What is meant by input and output work

An object with total mass mtotal = 16.2 kg is sitting at rest when it explodes into three pieces. One piece with mass m1 = 4.7 k

vichka [17]

Answer:

1.) 0 kgm/s

2) 6.3 kg

3) -0.0978 m/s

4)

5)

6)

An object with total mass mtotal = 16.2 kg is sitting at rest when it explodes into three pieces. One piece with mass m1 = 4.7 kg moves up and to the left at an angle of θ1 = 23° above the –x axis with a speed of v1 = 25.4 m/s. A second piece with mass m2 = 5.2 kg moves down and to the right an angle of θ2 = 28° to the right of the -y axis at a speed of v2 = 23.8 m/s.

2) What is the mass of the third piece?

3) What is the x-component of the velocity of the third piece?

4) What is the y-component of the velocity of the third piece?

5) What is the magnitude of the velocity of the center of mass of the pieces after the collision?

6) Calculate the increase in kinetic energy of the pieces during the explosion

Explanation:

Since explosions and collisions follow the law of conservation of Momentum.

1) Magnitude of the final momentum of the system = Magnitude of the initial momentum of the system

Since the body was initially at rest,

Magnitude of the initial momentum of the system = 0 kgm/s

Hence, Magnitude of the final momentum of the system is also equal to 0 kgm/s.

2) mass of the third piece

Sum of all the masses = 16.2 kg

4.7 + 5.2 + c = 16.2

c = 6.3 kg

3) doing an x-component balance on momentum

(4.7)×(-25.4 cos 23) + (5.2)×(23.8 cos 28) + (6.3)(v) = 0

-0.6163 + 6.3v = 0

v = -0.0978 m/s

Hope this Helps!!!

4 0

3 years ago

A car traveling 75 km/h slows down at a constant 0.50 m/s2 just by "letting up on the gas." calculate (a) the distance the car c

Serjik [45]

Solution:

At 1st convert km/h to m/s. 1 km = 1000 m, 1 h = 3600 s, 1 km/m = 1000/3600 = 5/18 m/s

Initial velocity = 75 * 5/18 = 20.8 m/s

The car’s velocity decreases from 20.8 m/s to 0 m/s at the rate of 0.5 m/s each second. We have the final velocity, initial velocity, and the acceleration.

Now according to the equation determine the distance.

vf^2 = vi^2 + 2 * a * d

a = -0.5 m/s^2

0 = 20.8^2 + 2 * -0.5 * d

so d = 431.64 m

since we have the final velocity, initial velocity, and the acceleration. Use the following equation to determine time.

vf = vi + a * t

0 = 20.8 – 0.5 * t

Solve for t = 41 seconds

(c) the distance travels by it during the first and fifth second are.

d = vi * t + ½ * a * t^2

d1 = 20.8 * 1 – ½ * 0.5 * 1^2 = 20.55 m

The easiest way to the distance for the 5th second is:

d = vi * t + ½ * a * t^2, a = -0.5

d5 = 20.8 * 5 – ½ * 0.5 * 5^2 = 91.5 m

d6 = 20.8 * 6 – ½ * 0.5 * 6^2 = 106.8m

d6 – d5 = 15.3 m

this is the required solution.

3 0

3 years ago

The following questions present a twist on the scenario above to test your understanding. Suppose another stone is thrown horizo

Ipatiy [6.2K]

The first part of the text is missing, you can find on google:

"A ball is thrown horizontally from the roof of a building 45 m. If it strikes the ground 56 m away, find the following values."

Let's now solve the different parts.

(a) 3.03 s

The time of flight can be found by analyzing the vertical motion only. The vertical displacement at time t is given by

$y(t) = h -\frac{1}{2}gt^2$

where

h = 45 m is the initial height

g = 9.8 m/s^2 is the acceleration of gravity

When y=0, the ball reaches the ground, so the time taken for this to happen can be found by substituting y=0 and solving for the time:

$0=h-\frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(45)}{9.8}}=3.03 s$

(b) 18.5 m/s

For this part, we need to analyze the horizontal motion only, which is a uniform motion at constant speed.

The horizontal position is given by

$x=v_x t$

where

$v_x$ is the horizontal speed, which is constant

t is the time

At t = 3.03 s (time of flight), we know that the horizontal position is x = 56 m. By substituting these numbers and solving for vx, we find the horizontal speed:

$v_x = \frac{x}{t}=\frac{56}{3.03}=18.5 m/s$

The ball was thrown horizontally: this means that its initial vertical speed was zero, so 18.5 m/s was also its initial overall speed.

(c) 35.0 m/s at 58.1 degrees below the horizontal

At the impact, we know that the horizontal speed is still the same:

$v_x = 18.5 m/s$

we need to find the vertical velocity. This can be done by using the equation

$v_y = u_y -gt$

where

$u_y =0$ is the initial vertical velocity

g is the acceleration of gravity

t is the time

Substituting t = 3.03 s, we find the vertical velocity at the time of impact:

$v_y = -(9.8)(3.03)=-29.7 m/s$

So the magnitude of the velocity at the impact (so, the speed at the impact) is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{18.5^2+(-29.7)^2}=35.0 m/s$

The angle instead can be found as:

$\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{-29.7}{18.5})=-58.1^{\circ}$

so, 58.1 degrees below the horizontal.

4 0

3 years ago

A 2300 kg truck has put its front bumper against the rear bumper of a 2500 kg suv to give it a push. with the engine at full pow

Snowcat [4.5K]

From the Newton’s First Law, we can see that acceleration is simply the ratio of Force over mass. In this case, mass is the sum of the mass of each car, that is:

mass = 2300 kg + 2500 kg = 4800 kg

So the formula is:

acceleration = Force / mass

acceleration = 18,000 N / 4800 kg

acceleration = 3.75 m/s^2

In 2 significant figures:

<span>acceleration = 3.8 m/s^2</span>

5 0

3 years ago

Read 2 more answers

One of the principal difficulties in establishing the theory of continental drift (the idea that tectonic plates drift around on

iragen [17]

Answer:

The energy that drives this movement is heat within the earth, this heat comes from two main sources which are;

1) residual heat left over from the formation of our planet billions of years ago.

2) The radioactive decay of naturally occurring chemical elements (notably uranium, thorium, and potassium ) in the earth releases energy in the form of heat. These two sources of heat warm Earth’s mantle and cause it to rise and sink, this rising and sinking creates a convectional movement of earth mantle.

4 0

4 years ago

What is meant by input and output work​

What is meant by input and output work