miniature spring-loaded, radio-controlled gun is mounted on an air puck. The gun's bullet has a mass of 5.00 g, and the gun and puck have a combined mass of 120 g. With the system initially at rest, the radio controlled trigger releases the bullet causing the puck and empty gun to move with a speed of 0.500 m/s. What is the bullet's speed

(a) The magnitude and direction of the net force on the crate while it is on the rough surface is 34.7 N in opposite direction.

(b) The net work done on the crate while it is on the rough surface is -22.6 J.

(c) The speed of the crate when it reaches the end is 0.5 m/s.

The magnitude and direction of the net force on the crate while it is on the rough surface is calculated as follows;

F (net) = F - Ff

where;

F (net) = F - μmg

where;

F (net) = 289 N  -   (0.359 x 92 X 9.8)

F (net) = -34.7 N

The negative sign indicates opposite direction to the applied force.

The net work done on the crate while it is on the rough surface is calculated as follows;

W = F(net) x L

where;

W = -34.7 x 0.65

W = -22.6 J

The speed of the crate when it reaches the end is calculated as follows;

acceleration of the crate = F(net) / m

a = -34.7 N / 92 kg

a = -0.377 m/s²

v² = u² + 2aL

v² = ( 0.855)²  +  ( 2 x -0.377 x 0.65)

v² = ( 0.855)²   - ( 2 x 0.377 x 0.65)

v² = 0.24

v = √ 0.24

v = 0.5 m/s

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Answer:

Explanation:

To find the final velocity of an object given it's initial velocity , time taken and it's acceleration we use the formula

where

v is the final velocity

u is the initial velocity

t is the time taken

a is the acceleration

From the question

u = 40 m/s

t = 10s

a = 5 m/s²

We have

v = 40 + 5(10) = 40 + 50

We have the final answer as

Hope this helps you

The cost of operation for an A.C for 10 hours per day for a month will be 71,100 francs.

Power is the amount of energy transferred or converted per unit time. The unit of power is the watt, equal to one joule per second. Power is a scalar quantity.

Cost of operation for 10 hours a day;

Daily consumption = 3kW x 10 hours

Daily Consumption = 30kW

Since 1kWH = 79 francs;

Daily consumption amount = 30 x 79 francs

Daily consumption amount = 2,370 francs

Therefore, the monthly consumption (using 30days) will be;

2,370 francs x 30 = 71,100 francs

In conclusion, 71,100 francs will be spent in a month (30 days) to run the 3kW rated A.C for 10 hours a day at 1kWH.

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Answer:

1.0 m/s^2

Explanation: happy to help :)

Answer: \(1\ m/s^2\)

Explanation:

Given

Masses of the block are \(m_1=1\ kg\)  and

\(m_2=2\ kg\)

Force applied by \(1\ kg\) block on \(2\ kg\) block is \(2\ N\)

From the free body diagram of \(2\ kg\) block, the net force on

\(\therefore m_2a=2\\\\\Rightarrow 2\times a=2\\\\\Rightarrow a=\dfrac{2}{2}\\\\\Rightarrow a=1\ m/s^2\)

Thus, the acceleration of two blocks is \(1\ m/s^2\)

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The acceleration of both points A and B can be determined from the given velocities and accelerations of the gear racks.

Point A:

x-component of acceleration = 0 m/s^2

y-component of acceleration = 6 m/s^2

Point B:

x-component of acceleration = 0 m/s^2

y-component of acceleration = -6 m/s^2

The acceleration of both points A and B can be determined from the given velocities and accelerations of the gear racks. The x-component of the acceleration of both points is equal to 0 m/s^2 since the gear racks are not moving horizontally. The y-component of the acceleration of both points is equal to 6 m/s^2 in the upward direction for point A, and -6 m/s^2 in the downward direction for point B. Therefore, the acceleration of point A is (0, 6) m/s^2 and the acceleration of point B is (0, -6) m/s^2.

The complete question is:

At a given instant, the gear racks have the velocities and accelerations shown.

Determine the acceleration of points A and B.

Enter the x and y components of the acceleration

the complete question is :

At a given instant, the gear racks have the velocities and accelerations shown ( attached below ).

Determine the acceleration of point A.

Enter the x and y components of the acceleration

Determine the acceleration of point B.

Enter the x and y components of the acceleration

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Answer:

The correct answer is option b

Explanation:

The range of a projectile is given by,

\(R=\frac{v^{2}sin 20 }{g} \)

Where

 v is the initial velocity with which the projectile was launched.

