A block slides on a horizontal surface with an initial speed of 10 m/s. If the block takes 2s to come to a stop, what is the coefficient of kinetic friction between the block and the surface? (The answer does not depend on the mass of the block.) (A) 0.1 (B) 0.2 (C) 0.4 (D) 0.5

Answer:

The net force on him is zero.

Explanation:

The velocity of skydiver is constant.

As we know that the acceleration is rate of change in velocity. So, here velocity os constant it means acceleration of skydiver is zero.

According to Newton's second law

Force acting on a body is equal to the product of mass and velocity of the body.

As acceleration is zero that means the net force acting on the body is zero.

Answer:

[tex]8.1^{\circ}, 171.9^{\circ}[/tex]

Explanation:

The magnitude of the magnetic force exerted on the moving electron is:

[tex]F=qvB sin \theta[/tex]

where here we have

[tex]F=1.40\cdot 10^{-16} N[/tex] is the magnitude of the force

[tex]q=1.6\cdot 10^{-19} C[/tex] is the magnitude of the electron charge

B = 1.23 T is the magnetic field intensity

[tex]\theta[/tex] is the angle between the direction of the electron's velocity and the magnetic field

Solving the equation for [tex]\theta[/tex], we find:

[tex]sin \theta = \frac{F}{qvB}=\frac{1.40\cdot 10^{-16}N}{(1.6\cdot 10^{-19} C)(5.06\cdot 10^3 m/s)(1.23 T)}=0.141[/tex]

which gives the following two angles:

[tex]\theta = 8.1^{\circ}\\\theta = 180^{\circ}-8.1^{\circ} = 171.9^{\circ}[/tex]

Answer:

D. 39 N m

Explanation:

m = mass of the weight used in crossfit workout = 7.0 kg

Force due to the weight used is given as

F = mg

F = (7.0) (9.8)

F = 68.6 N

d = distance of point of action of weight from shoulder joint = 0.57 m

τ = Torque about the shoulder joint due to the weight

Torque about the shoulder joint due to the weight is given as

τ = F d

Inserting the values

τ = (68.6) (0.57)

τ = 39 Nm

This statement is true.

Yes, supervisors can face disciplinary action for engaging in retaliation. They can be held accountable if they retaliate against an employee for reporting violations or participating in protected activities. Penalties can vary from warnings to termination.

True. Supervisors, just like any other employees, are subject to disciplinary action for engaging in retaliation. For instance, if a supervisor retaliates against an employee for reporting a violation of company policies or for engaging in protected activities like organizing or supporting a labor union, they can be held accountable. Disciplinary actions can range from written warnings to termination, depending on the severity of the retaliation. Therefore, it is essential for supervisors to respect the rights of the employees and abide by all workplace regulations to maintain a safe and fair environment.

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Answer:

sorry I've never took in this class before I will not be able to help you

She needs to move [tex]x[/tex] km in the east-west direction and [tex]y[/tex] km in the north-south direction so that

[tex]\sqrt{x^2+y^2}=2.17[/tex]

and

[tex]\tan29.6^\circ=\dfrac yx[/tex]

Solve the system to get

[tex]x=1.89\,\mathrm{km}[/tex]

[tex]y=1.07\,\mathrm{km}[/tex]

Answer:

Change in momentum is 1.1275 kg-m/s

Explanation:

It is given that,

Mass of the ball, m = 274 g = 0.274 kg

It hits the floor and rebounds upwards.

The ball hits the floor with a speed of 2.40 m/s i.e. u = -2.40 m/s  (-ve because the ball hits the ground)

It rebounds with a speed of 1.7 m/s i.e. v = 1.7 m/s (+ve because the ball rebounds in upward direction)

We have to find the change in the ball's momentum. It is given by :

[tex]\Delta p=p_f-p_i[/tex]

[tex]\Delta p=m(v-u)[/tex]

[tex]\Delta p=0.275\ kg(1.7\ m/s-(-2.4\ m/s))[/tex]

[tex]\Delta p=1.1275\ kg-m/s[/tex]

So, the change in the momentum is 1.1275 kg-m/s

The magnitude of the change in the ball's momentum when rebounding off the floor is 1.1275 kg·m/s, accounting for the change in direction during impact.

