Compute the voltage drop along a 24-m length of household no. 14 copper wire (used in 15-A circuits). The wire has diameter 1.628 mm and carries a 13-A current. The resistivity of copper is 1.68×10−8Ω⋅m.

Answer:

Resistance of the bulb is 2100 watts.

Explanation:

Given that,

Power of the bulb, P = 7.5 watts

Voltage, V = 125 volts

We have to find the resistance of the bulb. The power of an electrical appliance is given by the following formula as :

[tex]P=\dfrac{V^2}{R}[/tex]

[tex]R=\dfrac{V^2}{P}[/tex]

[tex]R=\dfrac{(125\ V)^2}{7.5\ W}[/tex]

R = 2083.34 ohms

or

R = 2100 ohms

Hence, the correct option is (d) "2100 ohms"

Answer:

Explanation:

h = height of the cliff

Consider upward direction as positive and downward direction as negative

Consider the motion of rock thrown straight up :

Y = vertical displacement = - h

v₀ = initial velocity = 8.63 m/s

a = acceleration = - 9.8 m/s²

t = time taken to hit the ground = 3 s

Using the equation

Y = v₀ t + (0.5) a t²

- h = (8.63) (3) + (0.5) (- 9.8) (3)²

h = 18.21 m

Consider the motion of rock thrown down :

Y' = vertical displacement = - 18.21

v'₀ = initial velocity = - 8.63 m/s

a' = acceleration = - 9.8 m/s²

t' = time taken to hit the ground = ?

Using the equation

Y' = v'₀ t' + (0.5) a' t'²

- 18.21 = (- 8.63) t' + (0.5) (- 9.8) t'²

t' = 1.2 s

Answer:

The radius of the disc is 2.098 m.

(e) is correct option.

Explanation:

Given that,

Moment of inertia I = 12100 kg-m²

Mass of disc m = 5500 kg

Moment of inertia :

The moment of inertia is equal to the product of the mass and square of the radius.

The moment of inertia of the disc is given by

[tex]I=\dfrac{mr^2}{2}[/tex]

Where, m = mass of disc

r = radius of the disc

Put the value into the formula

[tex]12100=\dfrac{5500\times r^2}{2}[/tex]

[tex]r=\sqrt{\dfrac{12100\times2}{5500}}[/tex]

[tex]r= 2.098\ m[/tex]

Hence, The radius of the disc is 2.098 m.

Answer:

Average net force, F = 15157.15 N

Explanation:

It is given that,

The mass of the car and riders is, [tex]m=3\times 10^3\ kg[/tex]

Initial speed of the car, u = 0

Final speed of the car, v = 43.4 m/s

Time, t = 8.59 seconds

We need to find the  average net force exerted on the car and riders by the magnets. It can be calculated using second law of motion as :

F = m a

[tex]F=m(\dfrac{v-u}{t})[/tex]

[tex]F=3\times 10^3\ kg\times (\dfrac{43.4\ m/s-0}{8.59\ s})[/tex]

F = 15157.15 N

So, the average net force exerted on the car and riders by the magnets. Hence, this is the required solution.

Final answer:

The average net force comes out to be 15,150 N.

Explanation:

The student has provided information about the Magic Mountain Superman ride, where riders are accelerated by magnets. To find the average net force exerted on the car and riders by the magnets, we can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration (F = m * a).

First, we determine the acceleration using the formula a = (v - u) / t, where 'v' is the final velocity, 'u' is the initial velocity (which is 0 since the car starts from rest), and 't' is the time taken to reach the final velocity.

Using the given data, the acceleration a = (43.4 m/s - 0 m/s) / 8.59 s = 5.05 [tex]m/s^2[/tex]. Now, we can use this acceleration to calculate the force with F = m * a. Substituting the values, we get [tex]F = 3.00 times 103 kg * 5.05 m/s^2 = 1.515 times 104[/tex]N. Hence, the average net force exerted by the magnets on the car and riders is 15,150 N.