 θ is the angle of the projection projectile.

 g is the acceleration due to gravity.

The range will be maximum when the term sin2θ assumes the maximum possible value.

The maximum possible value of sine of an angle is 1. Thus the maximum range is given by,

\(R=\frac{v^{2}x 1 }{g}\\ =\frac{v^{2} }{g} \)

Therefore the correct answer is option b, v²/g

The angular velocity of solid cylinder after 5s whose mass is 20kg with radius 0.4 m is subject to a force of 4.5 N at 30ﾟ above the horizontal and 0.17 m from it center is 1.195rad/s.

Given the mass of solid cylinder (M) = 20kg

The radius of cylinder (r) = 0.4m

The force exerted on cylinder (F) = 4.5N

The angle of force applied = 30°

The force applied from a distance (d) = 0.17m

Time after the velocity (t) = 5s

Let the angular velocity = ω

The angular velocity of the cylinder can be calculated using the equation of torque.

Torque (T) = Iα

Where I is the moment of inertia of the cylinder and α is the angular acceleration.

The moment of inertia of a cylinder is given by :

\(I = (1/2) mr^2\) such that:

\(Torque = (1/2) mr^2 * \alpha\)

\(T = (1/2) * (20 kg) * (0.4 m)^2 * \alpha\)

\(T = 1.6 kg m^2 * \alpha\)

The torque acting on the cylinder due to the applied force can be also calculated as follows where:

T = d * F * sin(θ)

T = 0.17 * 4.5 * sin 30°

T = 0.3825Nm

Now equating the magnitude of the torque to the torque calculated using the equation of torque, we get :

\(0.3825Nm = 1.6 kg m^2 * \alpha\)

\(\alpha = 0.239rad/s^2\)

Now, using the equation of angular acceleration, we can calculate the angular velocity after 5 seconds as:

ω = ω0 + αt where ω0 is the initial angular velocity = 0

α = ω/t

\(0.239rad/s^2 = \omega/5s\)

ω = 1.195rad/s

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Answer:

1 km

Explanation:

displacement =velocity ×time

displacement =40m/s ×25s

displacement =1000m equivalent to 1km

Answer:

The heat needed to raise the temperature of 1 gallon of water from 68°F to 40°C is approximately 28,265 Joules.

Explanation:

Answer:

soccer when the ball hits an unprepared player in the head. He also gave examples of concussions occurring when players accidentally knock their heads into other players while attempting to head the ball, particularly if they are attempting to flick the ball backwards.

Explanation:

Heading in soccer can increase your risk of concussions. Over time, repeated subconcussive injuries can also accumulate and cause brain damage.

Answer:

the pressure of the water at the given depth is 196,200 N/m².

Explanation:

Given;

density of the water, ρ = 1000 kg/m³

depth of the water, h = 20 m

acceleration due to gravity, g = 9.81 m/s²

The pressure at the given depth of the water is calculated as;

P = ρgh

P = 1000 x 9.81 x 20

P = 196,200 N/m²

Therefore, the pressure of the water at the given depth is 196,200 N/m².

When the teacher asked the student to make words out of the jumbled word gzeysktqix, the student was being tested on his ability to unscramble words. Unscrambling words is the process of taking a word or series of letters that are out of order and rearranging them to form a word that makes sense.

When trying to unscramble a word, it is important to look for any patterns that can help identify smaller words within the jumbled letters. This can help make the process easier and quicker. For example, in the jumbled word gzeysktqix, one might notice that the letters "sktqix" appear together.

This could indicate that these letters could potentially form a word. By looking at the remaining letters, one could notice that the letters "g", "z", "e", and "y" could also form smaller words. After some rearranging, the letters can be unscrambled to form the words "sky", "zig", "sex", and "yet". These are just a few examples, as there are likely many other words that can be formed from this jumbled word.

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Answer:

Explanation:

according to the graph at B the potential energy of the particle is 2J

therefore we can use the kinetic energy equation to calculate the particle's velocity or speed.

\(E_{k} =1/2mv^{2}\)

2J= 1/2*1/2kg*v^2

8=v^2

v= 2√2 ms-1

Answer:

A)  two voltages are equal, B) the two voltages are different

C) voltage is the same V = 0, D)  V_{battery} = V_{Lamp} > V_{wire}

Explanation:

For this exercise, it is asked to build a circuit with the plug, the lamp and the battery, the circuit must be a series circuit

Before connecting the battery voltage is Vo as they indicate that it is ideal there are no losses, the voltages of the other elements are zero.