To determine the magnitude of the change in the ball's momentum, you should first consider the initial and final momenta of the ball. Momentum is calculated as the product of mass and velocity. When the ball hits the floor, it has a downward momentum of (mass × velocity before hitting the floor). After rebounding, it has an upward momentum of (mass × velocity after rebounding). Since the problem states that up is in the positive direction, you will have to take into account the change in direction when calculating the change in momentum.

To calculate the magnitude of the change in momentum (Δp), you use the formula Δp = p_final - p_initial. Plugging in the values:

p_initial = mass × velocity before hitting = 0.275 kg × (-2.40 m/s) = -0.66 kg·m/s

p_final = mass × velocity after rebounding = 0.275 kg × 1.70 m/s = 0.4675 kg·m/s

Δp = p_final - p_initial = 0.4675 kg·m/s - (-0.66 kg·m/s) = 1.1275 kg·m/s

The negative sign for p_initial indicates that it was directed downwards. The magnitude of the change in momentum is simply the absolute value of Δp, which is 1.1275 kg·m/s.

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Answer:

The process in which heat flows by the mass movement of molecules from one place to another is C) convection.

Hope this helps :)

The process in which heat flows by the mass movement of molecules from one place to another is convection.

Heat transfer by convection is a method of heat transfer that involves the mass movement of molecules from one place to another.

Heat can also be transfered in the following methods;

Thus, the process in which heat flows by the mass movement of molecules from one place to another is convection.

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Final answer:

The proton's speed after 4 seconds will remain the same as its initial speed of 1.5 x 10^6 m/s.

Explanation:

To find the proton's speed 4 seconds after entering the magnetic field, we need to apply the right-hand rule. The magnetic force on a charged particle moving in a magnetic field is given by the equation F = qvBsin(θ), where F is the force, q is the charge, v is the velocity, B is the magnetic field, and θ is the angle between the velocity and the magnetic field. Since the proton's initial velocity vector makes an angle of 30° with the magnetic field, the force acting on the proton will be perpendicular to its velocity. Therefore, it will not change the proton's speed, but rather cause it to move in a circular path. As a result, the proton's speed after 4 seconds will remain the same as its initial speed of 1.5 x 10^6 m/s.

Answer:

F = 345.45 N

Explanation:

Angular acceleration of the disc is given as rate of change in angular speed

it is given by formula

[tex]\alpha = \frac{d\omega}{dt}[/tex]

[tex]\alpha = \frac{2\pi(0.600)}{2}[/tex]

[tex]\alpha = 1.88 rad/s^2[/tex]

now we know that moment of inertia of the solid uniform disc is given as

[tex]I = \frac{1}{2}mR^2[/tex]

[tex]I = \frac{1}{2}245(1.50)^2[/tex]

[tex]I = 275.625 kg m^2[/tex]

now we have an equation for torque as

[tex]\Tau = I\alpha[/tex]

[tex]r F = 275.625(1.88)[/tex]

[tex]F = \frac{275.625(1.88)}{1.50}[/tex]

[tex]F = 345.45 N[/tex]

Answer:

The intensity of this wave and energy is 6.3385 N/m² and 2.0509 J.

Explanation:

Given that,

Electric field amplitude E₀= 69.1 V/m

Area A= 0.0247 m²

Time t= 13.1 s

We need to calculate the intensity

Using formula of intensity

[tex]S=\dfrac{1}{2}c\epsilon_{0}E_{0}^2[/tex]

Where, c = speed of light

Put the value into the formula

[tex]S=\dfrac{1}{2}\times3\times10^{8}\times8.85\times10^{-12}\times(69.1)^2[/tex]

[tex]S=6.3385\ N/m^2[/tex]

(b). We need to calculate the energy

Using formula of energy

[tex]E=SAt[/tex]

Where, A = area

t = time

S = intensity

Put the value into the formula

[tex]E =6.3385\times0.0247\times13.1[/tex]

[tex]E =2.0509\ J[/tex]

Hence, The intensity of this wave and energy is 6.3385 N/m² and 2.0509 J.

