic energy. PE_gravity = m x g x h
PE_gravity = 5 kg x 9.8 m/s² x 0.31 m = 15.19 J.
Energy Loss Due to Friction: The mass encounters a frictional patch of length d = 0.8 m with a friction coefficient μ = 0.3. The work done by friction (which is energy lost) is calculated by:
W_friction = μ x m x g x cos(θ) x d.
Given θ = 18°, cos(18°) ≈ 0.9511.
W_friction = 0.3 x 5 kg x 9.8 m/s² x 0.9511 x 0.8 m = 11.16 J.
Total Mechanical Energy: The initial potential energy (spring + gravity) is:
8.624 J + 15.19 J = 23.814 J.
After accounting for the work done by friction, the remaining energy will be:
KE_final = 23.814 J - 11.16 J = 12.654 J.
Final Velocity: The kinetic energy at the bottom of the incline is converted into the final velocity.
Using the relation:
KE = 0.5 x m x vf².
12.654 J = 0.5 x 5 kg x vf²
vf² = (2 x 12.654 J) / 5 kg = 5.06 m²/s²
vf = √5.06 m²/s² ≈ 2.25 m/s.
Thus, the final velocity of the mass is approximately 2.25 m/s.
An unknown gas effuses 2.3 times faster than N2O4 at the same temperature. What is the identity of the unknown gas?
Answer:
The molar mass of the unknown gas is 17.3 g/mol. The molar mass matches that of ammonia (NH₃) the most (17 g/mol)
Explanation:
Let the unknown gas be gas 1
Let N₂O₄ gas be gas 2
Rate of effusion ∝ [1/√(Molar Mass)]
R ∝ [1/√(M)]
R = k/√(M) (where k is the constant of proportionality)₁₂
R₁ = k/√(M₁)
k = R₁√(M₁)
R₂ = k/√(M₂)
k = R₂√(M₂)
k = k
R₁√(M₁) = R₂√(M₂)
(R₁/R₂) = [√(M₂)/√(M₁)]
(R₁/R₂) = √(M₂/M₁)
R₁ = 2.3 R₂
M₁ = Molar Mass of unknown gas
M₂ = Molar Mass of N₂O₄ = 92.01 g/mol
(2.3R₂/R₂) = √(92.01/M₁)
2.3 = √(92.01/M₁)
92.01/M₁ = 2.3²
M₁ = 92.01/5.29
M₁ = 17.3 g/mol
The molar mass matches that of ammonia the most (17 g/mol)
The unknown gas in the system has been ammonia.
The rate of diffusion of the two gases has been proportional to the molar mass of the gases.
The ratio of the rate of two gases can be given as:
[tex]\rm \dfrac{RateA}{RateB}\;=\;\sqrt{\dfrac{Molar\;mass\[A}{Molar\;mass\;B} }[/tex]
The two gases can be given as:
Gas A = Nitrogen tetraoxide = 2.3x
Gas B = x
[tex]\rm \dfrac{2.3x}{x}\;=\;\sqrt{\dfrac{92.011}{m} }[/tex]
Mass of the unknown gas = 17.39 grams.
The mass has been equivalent to the mass of the Ammonia. Thus, the unknown gas in the system has been ammonia.
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An electric heater has a Nichrome heating element with a resistance of 9 Ω at 20oC. When 112 V are applied, the electric current heats the Nichrome wire to 1090oC. What is the operating wattage of this heater? (The temperature coefficient of resistivity of Nichrome is α = 0.0004 Co-1)
Answer:
975.28 W.
Explanation:
Using,
R' = R(1+αΔt)....................... Equation 1
Where R' = Resistance at the final temperature, R = Resistance at the initial temperature, α = temperature coefficient of resistivity of Nichorome, Δt = Temperature rise.
Given: R = 9 Ω, α = 0.0004/°C, Δt = 1090-20 = 1070 °C
Substitute into equation 1
R' = 9(1+0.0004×1070)
R' = 9(1.428)
R' = 12.862 Ω.
Note: Operating wattage of the heater means the operating power of the heater
The power of the heater is given as,
P = V²/R'...................... Equation 2
Where P = Operating wattage of the heater, V = Voltage, R' = Operating resistance.
Given: V = 112 V, R' = 12.862 Ω
Substitute into equation 2
P = 112²/12.862
P = 975.28 W.
