a 5.0-kg cart is moving horizontally at 6.0m/s. In order to change its speed to 10.0m/s, the net work done on the cart must be: 160J, 400J, 40J, 550J, 90J

Answer:

Gravity.

Explanation:

Gravity causes the child on a sled to slide down the hill.

Hope this helps!

Answer:

3.79 m/s^2

Explanation:

We know the small part loses 0.932 MJ of gravitational potential energy during its fall.

Potential energy = mass x gravitational field strength x height

Re-arranging to solve for gravitational field strength:

g = Potential energy/(mass x height)

Plugging in the given values:

g = 0.932 MJ / (1.8kg x 140km)

= 0.932 x 10^6 J / (1.8 x 1000kg x 140 x 1000m)

= 3.79 m/s^2

Therefore, the gravitational field strength of Mars is calculated to be 3.79 m/s^2.

A 0.5 kg soccer ball is kicked at 10 m/s. A goalie catches the ball and brings it to rest in 0.25 seconds then the force exerted on the ball by the goalie is 20N. Option C is correct.

Here, we must determine the change in momentum of the soccer ball. The momentum of an object is stated by the product of its mass and velocity. The mass of the soccer ball is 0.5 kg, and its initial velocity is 10 m/s. Therefore, the ball is conducted to be constant, and its final velocity is 0 m/s.

The change in momentum is computed by reducing the final momentum from the initial momentum. In this concern, the initial momentum is 0.5 kg × 10 m/s = 5 kg·m/s, and the final momentum is 0.5 kg × 0 m/s = 0 kg·m/s. Now, the change in momentum is 5 kg·m/s - 0 kg·m/s = 5 kg·m/s.

Next, we separate the change in momentum by the time taken to bring up the ball to rest, which is 0.25 seconds. Thus, the goalie's force exerted on the ball is 5 kg·m/s / 0.25 s = 20 N.

Therefore, the correct answer is C. 20 N.

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The correct Option is C. The force exerted on the ball by the goalie is 20 N.

The formula for the force exerted on an object is given by F = ma, where F is the force, m is the mass of the object and a is the acceleration.

The formula for acceleration is a = (v-u)/t, where v is the final velocity, u is the initial velocity and t is the time taken.

The acceleration is negative if the object is brought to rest.

So, for the given problem, the initial velocity of the soccer ball is 10 m/s and the final velocity is 0.

The time taken to bring it to rest is 0.25 s.

Therefore, the acceleration is given by:a = [tex](0 - 10)/0.25 = - 40 m/s^{2}[/tex]

Now, we can calculate the force exerted by the goalie using the formula: [tex]F = maF = 0.5 kg $\times$ (- 40 m/s^{2} ) = - 20 N[/tex]

We get a negative value for the force, which means that the force exerted is in the opposite direction to the motion of the ball.

However, the magnitude of the force is given by |-20 N| = 20 N.

So, the answer is option (C) 20 N.

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The velocity of the third piece is (81.25 m/s, -43.3 m/s).

To determine the velocity of the third piece, we can use the principle of conservation of momentum.

Given:

Mass of the first piece (m1) = 150 kg

Velocity of the first piece (v1) = 150 m/s (to the East)

Mass of the second piece (m2) = 100 kg

Velocity of the second piece (v2) = 200 m/s at a direction of south 60° West

Let's break down the velocities into their respective horizontal (x) and vertical (y) components.

For the first piece:

v1x = 150 m/s (since it's moving to the East)

v1y = 0 m/s (no vertical component)

For the second piece:

v2x = 200 m/s * cos(60°) = 200 m/s * 0.5 = 100 m/s (horizontal component)

v2y = -200 m/s * sin(60°) = -200 m/s * 0.866 = -173.2 m/s (vertical component, negative since it's moving downward)

Now, let's calculate the momentum of the first and second pieces:

The momentum of the first piece (p1) = m1 * v1

= 150 kg * 150 m/s

= 22,500 kg·m/s

The momentum of the second piece (p2) = m2 * v2

= 100 kg * (100 m/s, -173.2 m/s)

= (10,000 kg·m/s, -17,320 kg·m/s)

To find the total momentum after the explosion, we can add the momenta of the individual pieces:

Total momentum after the explosion = p1 + p2

= (22,500 kg·m/s, 0 kg·m/s) + (10,000 kg·m/s, -17,320 kg·m/s)

= (32,500 kg·m/s, -17,320 kg·m/s)

The total momentum after the explosion should also be equal to the momentum of the third piece:

The momentum of the third piece (p3) = m3 * v3

Given:

Mass of the third piece (m3) = 400 kg (calculated from the given mass of the bomb)

Let's assume the velocity of the third piece is (v3x, v3y).