Answer:

200 revolutions per minute = 20.9 rad/s

Explanation:

It is given that, a figure skater is rotating at a rate of 200 revolutions per minute. It is the angular velocity of the skater. We have to convert it into radian/second.

Revolution per minute can be converted to radian per minute as :

Since, 1 radian/second = 60/2π revolutions per minute

So, 1 revolution/minute = 2π/60 radian/second

200 revolution/minute = 200 × 2π/60 radian/second

200 revolution/minute = 20.9 radian/second

Hence, this is the required solution.

Answer:

The work done will be 57.15 J

Explanation:

Given that,

Mass = 12 kg

Distance = 3 m

Force = 22 N

Angle = 30°

We need to calculate the work done  

The work done is defined as,

[tex]W = Fd\cos\theta[/tex]

Where, F = force

d = displacement

Put the value into the formula

[tex]W=22\times3\times\cos30^{\circ}[/tex]

[tex]W=22\times3\times\dfrac{\sqrt{3}}{2}[/tex]

[tex]W = 57.15\ J[/tex]

Hence, The work done will be 57.15 J

Hello! My name is Zalgo and I am here to help you out on this concluding day. The answer would be C);lower. The reason it would be lower is because the hottest color of flames would be blue. Considering the way a start emits light is fire, this would be the most logical reason for it.

I hope that this helps! :P

"Stay Brainly and stay proud!" - Zalgo

(By the way, do you mind marking me as Brainliest? I'd greatly appreciate it! Thanks! X3)

a. The wheel accelerates uniformly, so its constant acceleration is equal to the average acceleration:

[tex]\alpha=\dfrac{0.20\frac{\rm rev}{\rm s}-0}{32.0\,\rm s}=0.0063\dfrac{\rm rev}{\mathrm s^2}[/tex]

b. Yes. Since

[tex]\alpha=\dfrac{\Delta\omega}{\Delta t}=\dfrac\omega{\Delta t}[/tex]

then multiplying [tex]\alpha[/tex] by 2 means we double the change in angular speed, but the wheel starts from rest so only the final angular speed [tex]\omega[/tex] gets doubled.

Answer:

not that high, because worms are never that efficient

The efficiency of a worm gear transmitted at a power of 15HP and at 1150 rpm with a pitch of 0.75 inches and pitch diameter of 3 inches is 77%.

What is Worm gear?

A worm gear is a type of gear that utilizes a spiral-threaded shaft to drive a toothed wheel. One of the six simple machines is the vintage worm gear. A worm gear is essentially a screw butted up against what appears to be a typical spur gear with slightly curved and inclined teeth.

Given: Power of gear(P) = 15 hp,

Torque (N) = 1150 rpm;

The pitch of gear (p) = 0.75;

Pitch diameter (D₁) = 3 inches;

Pressure angle (α) = 14 1 / [tex]2^{0}[/tex];

Friction coefficient  (μ) = 0.12

If m is the module of gear, z is the no of teeth, γ is the worm lead angle and l is the length of gear then

l = [tex]\pi[/tex]mz

l = 238.75               (m = 1 / p = 2)

And  tanγ = 1 / [tex]\pi[/tex]D₁

γ = 44.92

Now If the efficiency is η then ;

η =(cosα - μtanγ) / (cosα + μtanγ)

η = 77%

Therefore, the efficiency is 77%

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Answer:

Elastic potential energy, E = 2 J

Explanation:

It is given that,

Spring constant of the spring, k = 2500 N/m

The spring is stretched to a distance of 4 cm i.e. x = 0.04 m

We have to find the elastic potential energy possessed by the spring. A spring possessed elastic potential energy and it is given by:

[tex]E=\dfrac{1}{2}kx^2[/tex]

[tex]E=\dfrac{1}{2}\times 2500\ N/m\times (0.04\ m)^2[/tex]

E = 2 Joules.