The operating wattage for this heater can be calculated using Ohm's Law, resulting in approximately 1405.33 Watts assuming constant resistance. However, in reality, resistance alters with temperature, reflecting the importance of considering temperature effects in physics.
Explanation:The operating wattage for this electric heater, or the power (P), can be calculated using Ohm's Law where power equals voltage (V) times current (I), or P=IV. Because I = V/R, where R is resistance, the formula can also be written as P = V2/R. With the provided values, we have P = (112V)2 / 9Ω, which gives approximately 1405.33 Watts, assuming that the resistance remains constant over the temperature change.
However, in reality, the resistance changes with temperature according to the equation R = R0[1 + α(T - T0)] where R0 is the original resistance, α is the temperature coefficient of resistivity, T is the final temperature, and T0 is the initial temperature. Considering the provided values and the significant temperature increase, we would need to adjust the resistance for the increased temperature before calculating the power, underlining the importance of temperature effects in practical physics.
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A sample of nitrogen gas exerts a pressure of 9.80 atm at 32 C. What would its temperature be (in C) when its pressure is increased to 11.2 atm?
Answer:
T₂ = 111.57 °C
Explanation:
Given that
Initial pressure P₁ = 9.8 atm
T₁ = 32°C = 273 + 32 =305 K
The final pressure P₂ = 11.2 atm
Lets take the final temperature = T₂
We know that ,the ideal gas equation
If the volume of the gas is constant ,then we can say that
[tex]\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}[/tex]
[tex]T_2=\dfrac{P_2}{P_1}\times T_1[/tex]
Now by putting the values in the above equation ,we get
[tex]T_2=\dfrac{11.2}{9.8}\times 305\ K[/tex]
[tex]T_2=348.57\ K[/tex]
T₂ = 384.57 - 273 °C
T₂ = 111.57 °C
An electron is projected horizontally into the uniform electric field directed vertically downward between two parallelplates. The plates are 2.00 cm apart and are of length 4.00 cm. The initial speed of the electron is vi = 8.00 × 106 m/s. As it enters the region between the plates, the electron is midway between the two plates; as it leaves, the electron just misses the upperplate.a)What is the magnitude ofv the electric field?
Answer:
455N/C
Explanation:
from the question, the following data can be derived
Distance between plates=2cm=0.02m
length of plates=4cm=0.04m
initial speed of electron=8*10^6m/s
Note: the speed giving is the speed associated with the horizontal motion since it moves to cover the 4cm distance
we solve the equation component by component.
For the horizontal component, the time it takes to cover distance of 0.04m can be calculated as
[tex]time=\frac{distance }{velocity} \\t=\frac{0.04}{8*10^{6}}\\ t=5*10^{-9}secs[/tex]
this same time is used to cover the vertical distance which is midway between the plate,
Hence vertical distance covered is 0.02/2=0.01m
The acceleration in the vertical component can be calculated as
[tex]y=ut+1/2at^{2}\\u=0,\\y=0.01m\\a=\frac{2y}{t^{2}}\\ a=\frac{2*0.01}{5*10^{-9}}\\ a=8*10^{14}m/s^{2}[/tex]
since
F=qE
also F=ma
then
qE=ma
E=(ma)/q
m=mass of electron=9.1*10^-32kq
q=charge of electron=1.6*10^-19c
a=acceleration
if we substitute values
[tex]E=\frac{9.1*10^{-32}*8*10^{14}}{1.6*10^{-19}} \\E=455N/C[/tex]
To find the magnitude of the electric field when an electron is projected into a uniform electric field between two parallel plates, consider the forces acting on the electron.
An electron is projected horizontally into the uniform electric field directed vertically downward between two parallel plates.
The plates are 2.00 cm apart, and the initial speed of the electron is 8.00 × 10^6 m/s.
To calculate the magnitude of the electric field, you would need to consider the forces acting on the electron as it moves between the plates.
On the moon, the gravitational acceleration is approximately one-sixth that on the surface of the earth. A 5-kg mass is "weighed" with a beam balance on the surface of the moon.
a) What is the expected reading?
b) If this mass is weighed with a spring scale that reads correctly for standard gravity on earth, what is the reading?
Answer:
A. 8.175 N
B. 49.05 N
Explanation:
A.