Therefore, we have the equation:

(32,500 kg·m/s, -17,320 kg·m/s) = 400 kg * (v3x, v3y)

By equating the x and y components separately, we can solve for the velocity components of the third piece:

32,500 kg·m/s = 400 kg * v3x

-17,320 kg·m/s = 400 kg * v3y

Solving these equations, we find:

v3x = 81.25 m/s

v3y = -43.3 m/s

Therefore, the velocity of the third piece is approximately (81.25 m/s, -43.3 m/s).

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The second snowball should be thrown at an angle of approximately 48.196° with respect to the horizontal to arrive at the same point as the first snowball.

the second snowball should be thrown 4.582 seconds later in order for both to arrive at the same time.

To find the angle at which the second snowball should be thrown, we can use the fact that the horizontal displacement of both snowballs must be the same.

Let's first find the horizontal and vertical components of the velocity for the first snowball. The initial speed is 26.5 m/s, and the angle is 58.0° with respect to the horizontal.

The horizontal component of the velocity for the first snowball is given by:

V1x = V1 * cos(angle1)

    = 26.5 m/s * cos(58.0°)

    = 26.5 m/s * 0.530

    = 14.045 m/s

Now, let's find the vertical component of the velocity for the first snowball:

V1y = V1 * sin(angle1)

    = 26.5 m/s * sin(58.0°)

    = 26.5 m/s * 0.848

    = 22.472 m/s

Since the vertical acceleration is the same for both snowballs (gravity), the time it takes for both to arrive at the same point is the same. Therefore, we can use the time of flight of the first snowball to calculate the vertical displacement for the second snowball.

The time of flight for the first snowball can be calculated using the vertical component of velocity and the acceleration due to gravity:

t = (2 * V1y) / g

  = (2 * 22.472 m/s) / 9.8 m/s²

  ≈ 4.582 s

Now, let's find the vertical displacement for the second snowball:

Δy = V1y * t - (0.5 * g * t²)

    = 22.472 m/s * 4.582 s - (0.5 * 9.8 m/s² * (4.582 s)²)

    ≈ 103.049 m

To find the angle at which the second snowball should be thrown, we can use the horizontal displacement and the vertical displacement:

tan(angle2) = Δy / Δx

           = 103.049 m / (2 * 14.045 m/s * t)

           = 103.049 m / (2 * 14.045 m/s * 4.582 s)

           ≈ 1.085

Now, we can find the angle2 by taking the arctan of both sides:

angle2 ≈ arctan(1.085)

angle2 ≈ 48.196°

Therefore,

To find how many seconds later the second snowball should be thrown, we can simply use the time of flight of the first snowball, which is approximately 4.582 seconds.

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(a)The time taken for the pebble to reach the ground is approximately 2.01 seconds, and

(b) the horizontal distance traveled by the pebble is approximately 41.02 meters.

(c) The vertical distance traveled by the pebble is 55 meters.

(d) The initial vertical velocity of the pebble is 0 m/s because it is thrown horizontally.

(e) The vertical acceleration of the pebble is due to gravity and is approximately -9.8 m/s^2.

(f) The negative sign indicates that the pebble is moving downward.

a) To find the time taken for the pebble to reach the ground, we can use the equation for vertical motion:

h = (1/2)gt^2, where h is the vertical distance and g is the acceleration due to gravity.

Rearranging the equation, we have:

t = √((2h) / g), where t is the time taken.