Hence, the correct option is (d) " 2 Joules ".

Answer:

The block will be slide 0.36 m on the ice.

Explanation:

Given that,

Mass of arrow m₁= 80 g

Velocity of arrow u₁= 80 m/s

Mass of block m₂= 8.0 kg

Force F = 7.1 N

Using conservation of momentum

[tex]m_{1}u_{1}=m_{2}v_{2}[/tex]

[tex]80\times10^{-3}\times80=8.0\times v[/tex]

[tex]v =\dfrac{80\times10^{-3}\times80}{8.0}[/tex]

[tex]v = 0.8\ m/s[/tex]

The work done is equal to the change in kinetic energy

[tex]W=\Delta KE[/tex]

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

[tex]W=\dfrac{1}{2}\times8.0\times0.8^2[/tex]

[tex]W=2.56\ J[/tex]

We know that,

The work is defined as,

[tex]W = F\cdot d[/tex]

[tex]d = \dfrac{W}{F}[/tex]

[tex]d=\dfrac{2.56}{7.1}[/tex]

[tex]d =0.36\ m[/tex]

Hence, The block will be slide 0.36 m on the ice.

The solution to the question is found by applying the principles of conservation of momentum to calculate the velocity of the block after collision and then using the work-energy theorem to find the distance the block slides on ice.

The calculation to find the answer to your question requires the usage of conservation of momentum and the theory of work. Firstly, we apply conservation of momentum, using the formula: initial momentum = final momentum.  Momentum, p=m*v, where m is mass and v is velocity. The initial momentum is the momentum of the arrow just before it hits the block, equal to (0.080 kg * 80 m/s). The block of ice initially is at rest, so it has no momentum.

Post-collision, the arrow and block move together, so the final momentum is (8.080 kg * V), with V being the velocity we wish to calculate. Set the initial and final momentum equal and solve for V.

Now we have the velocity of the block and arrow post-collision. The block slides until brought to rest by friction. Here we use the work-energy theory, where the work is equal to the change in kinetic energy. The work done by the friction force is (friction force * distance), and the change in kinetic energy is (1/2)*m*V^2 - 0. Solve for distance to find the answer.

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The work done by the gravitational force on a ball that rises to a height and then returns to its original point is zero, as the work done during the ascent is equivalent but opposite to the work done during the descent.

The work done by the gravitational force on an object that rises to a height and then returns to its original point is zero. This is because the work done by the gravitational force when the ball rises is equal in magnitude but opposite in direction to the work done when the ball falls back down.

When the ball ascends, the work done by the gravitational force is negative since the displacement (upwards) is opposite to the direction of gravitational force (downwards). The formula to calculate the work done in this phase is W = -mg * h, where 'm' is the mass of the ball, 'g' is the acceleration due to gravity and 'h' is the height.

During the ball's descent, the work done is positive as the displacement (downwards) is in the same direction as the direction of gravitational force (downwards). The work done in this phase can be calculated using the same formula W = mg * h (noting that 'h' is a negative height as the ball is descending). When you add these together, the total work done by gravity over the full journey of the object is zero.

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Answer:

- 1.3 x 10⁻¹⁵ C/m

Explanation:

Q = Total charge on the circular arc = - 353 e = - 353 (1.6 x 10⁻¹⁹) C = - 564.8 x 10⁻¹⁹ C

r = Radius of the arc = 5.30 cm = 0.053 m

θ = Angle subtended by the arc = 48° deg = 48 x 0.0175 rad = 0.84 rad        (Since 1 deg = 0.0175 rad)

L = length of the arc

length of the arc is given as

L = r θ

L = (0.053) (0.84)

L = 0.045 m

λ = Linear charge density

Linear charge density is given as

[tex]\lambda =\frac{Q}{L}[/tex]

Inserting the values

[tex]\lambda =\frac{-564.8\times 10^{-19}}{0.045}[/tex]