Acceleration due to gravity (moon) = 1/6 * (acceleration due to gravity (earth)
Acceleration due to gravity (earth) = 9.81 m/s2
Acceleration due to gravity (moon) = 9.81/6
= 1.635 m/s
Weight. Fm = acceleration due to gravity * mass
= 1.635 * 5
= 8.175 N
B. Acceleration due to gravity (earth) * mass = Fe
= 9.81 * 5
= 49.05 N
How do astronomers use the Doppler effect to determine the velocities of astronomical objects?
Astronomers apply the Doppler effect because from there it is possible to obtain information about the change of light, which in turn affects the light spectrum and determines the movement of a body moving away or approaching us. The extent of the shift is directly proportional to the source's radial velocity relative to the observer.
The phenomenon that occurs to determine this process is linked to the wavelength. When the wave source moves towards you, the wavelength tends to decrease. This leads to a change in the color of the light moving towards the end of the spectrum, that is, towards the color blue. (It is really violet, but by convention the color blue was chosen as it is a more common color) When the source moves away from you and the wavelength lengthens, we call the color change a shift to red. Because the Doppler effect was first used with visible light in astronomy, the terms "blue shift" and "red shift" were well established.
Final answer:
Astronomers use the Doppler effect to calculate the velocities of stars and galaxies by observing changes in light wavelengths due to motion towards or away from the observer. It also helps in exoplanet detection and measuring a star's rotation speed by analyzing the broadened spectral lines.
Explanation:
Astronomers utilize the Doppler effect to determine the velocities of astronomical objects such as stars and galaxies. To calculate the radial velocity of an object, they require the speed of light, the original wavelength of the light emitted by the object, and the observed change in this wavelength due to the Doppler shift. This shift occurs because the object is moving relative to Earth—approaching objects cause a blue shift, where the wavelength shortens, while receding objects cause a red shift, where the wavelength lengthens.
The Doppler effect is also instrumental in exoplanet detection through stellar radial velocity measurements. When a planet orbits a star, it imparts a gravitational tug that causes the star to wobble slightly. This wobble changes the star's radial velocity, which can be detected as small shifts in the star's spectral lines, irrespective of the star's distance, as long as it can be observed with a high-resolution spectrograph.
Additionally, the Doppler effect helps measure the rotation speed of distant stars. By analyzing broadened spectral lines, which result from the spread of Doppler shifts due to the rotating star's edges moving towards and away from us, astronomers can infer how fast a star is spinning.
An unknown sample has a volume of 3.61 cm3 and a mass of 9.93 g. What is the density (g/cm3) of the unknown?
Answer:
2.75 g/cm³
Explanation:
given,
Volume of unknown sample, V = 3.61 cm³
mass of the sample, m = 9.93 g
density = ?
We know,
[tex]density = \dfrac{mass}{volume}[/tex]
[tex]\rho= \dfrac{9.93}{3.61}[/tex]
[tex]\rho = 2.75\ g/cm^3[/tex]
Hence, density of the unknown sample is equal to 2.75 g/cm³
The force, F, of the wind blowing against a building is given by where V is the wind speed, rho the density of the air, A the cross-sectional area of the building, and CD is a constant termed the drag coefficient. Determine the dimensions of the drag coefficient.
Answer:
dimensions of the drag coefficient is [tex][M^0 L^0 T^0][/tex]
Drag coefficient is a dimensionless quantity
Explanation:
force is given by[tex]F=\frac{C_{D} \rho V^2 A}{2}[/tex]
we get expression for drag coefficient [tex]C_{D} =\frac{2F}{\rho V^2 A}[/tex]
By substituting the dimensions of the F,V,A and density , we get
[tex]C_{D} =\frac{[F]}{[\rho ][V]^2[A]} \\C_{D} =\frac{[MLT^{-2}]}{[ML^{-3} ][L T^{-1}]^2[L^2]} \\C_{D} =\frac{[MLT^{-2}]}{[ML^{-3} ][L^2 T^{-2}][L^2]} \\C_{D} =\frac{[MLT^{-2}]}{[MLT^{-2}]}\\C_{D}=[M^0 L^0 T^0][/tex]
Drag coefficient is a dimensionless
The dimensions of the drag coefficient, CD, are kg/m.