A) Voltages battery

a) single battery Vo

b) the battery is connected to the lamp this takes something current, but the electromotive force does not change therefore the voltage is Vo; This is because the battery's energy comes from a chemical reaction of the elements inside it.

in summary the two voltages are equal

B) voltage Lamp

a) when the bulb is single, its voltage is zero V = 0

since it has no energy

b) when in the circuit V = V₀ - V_r

where V_r is the voltage across the wire, due to its resistance, the energy for these voltages is given by the battery

in this case the two voltages are different

C) the voltage across the wire

a) unconnected V = 0

b) connected V = V₀ - V_L

where V_L is the voltage across the lamp due to its resistance

we can write the wire voltage

                V_r = i R

as they indicate that the wire is ideal, its resistance is zero R = 0, consequently

               V_r = 0

the voltage is the same V = 0

D) circuit voltage

 For this part we must write Ohm's law for this circuit

            V₀ = V_L + V_r

where V₀ is the battery voltage, V_L and V_r are the voltage of the lamp and the wire, respectively.

The voltage of each element is

                   V_L = i R_L

                   V_r = i R

This is a series circuit so the current is also constant

                 V₀ = i (R_L + R)

therefore the voltage in the battery is the highest V₀

the voltage across the wire if it is ideal is zero V_r = 0, if it is ideal its resistance is zero

therefore the voltage on the lamp is equal to the voltage of the battery

          V_{battery} = V_{Lamp} > V_{wire}

Based on the information provided, there will be a net movement of water from the side with a lower concentration of solute to the side with a higher concentration of solute.

This process is known as osmosis, which is the movement of water across a selectively permeable membrane from an area of lower solute concentration to an area of higher solute concentration. The goal of osmosis is to reach an equilibrium, where the concentration of solute is the same on both sides of the membrane. In this case, the water will move from the side with a lower concentration of solute to the side with a higher concentration of solute until the concentrations are equal on both sides.

In summary, the net movement of water in the u-tube experiment will be from the side with a lower concentration of solute to the side with a higher concentration of solute in order to reach an equilibrium.

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Answer:

Explanation:

To find the focal length, we will use the following equation

Where do is the distance of the object and di is the distance of the image.

So, replacing do = 8.0 cm and di = 16.0 cm, we get

Therefore, the focal length is 5.3 cm

So, the final velocity of the ball when it is 2.101 above the ground is 13.99 m/s or can be rounded to 14 m/s.

Hi ! In this question, I will help you. This question will addopt the principle of final velocity in free fall. Free fall is vertical downward movement that occurs when any object dropped without initial velocity. In othee word, the object that falls is only affected by the presence of gravity and its initial high. In general, the final velocity in free fall can be expressed by this equation :

\( \boxed{\sf{\bold{v = \sqrt{2 \times g \times h}}}} \)

With the following condition :

We know that :

Note :

At this point (2.101 m above the ground), the object can still complete its movement up to exactly 0 m above the ground.

What was asked :

Step by Step

\( \sf{v = \sqrt{2 \times g \times \Delta h}} \)

\( \sf{v = \sqrt{2 \times g \times (h_1 - h_2)}} \)

\( \sf{v = \sqrt{2 \times 9.8 \times (12.09 - 2.101)}} \)

\( \sf{v = \sqrt{19.6 \times 9,989}} \)

\( \sf{v \approx \sqrt{195.78}} \)

\( \boxed{\sf{v = 13.99 \: m/s \approx 14 \: m/s}} \)

So, the final velocity of the ball when it is 2.101 above the ground is 13.99 m/s or can be rounded to 14 m/s.

Air can do work when it exerts a force on an object and causes it to undergo displacement. The ability of air to do work is evident in various phenomena, such as wind pushing sails, fans moving objects, and air pressure powering pneumatic systems.

Air can do work through its ability to exert a force over a distance. Work is defined as the transfer of energy that occurs when a force is applied to an object and it undergoes displacement in the direction of the force. When air is in motion, it possesses kinetic energy and can exert a force on objects in its path, thus performing work.

To understand how air can do work, we can consider the example of a moving fan. When a fan is turned on, the blades start to rotate, creating a flow of air. As the air moves, it carries kinetic energy. When the moving air encounters an object, such as a piece of paper, the air molecules collide with the paper's surface and exert a force on it. This force causes the paper to move and displaces it from its initial position.