Substituting the given values, we get:

t = √((2 * 55) / 9.8) ≈ 2.01 seconds.

b) The horizontal speed of the pebble remains constant throughout its motion. Therefore, the horizontal distance traveled by the pebble can be found by multiplying the horizontal speed by the time taken:

d = v0 * t, where d is the horizontal distance and v0 is the initial horizontal speed.

Substituting the given values, we have:

d = 20.5 * 2.01 ≈ 41.02 meters.

c) The vertical distance traveled by the pebble is given as 55 meters.

d) The initial vertical velocity of the pebble is 0 m/s because it is thrown horizontally.

e) The vertical acceleration of the pebble is due to gravity and is approximately -9.8 m/s^2.

f) The final vertical velocity of the pebble when it reaches the ground can be found using the equation:

v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.

Since the initial vertical velocity is 0 m/s and the acceleration due to gravity is -9.8 m/s^2, we have:

v = 0 + (-9.8) * 2.01 ≈ -19.8 m/s.

The negative sign indicates that the pebble is moving downward.

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a. The maximum height reached by the rocket is  1083 meters.

b.  The rocket reaches its maximum height 38 seconds after liftoff.

c.   The rocket is in the air for  1.09 seconds.

(b)

We will apply equation of motion :

v² = u² + 2aΔy

Δy = (v² - u²) / (2a)

Δy = (0 - 57.0²) / (2 * 1.50)

Δy = (-57.0)² / 3.00

Δy = 3,249 / 3.00

Δy = 1083 m

(c)

v = u + at

0 = u + at

t = -u / a

t = -57.0 / 1.50

t = 38 seconds

(d)

Δy = ut + (1/2)at²

140 = 57.0t + (1/2)(1.50)t²

(1/2)(1.50)t² + 57.0t - 140 = 0

t =  1.09 seconds.

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Answer:

C

Explanation:

If the tree is placed between the focus point F and the mirror in a concave mirror, the image of the tree will appear taller and on the same side of the mirror. Therefore, the correct answer is C. The image appears taller and on the same side of the mirror.

Answer:

5 m/s

Explanation:

We are given that 9000 kg of water flows through the pipe in 1 minute. Mass flow rate = mass/time

So, mass flow rate = 9000 kg / 1 minute = 150 kg/s

We know the cross sectional area of the pipe is 0.3 m2. From continuity equation, mass flow rate = density * area * velocity

So, 150 = 1000 * 0.3 * v (Density of water is approximately 1000 kg/m3)

Solving for v (velocity):

v = 150/(1000*0.3) = 5 m/s

Therefore, the speed of water in the pipe is 5 m/s.

Answer:

Hope this helps

Explanation:

Answer:

1. Concave mirror

2. Convex mirror

3. Concave mirror

4. Concave mirror

Explanation:

Concave mirror is placed near on an object it displays a virtual image

The graphic organizer suggests that the job of an astronaut is complex, demanding, and involves flight.

This conclusion can be drawn by examining the nature of the requirements listed in the graphic organizer. Firstly, the requirements listed in the organizer are numerous and encompass various aspects. They include educational qualifications, such as having a bachelor's degree in a relevant field, as well as specific experience, like piloting an aircraft.

These requirements highlight the complexity of the job and indicate that astronauts need to possess a diverse set of skills and knowledge. Additionally, the requirements for physical fitness and health demonstrate the demanding nature of the job.

Astronauts are expected to undergo rigorous physical training to ensure they can handle the physical stresses associated with space travel and the conditions they will encounter in space. This indicates that the job can be physically exhausting and requires individuals to be in excellent health.

Lastly, the inclusion of flight-related requirements, such as the need to pass a long-duration spaceflight physical and participate in aircraft flights, implies that the job of an astronaut involves actual flight experiences. This indicates that astronauts are directly involved in piloting spacecraft and are expected to have practical experience in flying.

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a) The speed of the 6.00 kg block after descending 0.800 m is 2.07 m/s.

b) We cannot calculate the work done by the friction force.

c) The net work done on the system of two blocks during the 0.800 m downward displacement of the 6.00 kg block is 29.13 J. The work done by gravity is 47.04 J, the work done by friction is unknown, and the work done by the tension in the rope is zero.