λ = - 1.3 x 10⁻¹⁵ C/m

Answer:

Required time, t = 0.84 seconds

Explanation:

It is given that,

Initial speed of an object, u = 22 m/s

Final velocity of an object, v = 27 m/s

Acceleration, a = 5.93 m/s²

We have to find the time required for an object to go a speed of 22 m/s to a speed of 27 m/s. It can be solved by using first equation of motion as:

[tex]v=u+at[/tex]

Where

t = time

[tex]t=\dfrac{v-u}{a}[/tex]

[tex]t=\dfrac{27\ m/s-22\ m/s}{5.93\ m/s^2}[/tex]

t = 0.84 seconds

Hence, the time required for an object is 0.84 seconds.

Answer:

1.5 x 10^-8 Tesla

Explanation:

E = 4.5 V/m

The relation between electric field and the magnetic field in electromagnetic wave is given by

c = E / B

Where, c be the velocity of light in vacuum, e be the electric field and B be the magnetic field

B = E / c

B = 4.5 / ( 3 x 10^8)

B = 1.5 x 10^-8 Tesla

The magnetic field of an electromagnetic wave at a specific point can be calculated using the known electric field and the speed of light. In this case, the magnitude of the magnetic field is 1.5 x 10^-8 Tesla (T).

In the context of an electromagnetic wave, the electric and magnetic fields are linked through Maxwell's Equations. Given that an electromagnetic wave propagates in free space, the ratio of the amplitudes of the electric field (E) to the magnetic field (B) is always equal to the speed of light (c), according to the equation c = E/B. Thus, when the magnitude of the electric field (E) is known, the magnetic field (B) can be calculated as B=E/c. In this case, the Electric field (E) is 4.50 V/m. Hence, using the value of speed of light c = 3x10^8 m/sec, we calculate the magnetic field (B) as follows: B = 4.50 V/m / 3x10^8 m/sec = 1.5 x 10^-8 Tesla (T).

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Answer:

Option (2)

Explanation:

The sun is a large astronomical body where there occurs the process of nuclear fusion. This process is responsible for the occurrence of two important things. Firstly, it helps in the conversion o f hydrogen atoms into helium, that fuels the energy of the sun, and secondly, it helps in the continuous conversion of matter into energy, that reaches the earth's surface and on which the living organisms are directly dependent on.

Thus, there occurs conversion of about 4 million tons of matter into energy every second.

Therefore, the correct answer is option (2).

Answer:

C

Explanation:

Centripetal acceleration is:

a = v² / r

First, convert km/h to m/s:

108 km/h × (1000 m / km) × (1 h / 3600 s) = 30 m/s

Therefore, the acceleration is:

a = (30 m/s)² / 300 m

a = 3 m/s²

Answer:

C

Explanation:

Positive acceleration describes an increase in speed; negative acceleration describes a decrease in speed.

Answer:

a) It takes 5.79 seconds for the grinding wheel to stop.

b) In an interval of 5.79 seconds it rotates 28.48 rad.

Explanation:

a) We have equation of motion v = u + at

  Here v = 0 rad/s, a = -1.70 rad/s², u = 94 rev/min = 9.84 rad/s

  Substituting

         0 = 9.84 - 1.70 x t

          t = 5.79 seconds.

It takes 5.79 seconds for the grinding wheel to stop.

b) We have equation of motion v² = u² + 2as

      v = 0 rad/s, a = -1.70 rad/s², u = 9.84 rad/s

  Substituting

         0² = 9.84² - 2 x 1.70 x s

          s = 28.48 rad

 So in an interval of 5.79 seconds it rotates 28.48 rad.

we have that from the Question, it can be said that   distance x from the left-hand end of the bar is

x=1.834m

From the Question we are told

A uniform metal bar is 5.00 m long and has mass 0.300 kg. The bar is pivoted on a narrow support that is 2.00 m from the left-hand end of the bar. What distance x from the left-hand end of the bar should an object with mass 0.900 kg be suspended so the bar is balanced in a horizontal position?