Answer:

[tex]0.6 \times 10 {}^{15} [/tex]

Explanation:

By using relation,

Speed of light = frequency × wavelength (in m)

Answer:

[tex]F = (3.9\times 10^3)\hat j - 2.8 \times 10^3\hat i[/tex]

Explanation:

Initial momentum of the ball is given as

[tex]P_i = mv_i[/tex]

[tex]P_i = 0.145 (40) = 5.8 kg m/s \hat i[/tex]

now final momentum of the ball is given as

[tex]P_f = 0.145(57) = 8.3 kg m/s \hat j[/tex]

now by the formula of force we have

[tex]F = \frac{P_f - P_i}{\Delta t}[/tex]

now we have

[tex]F = \frac{8.3 \hat j - 5.8 \hat i}{2.10 \times 10^{-3}}[/tex]

[tex]F = (3.9\times 10^3)\hat j - 2.8 \times 10^3\hat i[/tex]

The average vector force the ball exerts on the bat during their interaction is

[tex]\rm F = 3.9\times10^3\;\hat{j}-2.8\times10^3\;\hat{i}[/tex]

Given :

Initial Speed = 40 m/sec

Final Speed = 57 m/sec

Mass = 0.145 Kg

Time = 0.0021 sec

Solution :

Initial Momentum is,

[tex]\rm P_i = mv_i[/tex]

[tex]\rm P_i = 0.145\times 40 = 5.8\;Kg .m/sec\;\hat{i}[/tex]

Final momentum is,

[tex]\rm P_f = mv_f = 0.145\times57=8.3\; Kg.m/sec\; \hat{j}[/tex]

Now,

[tex]\rm F=\dfrac{P_f-P_i}{\Delta t}= \dfrac{8.3\hat{j}-5.8\hat{i}}{2.10\times10^-^3}[/tex]

[tex]\rm F = 3.9\times10^3\;\hat{j}-2.8\times10^3\;\hat{i}[/tex]

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Final answer:

In Physics, work is done when a force causes a displacement. Since the bag of groceries is moved horizontally at constant speed and there's no vertical displacement or horizontal force in the direction of motion, the work done on the bag is 0 joules.

Explanation:

The question involves the concept of work in Physics, specifically related to the work-energy principle. In this scenario, the bag of groceries is being carried across the room at a constant speed and constant height. According to the scientific definition of work, work is done when a force causes a displacement in the direction of the force. The formula for work is W = force x displacement. Here, the only forces doing work would be if the bag was lifted or if it was accelerated. Since the bag is being moved at a constant speed and not being lifted any further, there is no work done in the direction of motion, and the vertical height does not change. Therefore, despite the effort you feel you are exerting, the work done on the bag with regard to Physics is 0 joules

Answer choice: (B) 0.00J.

Answer:

442.5 rad

Explanation:

w₀ = initial angular velocity of the disk = 7.0 rad/s

α = Constant angular acceleration = 3.0 rad/s²

t = time period of rotation of the disk = 15 s

θ = angular displacement of the point on the rim

Angular displacement of the point on the rim is given as

θ = w₀ t + (0.5) α t²

inserting the values

θ = (7.0) (15) + (0.5) (3.0) (15)²

θ = 442.5 rad

(a) Only momentum conservation

In order to find the speed of the ball, momentum conservation is enough. In fact, we have the following equation:

[tex]m u + M U = m v + M V[/tex]

where

m is the mass of the baseball

u is the initial velocity of the baseball

M is the mass of the brick

U is the initial velocity of the brick

v is the final velocity of the baseball

V is the final velocity of the brick

In this problem we already know the value of: m, u, M, U, and V. Therefore, there is only one unknown value, v: so this equation is enough to find its value.

(b) 12.1 m/s

Using the equation of conservation of momentum written in the previous part:

[tex]m u + M U = m v + M V[/tex]

where we have:

m = 0.144 kg

u = +28.0 m/s (forward)

M = 5.25 kg

U = 0

V = +1.1 m/s (forward)

We can solve the equation to find v, the velocity of the ball:

[tex]v=\frac{mu-MV}{m}=\frac{(0.144 kg)(+28.0 m/s)-(5.25 kg)(+1.1 m/s)}{0.144 kg}=-12.1 m/s[/tex]

And the sign indicates that the ball bounces backward, so the speed is 12.1 m/s.