The work done by the air can be calculated using the equation:

Work = Force * Distance * cos(θ)

Where Force is the magnitude of the force exerted by the air, Distance is the displacement of the object, and θ is the angle between the direction of the force and the displacement.

In the case of air doing work on an object, the force exerted by the air is perpendicular to the direction of motion, resulting in θ = 90 degrees. Since cos(90) = 0, the equation simplifies to:

Work = Force * Distance * 0

Therefore, the work done by the air on the object is zero when the force exerted by the air is perpendicular to the displacement.

However, if the force exerted by the air is not perpendicular to the displacement, such as when blowing air at an angle to move an object, then work is performed. The air exerts a force on the object and causes it to move in the direction of the force, resulting in the transfer of energy.

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Answer:

The position of the object at \(t = 5\,s\) is 130.167 meters.

Explanation:

Let \(a(t) = 5\cdot t\,\left[\frac{m}{s^{2}} \right]\) the acceleration experimented by the object along the x-axis. We obtain the equation for the position of the object by integrating in acceleration formula twice:

Velocity

\(v(t) = \int {a(t)} \, dt\) (1)

\(v(t) = 5\int {t} \, dt\)

\(v(t) = \frac{5}{2}\cdot t^{2}+v_{o}\) (2)

Where \(v_{o}\) is the initial velocity of the object, measured in meters per second.

Position

\(s(t) = \int {v(t)} \, dt\) (3)

\(s(t) = \frac{5}{2}\int {t^{2}} \, dt+v_{o}\int \, dt\)

\(s(t) = \frac{5}{6}\cdot t^{3}+v_{o}\cdot t + s_{o}\) (4)

Where \(s_{o}\) is the initial position of the object, measured in meters per second.

If we know that \(s_{o} = 6\,m\), \(v_{o} = 4\,\frac{m}{s}\) and \(t = 5\,s\), then the position of the object is:

\(s(5) = \frac{5}{6}\cdot (5)^{3}+\left(4\right)\cdot (5)+6\)

\(s(5) = 130.167\,m\)

The position of the object at \(t = 5\,s\) is 130.167 meters.

Answer:

d is the awnser

Explanation:

MABY?!??

Answer:

hdhshsisjsbrtheisebvrtctvsjusyevevrvrg eggs haushehegehs

Student 1 would have a power 467 W and student 2 would have a power of 433 W. The correct option is the fourth option - Student 1 would have 467 W, and Student 2 would have 433 W of power.

From the question,

We are to calculate the power each student would have to climb the flight of stairs.

Power can be calculated using the formula

\(P = \frac{F \times d}{t}\)

Where

P is Power

F is the force

d is the distance

and t is the time

NOTE: The weight of the students represent the force

F = 700 N

d = 4 m

t = 6 s

∴ \(P = \frac{700 \times 4}{6}\)

P = 467 W

F = 650 N

d = 4 m

t = 6 s

∴ \(P = \frac{650 \times 4}{6}\)

P = 433 W

Hence, Student 1 would have a power 467 W and student 2 would have a power of 433 W. The correct option is the fourth option - Student 1 would have 467 W, and Student 2 would have 433 W of power.

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Answer:

a break in the circuit can also cause this problem

The equation of the progressive wave is y = 12 cos(40t - 4x)

The general wave equation is given by:

y = A sin(ωt - kx)

Where A is the amplitude, ω is the angular frequency = 2πf, f is the frequency, k is the wave number and y, x is the displacement.

Given the equation for a progressive wave is y=6 cos⁡(20t-4x). Hence:

The amplitude A = 6,

ω = 20 = 2πf

f = 20/2π = 3.183 Hz

Twice the amplitude = 2 * 6 = 12, twice the frequency = 2 * 3.183.

ω = 2π(3.183*2) = 40

Therefore the other progressive wave has an equation of:

y = 12 cos(40t - 4x)

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Answer:

about 42.35 m/s

Explanation:

Use the equation for accelerated motion (g), and with zero initial velocity that doesn't include time:

\(v_f^2=v_i^2+2\,a\,\Delta x\)

which for our case would reduce to:

\(v_f^2=v_i^2+2\,a\,\Delta x\\v_f^2=0+2\,9.8\,(91.5)\\v_f^2= 1793.4\\v_f=\sqrt{1793.4} \\v_f \approx 42.35\)

then the velocity just before hitting would be about 42.35 m/s