Two objects with masses of m1 = 3.70 kg and m2 = 5.70 kg are connected by a light string that passes over a frictionless pulley, as in the figure below. Answer parts a-c.

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 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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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.

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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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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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.

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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(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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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 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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