(c)  56.4 J, 13.7 J

The total mechanical energy before the collision is just equal to the kinetic energy of the baseball, since the brick is at rest; so:

[tex]E_i = \frac{1}{2}mu^2 = \frac{1}{2}(0.144 kg)(28.0 m/s)^2=56.4 J[/tex]

The total mechanical energy after the collision instead is equal to the sum of the kinetic energies of the ball and the brick after the collision:

[tex]E_f = \frac{1}{2}mv^2 + \frac{1}{2}MV^2 = \frac{1}{2}(0.144 kg)(12.1 m/s)^2 + \frac{1}{2}(5.25 kg)(1.1 m/s)^2=13.7 J[/tex]

(d)

A collision is:

- elastic when the total mechanical energy is conserved before and after the collision

- inelastic when the total mechanical energy is NOT conserved before and after the collision

In this problem we have:

- Energy before the collision: 56.4 J

- Energy after the collision: 13.7 J

Since energy is not conserved, this is an inelastic collision.

(e) No

As shown in part (c) and (d), the kinetic energy of the system is not conserved. This is due to the fact that in inelastic collisions (such as this one), there are some internal/frictional forces that act on the system, and that cause the dissipation of part of the initial energy of the system. This energy is not destroyed (since energy cannot be created or destroyed), but it is simply converted into other forms of energy (mainly heat and sound).

Answer:

The length of the rod should be

[tex]\frac{L}{4} \\ [/tex]

Explanation:

Period of simple pendulum is given by

[tex]T=2\pi\sqrt{\frac{l}{g}} \\ [/tex]

We have

[tex]\frac{T_1^2}{T_2^2}=\frac{l_1}{l_2}\\\\\frac{2^2}{1^2}=\frac{L}{l_2}\\\\l_2=\frac{L}{4} \\ [/tex]

The length of the rod should be

[tex]\frac{L}{4} \\ [/tex]

Getting a promotion is NOT an example of retaliation.

C. Getting A Promotion

the action of harming someone because they have harmed oneself; revenge.

An example of to retaliate is for a person to punch someone who has hit him. (intransitive) To do something harmful or negative to get revenge for some harm; to fight back or respond in kind to damage or affront. Johny insulted Peter to retaliate for Peter's acid remark earlier.

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To find the magnitude of the net force acting on the kayak, we will use the following kinematic equation derived from Newton's second law of motion:

v2 = u2 + 2as
Where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance over which the acceleration occurs.
We know that the kayak starts from rest, so u = 0.
The final velocity v is 0.680 m/s and the distance s is 0.428 m. Rearranging the equation for acceleration a, we get:

a = (v2 - u2) / (2s)

By plugging in the given values, we find that the acceleration a is:
a = (0.6802 - 02) / (2  imes 0.428) = 0.6802 / 0.856 = 0.541 m/s2

Now, using Newton's second law (Force = mass x acceleration), we find the net force F:
F = mass x a
F = 82.7 kg x 0.541 m/s2 = 44.7397 N

The magnitude of the net force acting on the kayak is approximately 44.74 N.

Answer:

Pendulum B

Explanation:

The time period of a pendulum is given by :

[tex]T=2\pi\sqrt{\dfrac{L}{g}}[/tex]

Case 1.

Mass, m = 200 g = 0.2 kg

Length of string, l = 1 m

Time, [tex]T_1=2\pi\sqrt{\dfrac{1\ m}{9.8\ m/s^2}}[/tex]

T₁ = 2.007 Seconds

Since, [tex]f=\dfrac{1}{T_1}[/tex]

[tex]f_1=\dfrac{1}{2.007}[/tex]

f₁ = 0.49 Hz

Case 2.

Mass, m = 400 g = 0.4 kg

Length of string, l = 0.5 m

Time, [tex]T_2=2\pi\sqrt{\dfrac{0.5\ m}{9.8\ m/s^2}}[/tex]

T₂ = 1.41 seconds

[tex]f₂=\dfrac{1}{T_2}[/tex]

[tex]f₂=\dfrac{1}{1.41}[/tex]

f₂ = 0.709 seconds

Hence, pendulum B have highest frequency of vibration.