(a) The work done on the 6.00 kg block by gravity can be calculated using the formula:

Work_gravity = force_gravity * displacement * cos(theta),

where force_gravity is the weight of the block, displacement is the downward displacement of the block, and theta is the angle between the force and displacement vectors (which is 0 degrees in this case).

The weight of the block is given by:

force_gravity = mass * acceleration_due_to_gravity = 6.00 kg * 9.8 m/s^2 = 58.8 N.

Plugging in the values, we get:

Work_gravity = 58.8 N * 0.800 m * cos(0) = 47.04 J.

The work done on the 6.00 kg block by the tension in the rope is given by:

Work_tension = tension * displacement * cos(theta).

Plugging in the values, we get:

Work_tension = 37.0 N * 0.800 m * cos(180) = -29.6 J.

The negative sign indicates that the tension is in the opposite direction of the displacement.

Using the work-energy theorem, we can find the speed of the 6.00 kg block after descending 0.800 m:

Work_net = change_in_kinetic_energy.

Since the block starts from rest, its initial kinetic energy is zero. Therefore:

Work_net = Final_kinetic_energy - Initial_kinetic_energy = 1/2 * mass * velocity^2.

Solving for velocity, we get:

velocity = sqrt(2 * Work_net / mass).

The net work done on the block is the sum of the work done by gravity and the tension:

Work_net = Work_gravity + Work_tension = 47.04 J - 29.6 J = 17.44 J.

Plugging in the values, we get:

velocity = sqrt(2 * 17.44 J / 6.00 kg) = 2.07 m/s.

Therefore, the speed of the 6.00 kg block after descending 0.800 m is 2.07 m/s.

(b) The total work done on the 8.00 kg block during the 0.800 m displacement can be calculated using the work-energy theorem:

Work_net = change_in_kinetic_energy.

Since the 8.00 kg block is not moving vertically, its initial and final kinetic energies are zero. Therefore:

Work_net = Final_kinetic_energy - Initial_kinetic_energy = 0.

The work done on the 8.00 kg block by the tension in the rope is given by:

Work_tension = tension * displacement * cos(theta).

Plugging in the values, we get:

Work_tension = 37.0 N * 0.800 m * cos(0) = 29.6 J.

The work done on the 8.00 kg block by the friction force can be calculated using the formula:

Work_friction = force_friction * displacement * cos(theta),

where force_friction is the frictional force on the block. However, the problem statement does not provide the value of the friction force. Therefore, we cannot calculate the work done by the friction force.

(c) The net work done on the system of two blocks during the 0.800 m displacement of the 6.00 kg block can be found using the work-energy theorem:

Work_net = change_in_kinetic_energy.

Since the system starts from rest, the initial kinetic energy of the system is zero. Therefore:

Work_net = Final_kinetic_energy - Initial_kinetic_energy = 1/2 * (6.00 kg + 8.00 kg) * velocity^2.

Simplifying, we get:

Work_net = 1/2 * 14.00 kg * velocity^2.

Using the value of velocity calculated in part (a), we get:

Work_net = 1/2 * 14.00 kg * (2.07 m/s)^2 = 29.13 J.

The work done on the system of two blocks by gravity is the sum of the work done on the individual blocks by gravity:

Work_gravity_system = Work_gravity_6kg + Work_gravity_8kg = 47.04 J + 0 J = 47.04 J.

The work done on the system of two blocks by the tension in the rope is the sum of the work done on the individual blocks by the tension:

Work_tension_system = Work_tension_6kg + Work_tension_8kg = -29.6 J + 29.6 J = 0 J.

Therefore, the net work done on the system of two blocks during the 0.800 m downward displacement of the 6.00 kg block is 29.13 J. The work done by gravity is 47.04 J, the work done by friction is unknown, and the work done by the tension in the rope is zero.

Note: The calculations for part (b) and (c) were based on the given information, but the value of the friction force was not provided in the problem statement.

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Answer:

[tex]\displaystyle \frac{M + m}{m}\, \sqrt{2\, x\, u\,g}[/tex].

Explanation:

This question can be solved in the following steps:

Assume that the table is level. The normal force on the block would be equal to the weight of the block in magnitude [tex](M + m)\, g[/tex], but opposite in direction. As the block slows down, the only unbalanced force on the block would be friction [tex](-u\, (M + m)\, g)[/tex] (negative since this force is opposite to the direction of motion.)