Generally the equation for Balanced torque is mathematically given as

T=0.3/5*3

T=0.18kg

Therefore  

The clockwise torque is

T_c=0.18*1.5g

And

The anti-clockwise torque is

T=0.3/5*2g

T=0.12

Hence

0.18*1.5g=0.3/5*2g+0.9kxg

Therefore

x=1.834m

Therefore

distance x from the left-hand end of the bar is

x=1.834m

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The position of the 0.9 kg mass to balance the metal bar is 1.83 m.

The given parameters;

The center of gravity of the metal bar = 2.5 m

A sketch of weight on the metal bar;

|-----------2 m-----------|--- 0.5 m-----|

0---------------------------Δ--------------------------------------------------

   P ↓|-------- x----------|                     ↓

    0.9 kg                                   0.3 kg

Take moment about the pivot point;

0.9x = 0.3(0.5)

0.9x = 0.15

[tex]x = \frac{0.15}{0.9} \\\\x = 0.167 \ m[/tex]

2 m - P = 0.167 m

P = 2m - 0.167 m

P = 1.83 m

Thus, the position of the 0.9 kg mass to balance the metal bar is 1.83 m.

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Answer:

Part a)

U = 31.25 J

Part b)

U = 312.5 J

Explanation:

Part A)

A spring that requires a force of 50 N to be stretched 0.2 m from its equilibrium position.

So here we have

[tex]F = kx[/tex]

[tex]50 = k(0.2)[/tex]

k = 250 N/m

now the energy stored in the spring is given by

[tex]U = \frac{1}{2}kx^2[/tex]

[tex]U = \frac{1}{2}(250)(0.5)^2[/tex]

[tex]U = 31.25 J[/tex]

Part B)

A spring that requires 50 J of work to be stretched 0.2 m from its equilibrium position.

So here we know the formula of spring energy as

[tex]U = \frac{1}{2}kx^2[/tex]

[tex]50 = \frac{1}{2}k(0.2)^2[/tex]

[tex]k = 2500 N/m[/tex]

now by the formula of energy stored in spring

[tex]U = \frac{1}{2}kx^2[/tex]

[tex]U = \frac{1}{2}(2500)(0.5)^2[/tex]

[tex]U = 312.5 J[/tex]

Final answer:

The work required to stretch the first spring 0.5 m from its equilibrium position, given a spring constant of 250 N/m, is 31.25 J. For the second spring, with a spring constant of 2500 N/m, the work required is 312.5 J.

Explanation:

To calculate the work required to stretch the springs mentioned in the student's question, we use Hooke's law which is given by W = ½ k x², where W is the work done, k is the spring constant, and x is the displacement of the spring from its equilibrium position.

Part a

Given that a force of 50 N stretches the spring 0.2 m, we first determine the spring constant using F = k x:
k = F / x = 50 N / 0.2 m = 250 N/m.
Then, we calculate the work done to stretch the spring 0.5 m from its equilibrium position using W = ½ k x² = ½ × 250 N/m × (0.5 m)² = 31.25 J.

Part b

If 50 J of work is required to stretch the spring 0.2 m, the spring constant can be obtained by rearranging the work formula:
50 J = ½ k (0.2 m)², solving for k, yields k = 50 J / (0.5 × (0.2 m)²) = 2500 N/m.
Finally, calculating the work needed to stretch the spring 0.5 m from its equilibrium using W = ½ k x² = ½ × 2500 N/m × (0.5 m)² = 312.5 J.