Answer:

570 N

Explanation:

Draw a free body diagram on the rider.  There are three forces: tension force 15° below the horizontal, drag force 30° above the horizontal, and weight downwards.

The rider is moving at constant speed, so acceleration is 0.

Sum of the forces in the x direction:

∑F = ma

F cos 30° - T cos 15° = 0

F = T cos 15° / cos 30°

Sum of the forces in the y direction:

∑F = ma

F sin 30° - W - T sin 15° = 0

W = F sin 30° - T sin 15°

Substituting:

W = (T cos 15° / cos 30°) sin 30° - T sin 15°

W = T cos 15° tan 30° - T sin 15°

W = T (cos 15° tan 30° - sin 15°)

Given T = 1900 N:

W = 1900 (cos 15° tan 30° - sin 15°)

W = 570 N

The rider weighs 570 N (which is about the same as 130 lb).

The weight of the rider is 566.89 N.

The given parameters;

Let the weight of the rider = W

Apply Newton's second law of motion, to determine the net force in horizontal and vertical direction.

F = ma

Sum of the forces in horizontal direction is calculated as follows;

[tex]\Sigma F_x = 0\\\\Fcos(30) - Tcos(15) = 0\\\\0.866 F - 0.965T = 0\\\\0.866F = 0.965T\\\\F = \frac{0.965T}{0.866} , \ \ T = 1900 \ N\\\\F = \frac{0.965 \times 1900}{0.866} \\\\F = 2,177.21 \ N[/tex]

Sum of the forces in the vertical direction is calculated as follows;

[tex]\Sigma F_y = 0\\\\Fsin(30) - W -Tsin(15) = 0\\\\W = Fsin(30)- Tsin(15)\\\\W = (2117.21 \times 0.5) - (1900\times 0.2588)\\\\W = 566.89 \ N[/tex]

Thus, the weight of the rider is 566.89 N.

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Answer:

     Acceleration is 0.24 m/s² at 104.44° to positive x-axis.

Explanation:

Refer the figure given.

Let us take component i along positive X axis and component j along positive Y axis.

50 N force can be resolved in to 50 cos 49 i + 50 sin 49 j

        F1 = 32.80 i + 37.74 j

55 N force can be resolved in to 55 cos 220 i + 55 sin 220 j

        F2 = -42.13 i -35.35 j

Total force

       F = F1 + F2 = 32.80 i + 37.74 j -42.13 i -35.35 j = -9.33 i + 2.39 j

We have

      F = ma = 40a =-9.33 i + 2.39 j

Acceleration

      a =-0.233 i + 0.060 j

      [tex]\texttt{Magnitude}=\sqrt{(-0.233)^2+(0.060)^2}=0.24m/s^2[/tex]

      [tex]\texttt{Direction},\theta =tan^{-1}\left ( \frac{0.060}{-0.233} \right )=104.44^0[/tex]

     Acceleration is 0.24 m/s² at 104.44° to positive x-axis.

Answer:

26.7 m/s

Explanation:

The speed of the car is

v = 60 mi/h

We know that

1 mile = 1.6 km = 1600 m

1 h = 60 min = 3600 s

So we can convert the speed from mi/h into m/s by multiplying by the following factor:

[tex]v = 60 \frac{mi}{h} \cdot \frac{1600 m/mi}{3600 s/h}=26.7 m/s[/tex]

Answer: [tex]4.068(10)^{13} km[/tex]

Explanation:

A light year is a unit of length and is defined as "the distance a photon would travel in vacuum during a Julian year at the speed of light at an infinite distance from any gravitational field or magnetic field. "

In other words: It is the distance that the light travels in a year.

This unit is equivalent to [tex]9.461(10)^{12}km[/tex], which mathematically is expressed as:

[tex]1Ly=9.461(10)^{12}km[/tex]

Doing the conversion:

[tex]4.3Ly.\frac{9.461(10)^{12}km}{1Ly}=4.068(10)^{13}km[/tex]

The Answer: The star closest to our sun is 4.2 light years away from the sun.

Explanation:

Answer:

Speed of the liquid in the second segment = 0.61 m/s

Explanation:

This discharge is constant.