The acceleration of the block would be:

[tex]\begin{aligned} a &= \frac{(\text{net force})}{(\text{mass})} \\&= \frac{-u\, (M + m)\, g}{M + m} \\ &= (-u\, g)\end{aligned}[/tex].

Apply the following SUVAT equation to find the velocity [tex]v_{i}[/tex] of the block right after the collision:

[tex]\displaystyle {v_{2}}^{2} - {v_{1}}^{2} = 2\, a\, x[/tex],

Where:

Rearrange and solve for [tex]v_{1}[/tex], the velocity right after collision:

[tex]\begin{aligned}v_{1} &= \sqrt{{v_{2}}^{2} - 2\, a\, x} \\ &= \sqrt{0^{2} - 2\, (-u\, g)\, x} \\ &= \sqrt{2\, x\, u\, g}\end{aligned}[/tex].

Apply the conservation of momentum to find the velocity of the bullet before the collision. Right after the collision, sum of momentum would be:

[tex](M + m)\, \sqrt{2\, x\, u\, g}[/tex].

Right before the collision, sum of momentum would be:

[tex]m\, v_{0}[/tex].

By the conservation of momentum:

[tex]m\, v_{0} = (M + m)\, \sqrt{2\, x\, u\, g}[/tex].

Rearrange and solve for [tex]v_{0}[/tex]:
[tex]\displaystyle v_{0} = \frac{M + m}{m}\, \sqrt{2\, x\, u\, g}[/tex].

The initial speed v0 of the bullet in terms of M, m, u, g, and d is identified by the equation v0 = (M + m) * [tex]\sqrt{((2 * u * g * d * m) / (M + m))} /m[/tex].

To find the initial speed v0 of the bullet in terms of M, m, u, g, and d, we can apply the principles of conservation of momentum and energy.

First, let's consider the conservation of momentum. Before the collision, the momentum of the bullet is given by m * v0 (where v0 is the initial velocity of the bullet), and the momentum of the wooden block is zero since it is initially at rest. After the collision, the combined system of the bullet and block moves together, so their momentum is (M + m) * V (where V is the common final velocity of the bullet and block). Since momentum is conserved, we have:

m * v0 = (M + m) * V

Next, let's consider energy conservation. The work done by the friction force over the distance d is given by the product of the force of friction and the distance d. The work done by friction is equal to the initial kinetic energy of the bullet-block system, which is (1/2) * (M + m) * V^2. Thus, we have:

(1/2) * (M + m) * V² = u * (M + m) * g * d

Now we can solve these two equations simultaneously to find the initial velocity v0. Rearranging the first equation, we have:

v0 = (M + m) * V / m

Substituting this expression for v0 into the second equation, we get:

(1/2) * (M + m) * [(M + m) * V / m]² = u * (M + m) * g * d

Simplifying and solving for V, we obtain:

V = [tex]\sqrt{((2 * u * g * d * m) / (M + m))}[/tex]

Finally, substituting this expression for V back into the first equation, we can find v0:

v0 = (M + m) * [tex]\sqrt{(2 * u * g * d * m) / (M + m)}[/tex] / m

Therefore, the initial speed v0 of the bullet in terms of M, m, u, g, and d is given by the above equation.

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Given that the initial velocity at which the ball is thrown vertically upward is 15m/s. Let us also assume that the value of acceleration due to gravity (g) = 9.8m/s² and in this case, the value will be -9.8m/s² as the ball is moving against gravity.

a) To calculate how high the ball rises, we can use the kinematic equation:

v² = u² + 2gs......(i)

where v ⇒ final velocity

u ⇒ initial velocity

g ⇒ acceleration and,

s ⇒ displacement (the height)

The final velocity will be 0 when the ball reaches its maximum height.

Substituting the values in equation (i), we get

0² = 15² + (2*-9.8*s)

0 = 225 - 19.6s

Thus, s = 225/19.6 = 11.48 m.