(a) 0.165 m/s

The total initial momentum of the astronaut+capsule system is zero (assuming they are both at rest, if we use the reference frame of the capsule):

[tex]p_i = 0[/tex]

The final total momentum is instead:

[tex]p_f = m_a v_a + m_c v_c[/tex]

where

[tex]m_a = 125 kg[/tex] is the mass of the astronaut

[tex]v_a = 2.50 m/s[/tex] is the velocity of the astronaut

[tex]m_c = 1900 kg[/tex] is the mass of the capsule

[tex]v_c[/tex] is the velocity of the capsule

Since the total momentum must be conserved, we have

[tex]p_i = p_f = 0[/tex]

so

[tex]m_a v_a + m_c v_c=0[/tex]

Solving the equation for [tex]v_c[/tex], we find

[tex]v_c = - \frac{m_a v_a}{m_c}=-\frac{(125 kg)(2.50 m/s)}{1900 kg}=-0.165 m/s[/tex]

(negative direction means opposite to the astronaut)

So, the change in speed of the capsule is 0.165 m/s.

(b) 520.8 N

We can calculate the average force exerted by the capsule on the man by using the impulse theorem, which states that the product between the average force and the time of the collision is equal to the change in momentum of the astronaut:

[tex]F \Delta t = \Delta p[/tex]

The change in momentum of the astronaut is

[tex]\Delta p= m\Delta v = (125 kg)(2.50 m/s)=312.5 kg m/s[/tex]

And the duration of the push is

[tex]\Delta t = 0.600 s[/tex]

So re-arranging the equation we find the average force exerted by the capsule on the astronaut:

[tex]F=\frac{\Delta p}{\Delta t}=\frac{312.5 kg m/s}{0.600 s}=520.8 N[/tex]

And according to Newton's third law, the astronaut exerts an equal and opposite force on the capsule.

(c) 25.9 J, 390.6 J

The kinetic energy of an object is given by:

[tex]K=\frac{1}{2}mv^2[/tex]

where

m is the mass

v is the speed

For the astronaut, m = 125 kg and v = 2.50 m/s, so its kinetic energy is

[tex]K=\frac{1}{2}(125 kg)(2.50 m/s)^2=390.6 J[/tex]

For the capsule, m = 1900 kg and v = 0.165 m/s, so its kinetic energy is

[tex]K=\frac{1}{2}(1900 kg)(0.165 m/s)^2=25.9 J[/tex]

The astronaut's push off leads to a change in the speed of the space capsule of -0.164 m/s. The average force exerted is approximately 416.67N. The final kinetic energy of the astronaut and the space capsule are 156.25J and 25628.1J respectively.

This problem covers the principle of conservation of momentum. The astronaut and the space capsule constitute a closed system, where the total momentum before and after the push must be equal.

(a) Using the principle of conservation of momentum (initial momentum = final momentum), we can calculate the change in speed of the space capsule when the astronaut pushes off. We start with the equation m1V1 + m2V2 = 0, where m1 is the astronaut's mass and V1 is her speed, and m2 is the space capsule's mass and V2 is its velocity. Solving for V2 gives us a change in speed of the space capsule of -0.164m/s (which is in the opposite direction to the astronaut's motion).

(b) The average force exerted can be calculated by changing momentum over time (Force = Change in momentum / Time). Here we obtain approximately 416.67N.

(c) The final kinetic energy of each object is K = 1/2*m*v^2. For the astronaut, this is approximately 156.25J and for the space capsule, this is 25628.1J.

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Answer:

An electric motor is a device that changes electrical energy into mechanical energy. This change occurs due to the interaction between the magnetic field of magnets and the magnetic field due to the  electric current in the  loop. The interaction between the two produces a torque that makes the loop rotate on a shaft.

An electric motor is electric machinery that converts electrical energy supplied to it to mechanical energy.

It works on the principle of applying a magnetic field in electromagnetism.

A current-carrying loop is exposed to a magnetic field which results in torque.

The torque formed rotates the coil which is then followed by rotation of propellers as the current passes through the loop.

Explanation

An electric motor consist of a rotor- a moving part, commutator, brushes, axle, field-magnet, power supply and a stator- a part going around the rotor.

Answer:

1.974 g

Explanation:

Electrochemical equivalent of copper, z = 0.000329 g/C

I = 10 A

t = 10 minutes = 10 x 60 = 600 seconds

By the use of Farady's law of electrolysis

m = z I t

m = 0.000329 x 10 x 600

m = 1.974 g