That is

        Q₁ = Q₂

       A₁v₁ = A₂v₂

       [tex]\frac{\pi d_1^2}{4}\times v_1=\frac{\pi d_2^2}{4}\times v_2\\\\\frac{\pi \times 1.7^2}{4}\times 4.7=\frac{\pi \times 4.7^2}{4}\times v_2\\\\v_2=0.61m/s[/tex]

Speed of the liquid in the second segment = 0.61 m/s

Answer:

Explanation:

The escape velocity on a planet is directly proportional to the square root of acceleration due to gravity and the radius of the planet.

According to the question acceleration due to gravity that means g is same but there is no idea about the radius.

So if radius is change then escape velocity at that planet may be more or less depending on the radius.

If the radius is also same then the escape velocity is same.

Answer:

Speed of water at the top of fall = 5.40 m/s

Explanation:

We have equation of motion

[tex]v^2=u^2+2as[/tex]

Here final velocity, v = 26 m/s

a = acceleration due to gravity

[tex]a=9.8m/s^2 \\ [/tex]

displacement, s = 33 m

Substituting

[tex]26^2=u^2+2\times 9.8 \times 33\\\\u^2=29.2\\\\u=5.40m/s \\ [/tex]

Speed of water at the top of fall = 5.40 m/s

Explanation:

At the maximum height, the ball's velocity is 0.

v² = v₀² + 2a(x - x₀)

(0 m/s)² = (12.3 m/s)² + 2(-9.80 m/s²)(x - 0 m)

x = 7.72 m

The ball reaches a maximum height of 7.72 m.

The times where the ball passes through half that height is:

x = x₀ + v₀ t + ½ at²

(7.72 m / 2) = (0 m) + (12.3 m/s) t + ½ (-9.8 m/s²) t²

3.86 = 12.3 t - 4.9 t²

4.9 t² - 12.3 t + 3.86 = 0

Using quadratic formula:

t = [ -b ± √(b² - 4ac) ] / 2a

t = [ 12.3 ± √(12.3² - 4(4.9)(3.86)) ] / 9.8

t = 0.368, 2.14

The ball reaches half the maximum height after 0.368 seconds and after 2.14 seconds.

The maximum height the ball reaches above where it leaves your hand is 7.72 m

The time taken for the ball to pass half of its maximum height moving upwards is 2.14s and 0.37 s when moving downwards.

The given parameters;

initial velocity, u = 12.3 m/s

acceleration due to gravity, g = 9.8 m/s²

The maximum height the ball reaches above where it leaves your hand is calculated as;

[tex]v_f^2 = v_0^2 - 2gh\\\\at \ maximum \ height \ final \ velocity \ v_f = 0\\\\2gh = v_0 ^2\\\\h = \frac{v_0^2}{2g} \\\\h = \frac{(12.3)^2}{2(9.8)} \\\\h = 7.72 \ m[/tex]

The time for the ball to reach half of the maximum height is calculated as;

[tex]half \ of \ the \ maximum \ height = \frac{7.72}{2} = 3.86 \ m[/tex]

[tex]h = v_0t - \frac{1}{2} gt^2\\\\3.86 = 12.3t - 0.5\times 9.8t^2\\\\3.86 = 12.3t - 4.9t^2\\\\4.9t^2- 12.3t + 3.86=0\\\\solve \ the \ quadratic \ equation \ using \ formula \ method;\\\\a = 4.9, \ \ b = -12.3, \ \ c = 3.86\\\\t = \frac{-b \ \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\t = \frac{-b \ \ +/- \ \ \sqrt{(-12.3)^2 - 4(4.9\times 3.86)} }{2(4.9)} \\\\t = 2.14 \ s \ \ or \ \ 0.37 \ s[/tex]

The time taken for the ball to pass half of its maximum height moving upwards is 2.14s and 0.37 s when moving downwards.