Therefore, the ball rises approximately 11.48 meters.

b) To find the time taken to reach the highest point, we can use the kinematic equation,

v = u + gt......(ii)

where t = time

Substituting the values in equation (ii)

0 = 15 - 9.8*t

t = -15/ -9.8 = 1.53 seconds

Thus, the time taken to reach the highest point = 1.53 seconds.

c) To find the time taken for the ball to hit the ground after it reaches its highest point, we can use the equation,

s = ut +1/2gt².....(iii)

As the ball is moving downwards, the initial velocity, u will be 0m/s.

Thus, substituting the values in equation (iii), we get

11.48 = 0*t + 1/2*9.8*t²

11.48 = 4.9t²

t² = 2.34

Therefore t = 1.53 seconds

Thus, the time taken for the ball to hit the ground is 1.53 seconds.

d)  To find the velocity at which the ball returns to the level from which it started, we can use the equation

v = u+ gt.....(iv)

v = 0 + 9.8*1.53

Thus, v = 14.99 ≅ 15 m/s

Therefore, the velocity when it returns to the level from which it started is 15m/s.

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The six digit grid coordinates for the windtee  should be 100049.

The United States military and NATO both utilize the military grid reference system (mgrs) as their geographic reference point.

When utilizing the geographic grid system, one must indicate whether coordinates are east (e) or west (w) of the prime meridian and either north (n) or south (s) of the equator.

If hill 192 is located midway between grid lines 47 and 48 and the grid line is 47, the coordinate would be 750.

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The function of the power supply is to provide electrical energy to the device or system that needs it. The power supply converts the incoming voltage from the power source into a form that is usable by the device, such as DC voltage.

The readout is a device or component that displays data or information to the user. The readout could be a simple LED display or a complex graphical display.

A peripheral is a device or component that connects to a computer or other electronic device to provide additional functionality. Examples of peripherals include printers, scanners, and external hard drives.

A microcomputer is a type of computer that is designed to fit on a single microchip. Microcomputers are found in a wide range of devices, including smart phones, tablets, and embedded systems.

A transducer is a device that converts one form of energy to another. In electronics, transducers are commonly used to convert electrical energy into mechanical energy, or vice versa.

The processor is the central component of a computer or electronic device. The processor is responsible for executing instructions and controlling the other components of the system. The performance and capabilities of a device are largely determined by the speed and power of the processor.

(a) The initial speed of the ball is 60.4 m/s

(b) The time of motion of the ball is 1.92 seconds.

(c) The velocity component of the ball is , x - component = 48.24 m/s

and y - component = 36.35 m/s.

(d) The speed of the ball as it reaches the wall is 64.8 m/s.

(a) The initial speed of the ball is calculated as;

t = √ (2h/g)

where;

t = √ (2 x 18 / 9.8 )

t = 1.92 s

v = d / t

v = 116 m / 1.92 s

v = 60.4 m/s

(b) The time of motion of the ball is 1.92 seconds.

(c) The velocity component of the ball is calculated as;

x - component = 60.4 m/s x cos (37) = 48.24 m/s

y - component = 60.4 m/s x sin (37) = 36.35 m/s

(d) The speed of the ball as it reaches the wall is calculated as;

v = vi + gt

where;

the time to travel 1 m high = √ (2 x 1 / 9.8 )

t = 0.45 s

v = 60.4 m/s  + 0.45(9.8)

v = 64.8 m/s

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Option A. The angle between the force and displacement is 180°, the maximum work is considered to be done by the force.

Work is defined as the product of the force applied to an object and the displacement of the object in the direction of the force. Mathematically, work (W) is given by the equation:

W = F * d * cos(theta)

Where

F = magnitude of the force

d = magnitude of the displacement

theta = angle between the force and displacement vectors.

In order to maximize the work done by a force, we need to maximize the value of the cosine of the angle theta. The cosine function reaches its maximum value of 1 when the angle theta is 0° or 180°.

When the angle between the force and displacement is 0° (option E), the force and displacement vectors are perfectly aligned in the same direction. In this case, the work done is maximized. Therefore, the correct answer is option A.

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The force that acts on a projectile in the horizontal direction is Gravitational force.