Explanation:The dimensions of the drag coefficient, CD, can be determined by examining the equation for force, F, of the wind blowing against a building. In this equation, the dimensions for force are mass x acceleration, which are kg x m/s^2. On the other side of the equation, the wind speed, V, has dimensions of m/s, the density, rho, has dimensions of kg/m^3, and the cross-sectional area, A, has dimensions of m^2. Therefore, in order for the equation to be balanced, the dimensions of the drag coefficient must be kg/m.
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They are then fixed at positions that are 4.30 x 10-11 m apart. What is EPEfinal - EPEinitial, which is the change in the electric potential energy?
Answer:
The change in the electric potential energy is [tex]-3.2\times10^{-16}\ J[/tex]
Explanation:
Given that,
Distance [tex]d=4.30\times10^{-11}\ m[/tex]
suppose, Two particles with charges +6 e and -10 e are initially very far apart
We need to calculate the change in the electric potential energy
Using formula of energy
[tex]\text{electric potential energy}=\text{final electric potential energy-initial electric potential energy}[/tex]
[tex]EPE=EPE_{f}-EPE_{i}[/tex]
Here, initial electric potential energy= 0
final electric potential energy [tex]EPE_{f}=\dfrac{kq_{1}q_{2}}{r_{2}^2}[/tex]
Put the value into the formula
[tex]EPE=\dfrac{kq_{1}q_{2}}{r_{2}^2}+0[/tex]
Put the value into the formula
[tex]EPE=\dfrac{9\times10^{9}\times6\times1.6\times10^{-19}\times(-10\times1.6\times10^{-19})}{(4.30\times10^{-11})}[/tex]
[tex]EPE=-3.2\times10^{-16}\ J[/tex]
Hence, The change in the electric potential energy is [tex]-3.2\times10^{-16}\ J[/tex]
This is an incomplete question, here is a complete question.
Two particles with charges +6 e⁻ and -10 e⁻ are initially very far apart (effectively an infinite distance apart). They are then fixed at positions that are 4.30 × 10⁻¹¹ m apart. What is EPE(final) - EPE(initial), which is the change in the electric potential energy?
Answer : The change in the electric potential energy is, [tex]-2.92\times 10^{-6}J[/tex]
Explanation : Given,
Formula used for electric potential energy of the two charges when they are separated is:
[tex]EPE=\frac{1}{4\pi \epsilon_0}\times {\frac{q_1\times q_2}{r^2}[/tex]
[tex]EPE=\frac{k\times q_1\times q_2}{r^2}[/tex]
where,
EPE = electric potential energy
k = [tex]\frac{1}{4\pi \epsilon_0}=8.99\times 10^9[/tex]
[tex]q_1[/tex] = charge on 1st particle = +6 e⁻ = [tex]6\times 10^{-19}C[/tex]
[tex]q_2[/tex] = charge on 2nd particle = -10 e⁻ = [tex]-10\times 10^{-19}C[/tex]
r = distance between two charges = [tex]4.30\times 10^{-11}m[/tex]
Now put all the given values in the above formula, we get:
[tex]EPE=\frac{(8.99\times 10^9)\times (6\times 10^{-19})\times (-10\times 10^{-19})}{(4.30\times 10^{-11})^2}[/tex]
[tex]EPE=-2.92\times 10^{-6}J[/tex]
Initially EPE = 0 J
Thus, [tex]EPE_{final}-EPE_{initial}=-2.92\times 10^{-6}J[/tex]
The positive sign indicate the attractive force and negative sign indicate the repulsive force.
Thus, the change in the electric potential energy is, [tex]-2.92\times 10^{-6}J[/tex]
Two balls, made of different materials, are rubbed against each other, resulting in 0.30 nC of charge moving from one ball to the other. The balls are then held 0.90 m apart. What is the magnitude of the dipole moment of the two balls?
Answer:
[tex]5.4\times 10^{-10}C-m[/tex]
Explanation:
We are given that
Charge=[tex]q=0.30 nC=0.3\times 10^{-9} C[/tex]
[tex]1 nC=10^{-9}C[/tex]
Distance between two balls=l=0.90 m
We have to find the magnitude of dipole moment of the two balls.