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Answer:

137 J

Explanation:

Momentum is conserved, so:

(2.10)(13.0) + (6.50)(-4.10) = (2.10)(-8.20) + (6.50) v

v = 2.75 m/s

Energy before collision:

E = 1/2 (2.10) (13.0)² + 1/2 (6.50) (4.10)²

E = 232 J

Energy after collision:

E = 1/2 (2.10) (8.20)² + 1/2 (6.50) (2.75)²

E = 95.2 J

Energy lost in collision:

232 J - 95.2 J = 137 J

Final answer:

A total of 161.488 Joules of kinetic energy is lost in this collision.

Explanation:

The subject question involves a head-on inelastic collision between two balls of different masses traveling towards each other at different speeds and asks for the amount of kinetic energy lost in the collision. To find the kinetic energy lost, we first calculate the initial and final kinetic energies and then take their difference.

Initial kinetic energy (Eki) is the sum of the kinetic energies of the two balls before the collision:

Kinetic energy of the 2.10 kg ball: [tex](1/2) * 2.10 kg * (13.0 m/s)^2[/tex]

Kinetic energy of the 6.50 kg ball: [tex](1/2) * 6.50 kg * (4.1 m/s)^2[/tex]

Final kinetic energy (Ekf) involves only the kinetic energy of the 2.10 kg ball, as the final velocity of the 6.50 kg ball is not given:

Kinetic energy of the 2.10 kg ball: (1/2) * 2.10 kg * [tex](8.20 m/s)^2[/tex]

Kinetic energy lost during the collision is calculated as:

Elost = Eki - Ekf

After calculation:

Initial kinetic energy, Eki = (1/2) * 2.10 kg * [tex](13.0 m/s)^2[/tex] + (1/2) * 6.50 kg * [tex](4.1 m/s)^2[/tex]

Final kinetic energy, Ekf = (1/2) * 2.10 kg * [tex](8.20 m/s)^2[/tex]

Kinetic energy lost, Elost = Eki - Ekf

Plugging the numbers in:

Eki = 0.5 * 2.10 kg * [tex](13.0 m/s)^2[/tex] + 0.5 * 6.50 kg * [tex](4.1 m/s)^2[/tex]
= (0.5 * 2.10 * 169) + (0.5 * 6.50 * 16.81)
= 177.45 J + 54.64 J
= 232.09 J

Ekf = 0.5 * 2.10 kg * [tex](8.20 m/s)^2[/tex]
= (0.5 * 2.10 * 67.24)
= 70.602 J

Therefore, kinetic energy lost, Elost = 232.09 J - 70.602 J = 161.488 J

A total of 161.488 Joules of kinetic energy is lost in this collision.

Answer:

Weight of gasoline = 2039.93 kN

Volume = 80783.80 gallon

Explanation:

Volume = Base area x Depth

[tex]\texttt{Base area = }\frac{\pi d^2}{4}=\frac{\pi \times 25^2}{4}=490.88ft^2[/tex]

Depth = 22 ft

Volume = 490.88 x 22 = 10799.22 ft³ = 10799.22 x 0.0283168 = 305.8 m³

Density of gasoline = 680 kg/m³

Mass of gasoline = 305.8 x 680 = 207943.65 kg

Weight of gasoline = 207943.65 x 9.81 = 2039.93 kN

We have 1 m³ = 264.172 gallon

            305.8 m³ = 305.8 x 264.172 = 80783.80 gallon

Answer:

Work done by the frictional force is [tex]3.41\times 10^5\ J[/tex]

Explanation:

It is given that,

Mass of the car, m = 1000 kg

Initial velocity of car, u = 26.1 m/s

Finally, it comes to rest, v = 0

We have to find the work done by the frictional forces. Work done is equal to the change in kinetic energy as per work - energy theorem i.e.

[tex]W=k_f-k_i[/tex]

[tex]W=\dfrac{1}{2}m(v^2-u^2)[/tex]

[tex]W=\dfrac{1}{2}\times 1000\ kg(0^2-(26.1\ m/s)^2)[/tex]

W = −340605 J

or

[tex]W=3.41\times 10^5\ J[/tex]

Hence, the correct option is (a).

The work done by frictional forces to slow a 1000 kg car from 26.1 m/s to rest is 3.41 x 10^5 J (a). This value is derived from the car's initial kinetic energy, which is lost due to friction.