A projectile is an object upon which the only force is gravity. Gravity acts to influence the vertical motion of the projectile, thus causing a vertical acceleration. The horizontal motion of the projectile is the result of the tendency of any object in motion to remain in motion at constant velocity.

Due to the absence of horizontal forces, a projectile remains in motion with a constant horizontal velocity. Horizontal forces are not required to keep a projectile moving horizontally. Hence, The only force acting upon a projectile is gravity.

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(a) The tension in the inclined cable (T1) and horizontal cable (T2) supporting the cat burglar is equal. The tension in the vertical cable (T3) is 524 N.

(b) If the horizontal cable is reattached higher up, the tension in the inclined cable (T1) would increase.

(a) To find the tension in each cable supporting the 524-N cat burglar, we'll consider the forces acting on the system. Let's denote the tension in the inclined cable as T1, the tension in the horizontal cable as T2, and the tension in the vertical cable as T3. The angle between the inclined cable and the vertical cable is given as θ.

In the vertical direction, the tension in the vertical cable T3 balances the weight of the cat burglar:

T3 - 524 N = 0

T3 = 524 N

In the horizontal direction, the tension in the inclined cable T1 can be expressed as:

T1 * cos(θ) = T2

Now, we need to determine the value of θ to calculate T1 and T2. Let's assume that θ is the given angle of θ = 0.

Substituting the angle and rearranging the equation, we have:

T1 = T2 / cos(θ)

T1 = T2 / cos(0)

T1 = T2 / 1

T1 = T2

So, the tension in the inclined cable (T1) is equal to the tension in the horizontal cable (T2).

Therefore, the tension in each cable is as follows:

T1 (inclined cable) = T2 (horizontal cable)

T1 = T2

T3 (vertical cable) = 524 N

(b) If the horizontal cable were reattached higher up on the wall, the tension in the inclined cable (T1) would increase.

The correct answer is option A.  

This is because reattaching the horizontal cable at a higher point on the wall would increase the horizontal component of the tension, resulting in a larger tension in the inclined cable. The tension in the vertical cable (T3) would remain the same as it is independent of the position of the horizontal cable.

In summary, the tension in the inclined cable (T1) and the horizontal cable (T2) are equal, and their value depends on the angle θ. The tension in the vertical cable (T3) is 524 N. If the horizontal cable were reattached higher up on the wall, the tension in the inclined cable would increase, while the tension in the vertical cable would remain the same.

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The velocity of the third piece is v₃ = -12500 kg·m/s / m₃

The law of conservation of momentum states that the total momentum before the explosion is equal to the total momentum after the explosion.

velocity of the third piece =  v₃.

The total initial momentum before the explosion = 0

The total final momentum after the explosion= 0

Initial momentum = 0 kg·m/s (since the bomb is at rest)

Final momentum = m₁v₁ + m₂v₂ + m₃v₃

m₁ = mass of the first piece = 150 kg

v₁ = velocity of the first piece = 150 m/s (to the east)

m₂ = mass of the second piece = 100 kg

v₂ = velocity of the second piece = 200 m/s (south 60° west)

m₃ = mass of the third piece = unknown

v₃ = velocity of the third piece = unknown

0 = (150 kg)(150 m/s) + (100 kg)(200 m/s)(cos(60°)) + (m₃)(v₃)

final momentum = 0 and hence  v₃ is found as :

0 = 22500 kg·m/s - 10000 kg·m/s + (m₃)(v₃)

-12500 kg·m/s = (m₃)(v₃)

v₃ = -12500 kg·m/s / m₃

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a.  the tension in the rope is  91.5 N.

b.   the moment of inertia of the wheel is  0.1008 kg⋅m².

c.  the angular speed of the wheel 2.30 s after it begins rotating is  38.34 rad/s.

(a)

The tension in the rope can be found by considering the forces acting on the object.

ma = mg*sin(θ) - T

(14.0 kg)(2.00 m/s²)

= (14.0 kg)(9.8 m/s²)*sin(37°) - T

T = (14.0 kg)(9.8 m/s²)*sin(37°) - (14.0 kg)(2.00 m/s²)

T =  91.5 N

(b)

The moment of inertia of a wheel:

I = (1/2)MR²

I = (1/2)(14.0 kg)(0.12 m)²

I = 0.1008 kg⋅m²

(c)

The angular acceleration of the wheel:

α = a/R

α = angular acceleration,

a = linear acceleration of the object,

R =  radius of the wheel.