We know that
Dipole moment=[tex]\mid p\mid=2lq[/tex]
Where q= Charge
l=Distance between two bodies
Using the formula
Magnitude of dipole moment=[tex]\mid P\mid=2\times 0.3\times 10^{-9}\times 0.9=5.4\times 10^{-10}C-m[/tex]
Hence, the magnitude of the dipole moment of the two balls=[tex]5.4\times 10^{-10}C-m[/tex]
An object is attached to the lower end of a 32-coil spring that is hanging from the ceiling. fie spring stretches by 0.160 m. The spring is then cut into two identical springs of 16 coils each. As the drawing shows, each spring is attached between the ceiling and the object. By how much does each spring stretch
Answer:
0.080 m
Explanation:
According to Hooke's law, a spring with stiffness k will stretch a distance of Δx when a force F is applied:
F = k Δx
If we say the weight of the object is W, then the stiffness of the original spring is:
W = k (0.160 m)
k = W / 0.160
When the spring is cut in half, the stiffness of each new spring is the same as the original. This time, the weight of the object is evenly distributed between each spring, so the force on each is W/2.
F = k Δx
W/2 = (W/0.160) Δx
1/2 = Δx / 0.160
Δx = 0.080
Each spring stretches 0.080 meters.
A 3-kW resistance heater in a water heater runs for 3 hours to raise the water temperature to the desired level. Determine the amount of electric energy used in both kWh and kJ. The amount of electricity used, in kWh, is kWh. The amount of electricity used, in kJ, is
Answer:
Energy, 9 kWh or 32400 kJ
Explanation:
Given that,
The power of heater, P = 3 kW
It runs for 3 hours to raise the water temperature to the desired level. We need to find the amount of electric energy used. We know that the electrical power of an object is given by total energy delivered per unit time. It is given by :
[tex]P=\dfrac{E}{t}[/tex]
[tex]E=P\times t[/tex]
[tex]E=3\ kW\times 3\ h[/tex]
E = 9 kWh
Since, 1 kWh = 3600 kJ
E = 32400 kJ
So, the amount of electric energy used is 9 kWh or 32400 kJ. Hence, this is the required solution.
The amount of electric energy used by a 3-kW resistance heater running for 3 hours would be 9kWh, which is equivalent to 32,400 kJ.
Explanation:To determine the amount of electric energy used, we use the formula E = Pt, where E represents energy, P is power, and t is time. Here, the power used is 3kW (or kilowatts) and time is 3 hours. So, E = 3kW * 3 hours = 9 kWh (kilowatt-hours).
Moving forward, 1 kWh is equal to 3600 kilojoules (kJ). Therefore, to convert the energy we obtained in kilowatt-hours to kilojoules, we multiply it by 3600. This amounts to: 9 kWh * 3600 kJ/kWh = 32,400 kJ.
Therefore, the amount of electric energy used is 9 kWh or 32,400 kJ.
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A 3" diameter germanium wafer that is 0.020" thick at 300K has 2.272*10^17 Arsenic atoms added to it. What is the resistivity of the wafer? Germanium has 4.42*10^22 atoms/cc, electron and hole mobilities are 3900 and 1900 cm^2/(V*s). What is the resistivity of the Ge in ohm*microns? From other similar questions I see they are not tking into consideration the volume of the wafer plus the unit conversion.
Explanation:
Formula to calculate resistivity is as follows.
[tex]\rho = \frac{1}{qN \mu}[/tex]
= [tex]\frac{1}{(1.6 \times 10^{-19}) \times 2.72 \times 10^{17} \times 3900}[/tex] ohm/cm
= [tex]5.89 \times 10^{-3} ohm/cm[/tex]
As germanium is an intrinsic conductor. Hence, resistivity of Ge is as follows.
[tex]\rho_{1} = \frac{1}{2qN_{o}\sqrt{\mu_{e}\mu_{r}}}[/tex]
= [tex]\frac{1}{2q(N_{A})^{\frac{1}{2}}\sqrt{\mu_{e}\mu_{r}}}[/tex]
= [tex]\frac{1}{2 \times (1.6 \times 10^{-19}) \times (\sqrt{4.42 \times 10^{22}})\sqrt{(3900)(1900)}}[/tex] ohm/cm
= 0.546 [tex]ohm (\mu m)^{-1}[/tex]
Thus, we can conclude that resistivity of the Ge is 0.546 [tex]ohm (\mu m)^{-1}[/tex].