The question is asking how much work the frictional forces need to do to bring a 1000 kg car to rest, from an initial speed of 26.1 m/s. In such a scenario, these forces would be working against the car's kinetic energy.

The car's initial kinetic energy can be calculated using the formula 1/2 m v^2 (m = mass, v = speed). So the kinetic energy = 1/2 * 1000 kg * (26.1 m/s)^2 = 3.41 x 10^5 J. Since work done by friction is equal to the change in kinetic energy, and the car is brought to rest, the total work done by frictional forces is 3.41 x 10^5 J. Therefore, the correct answer is (a) 3.41 x 10^5 J.

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Explanation:

This situation is a good example of the projectile motion or parabolic motion, in which the travel of the bullet has two components: x-component and y-component. Being their main equations as follows:

x-component:

[tex]x=V_{o}cos\theta t[/tex]   (1)

Where:

[tex]V_{o}=200m/s[/tex] is the bullet's initial speed

[tex]\theta=0[/tex] because we are told the bullet is shot horizontally

[tex]t[/tex] is the time since the bullet is shot until it hits the ground

y-component:

[tex]y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}[/tex]   (2)

Where:

[tex]y_{o}=1.5m[/tex]  is the initial height of the bullet

[tex]y=0[/tex]  is the final height of the bullet (when it finally hits the ground)

[tex]g=9.8m/s^{2}[/tex]  is the acceleration due gravity

Now, for the first part of this problem, the time the bullet elapsed traveling, we will use equation (2) with the conditions given above:

[tex]0=1.5m+200m/s.sin(0) t-\frac{9.8m/s^{2}.t^{2}}{2}[/tex]   (3)

[tex]0=1.5m-\frac{9.8m/s^{2}.t^{2}}{2}[/tex]   (4)

Finding [tex]t[/tex]:

[tex]t=\sqrt{\frac{1.5m(2)}{9.8m/s^{2}}}[/tex]   (5)

Then we have the time elapsed before the bullet hits the ground:

[tex]t=0.553s[/tex]   (6)

For the second part of this problem, we are asked to find how far does the bullet traveled horizontally. This means we have to use the equation (1) related to the x-component:

[tex]x=V_{o}cos\theta t[/tex]   (1)

Substituting the knonw values and the value of [tex]t[/tex] found in (6):

[tex]x=200m/s.cos(0)(0.553s)[/tex]   (7)

[tex]x=200m/s(0.553s)[/tex]   (8)

Finally:

[tex]x=110.656m[/tex]  

The bullet will hit the ground after approximately 0.553 seconds, and in that time, it will travel horizontally around 110.6 meters.

To determine how much time elapses before the bullet hits the ground (1.5 m drop) when fired horizontally at 200 m/s, we use the equation for free fall motion h = (1/2)gt², where h is the height, g is the acceleration due to gravity (approximately 9.8 m/s²), and t is the time in seconds. Solving for time t, we find that the bullet will hit the ground after approximately 0.553 seconds.

For part (b), to find how far the bullet travels horizontally, we simply multiply the time in the air by the bullet's initial horizontal velocity. This gives a horizontal distance of 200 m/s * 0.553 s = 110.6 meters. Therefore, the bullet travels roughly 110.6 meters before hitting the ground.

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Answer:

The radius of a circular path of a charged particle orbit depends on the charge and velocity of the particle.

Explanation:

Answer:

The work done by the friction is 55.8 J.

Explanation:

Given that,

Height h = 7 m

Mass of the object = 3 kg

Angle = 30°

Speed = 10 m/s

The initial potential energy of the object is converted in to the kinetic energy at the bottom of the incline and  the work done by the friction

[tex]mgh=\dfrac{1}{2}mv^2+W[/tex]

Where, m = mass of the object

v = final velocity

g = acceleration due to gravity

h = height

Put the value in the equation

[tex]3\times9.8\times7=\dfrac{1}{2}\times3\times(10)^2+W[/tex]

[tex]W = (205.8-150)\ J[/tex]

[tex]W=55.8\ J[/tex]

Hence, The work done by the friction is 55.8 J.

The specific heat of silver is 0.235 J/g⋅∘C. It would take 39.1 J of energy to raise the temperature of 9.00 g of silver by 18.3 ∘C.