α = (2.00 m/s²)/(0.12 m)

α = 16.67 rad/s²

The angular speed (ω) of the wheel after time t is :

ω = ω₀ + αt

ω = 0 + (16.67 rad/s²)(2.30 s)

ω = 38.34 rad/s

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The amount of heat rejected to the room by the ideal refrigerator can be calculated using the Carnot efficiency. With the given temperatures and heat absorbed, the heat rejected to the room is 225 J.

To calculate the amount of heat rejected to the room by the ideal refrigerator, we can use the Carnot efficiency, which is given by the formula:

Efficiency = 1 - ([tex]T_c_o_l_d[/tex] / [tex]T_h_o_t[/tex])

where[tex]T_c_o_l_d[/tex]is the temperature of the cold reservoir (freezer compartment) and [tex]T_h_o_t[/tex] is the temperature of the hot reservoir (room temperature).

Given:

[tex]T_c_o_l_d[/tex] = -9 °C (converted to Kelvin: 264 K)

[tex]T_h_o_t[/tex]= 25 °C (converted to Kelvin: 298 K)

Heat absorbed from the freezer compartment ([tex]Q_c_o_l_d[/tex] = 120 J

First, we calculate the Carnot efficiency:

Efficiency = 1 - (264 K / 298 K)

Efficiency ≈ 0.1134

The Carnot efficiency represents the ratio of heat transferred from the cold reservoir to the work done by the refrigerator. Since the refrigerator is operating in reverse, the work done is equal to the heat absorbed from the freezer compartment ([tex]Q_c_o_l_d[/tex]).

[tex]Q_c_o_l_d[/tex] = 120 J

Now, we can calculate the heat rejected to the room ([tex]Q_h_o_t[/tex]) using the equation:

[tex]Q_h_o_t[/tex] = Efficiency * [tex]Q_c_o_l_d[/tex]

[tex]Q_h_o_t[/tex] ≈ 0.1134 * 120 J

[tex]Q_h_o_t[/tex] ≈ 13.61 J

Therefore, the amount of heat rejected to the room by the ideal refrigerator is approximately 13.61 J.

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The acceleration of the two blocks is[tex]2.14 m/s^{2[/tex]} and the distance does block M2 move in 2.00 s is 4.27 m.

Now we need to find the acceleration of the two blocks and the distance does block M2 move in 2.00 s.

We know that: mass of M1, m1 = 5.00 kg mass of M2, m2 = 19.0 kgθ = 25.0°Taking upward direction as positive for block M1 and downwards as positive for block M2.

Therefore, we can write the following equation of motion for the two blocks:

For M2: m2g - T = m2a ...(1)

For M1: T - m1g = m1a ...(2)

We can see from the figure that M2 is on an inclined plane making an angle θ with the horizontal.

We can resolve the weight of M2 into two components:

Perpendicular to the plane = m2gcosθParallel to the plane = m2gsinθ

The component parallel to the plane will tend to make the block move downwards.

Therefore, the effective weight will be:

mg = m2gsinθ ...(3)

From equation (1) we can write:

T = m2g - m2a ...(4)

Substituting equation (4) in equation (2), we get:

m2g - m2a - m1g = m1a ...(5)

On solving equation (5), we get the acceleration as:

a = g(m2sinθ - m1) / (m1 + m2)

On substituting the given values, we get:

[tex]a = 2.14 m/s^{2}[/tex]

The distance moved by M2 in 2 seconds can be found out using the formula:[tex]s = ut + \frac{1}{2} at^{2}[/tex]

Here, initial velocity, u = 0m/s Time, t = 2s Acceleration, [tex]a = 2.14 m/s^{2}[/tex]

On substituting these values, we get the distance travelled by M2 as: s = 4.27 m

Therefore, the acceleration of the two blocks is [tex]2.14 m/s^{2}[/tex]. And the distance does block M2 move in 2.00 s is 4.27 m.

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