The specific heat of silver can be calculated using the information given. The molar heat capacity of silver is 25.35 J/mol⋅∘C. To find the specific heat, we need to convert the molar heat capacity to g⋅∘C. The molar mass of silver is 107.87 g/mol. Therefore, the specific heat of silver is 0.235 J/g⋅∘C.

To calculate the amount of energy required to raise the temperature of 9.00 g of silver by 18.3 ∘C, we can use the formula Q = mcΔT, where Q is the energy, m is the mass, c is the specific heat, and ΔT is the change in temperature. Plugging in the values, we get Q = (9.00 g)(0.235 J/g⋅∘C)(18.3 ∘C) = 39.1 J.

Therefore, it would take 39.1 J of energy to raise the temperature of 9.00 g of silver by 18.3 ∘C.

Answer:

It feels like it has more weight

The particle has constant acceleration according to

[tex]\vec a(t)=2\,\vec\imath-4\,\vec\jmath-2\,\vec k[/tex]

Its velocity at time [tex]t[/tex] is

[tex]\displaystyle\vec v(t)=\vec v(0)+\int_0^t\vec a(u)\,\mathrm du[/tex]

[tex]\vec v(t)=\vec v(0)+(2\,\vec\imath-4\,\vec\jmath-2\,\vec k)t[/tex]

[tex]\vec v(t)=(v_{0x}+2t)\,\vec\imath+(v_{0y}-4t)\,\vec\jmath+(v_{0z}-2t)\,\vec k[/tex]

Then the particle has position at time [tex]t[/tex] according to

[tex]\displaystyle\vec r(t)=\vec r(0)+\int_0^t\vec v(u)\,\mathrm du[/tex]

[tex]\vec r(t)=(3+v_{0x}t+t^2)\,\vec\imath+(6+v_{0y}t-2t^2)\,\vec\jmath+(9+v_{0z}t-t^2)\,\vec k[/tex]

At at the point (3, 6, 9), i.e. when [tex]t=0[/tex], it has speed 8, so that

[tex]\|\vec v(0)\|=8\iff{v_{0x}}^2+{v_{0y}}^2+{v_{0z}}^2=64[/tex]

We know that at some time [tex]t=T[/tex], the particle is at the point (5, 2, 7), which tells us

[tex]\begin{cases}3+v_{0x}T+T^2=5\\6+v_{0y}T-2T^2=2\\9+v_{0z}T-T^2=7\end{cases}\implies\begin{cases}v_{0x}=\dfrac{2-T^2}T\\\\v_{0y}=\dfrac{2T^2-4}T\\\\v_{0z}=\dfrac{T^2-2}T\end{cases}[/tex]

and in particular we see that

[tex]v_{0y}=-2v_{0x}[/tex]

and

[tex]v_{0z}=-v_{0x}[/tex]

Then

[tex]{v_{0x}}^2+(-2v_{0x})^2+(-v_{0x})^2=6{v_{0x}}^2=64\implies v_{0x}=\pm\dfrac{4\sqrt6}3[/tex]

[tex]\implies v_{0y}=\mp\dfrac{8\sqrt6}3[/tex]

[tex]\implies v_{0z}=\mp\dfrac{4\sqrt6}3[/tex]

That is, there are two possible initial velocities for which the particle can travel between (3, 6, 9) and (5, 2, 7) with the given acceleration vector and given that it starts with a speed of 8. Then there are two possible solutions for its position vector; one of them is

[tex]\vec r(t)=\left(3+\dfrac{4\sqrt6}3t+t^2\right)\,\vec\imath+\left(6-\dfrac{8\sqrt6}3t-2t^2\right)\,\vec\jmath+\left(9-\dfrac{4\sqrt6}3t-t^2\right)\,\vec k[/tex]

To find the equation for the position vector of the particle, use the formula r(t) = r(0) + v(0) * t + (1/2) * a * t^2.

To find the equation for the position vector of the particle, use the following steps:

For the given problem, the equation for the position vector of the particle is: r(t) = (3 + 4t)i + (6 - 4t + 2t^2)j + (9 - 2t + t^2)k.