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3 years ago

9

A 0.45 kg soccer ball changes its velocity by 20.0 m/s due to a force applied to it in 0.10 seconds. What force was necessary fo

r this change in velocity?

2 answers:

Illusion [34]3 years ago

8 0

Know that change in momentum = impulse (Δp = F*t), momentum is mass*velocity, and impulse is force*time

Δp = F*t
$m_}}v__{f}} - m_}}v__i}} = Ft \\ 
m(v__f}}-v__i}}) = Ft \\ 
(0.45kg)(20 m/s) = F(0.10s) \\ 
F = 90N$ .

frutty [35]3 years ago

6 0

Answer:

90 N

Explanation:

The force applied to the ball is given by:

$F=\frac{\Delta p}{\Delta t}$

where

$\Delta p$ is the change in momentum of the ball

$\Delta t$ is the time taken

The change on momentum of the ball is:

$\Delta p=m\Delta v=(0.45 kg)(20 m/s)=9 kg m/s$

So, the force applied is

$F=\frac{9 kg m/s}{0.10 s}=90 N$

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In an electric field, 0.90 joule of work is required to bring 0.45 coulomb of charge from point a to point

jarptica [38.1K]

The difference in electric potential energy between the two points is
$\Delta U = q \Delta V$
where q is the magnitude of the charge and $\Delta V$ is the electric potential difference.

But for energy conservation, the difference in electric potential energy $\Delta U$ between the two points is equal to the work done to move the charge between A and B:
$W=\Delta U$
so we have
$W=q \Delta V$

and by substituting the numbers of the problem, we find the value of $\Delta V$ :
$\Delta V = \frac{W}{q}= \frac{0.90 J}{0.45 C}=2 V$

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3 years ago

A rocket burns fuel to create hot gases that explode violently out of the rocket engine. This explosion creates thrust. Thrust i

Lerok [7]

Answer:

Thrust due to fuel consumption must overcome gravitational force from the Earth to send the rocket up into space.

Explanation:

From the concept of Escape Velocity, derived from Newton's Law of Gravitation, definition of Work, Work-Energy Theorem and Principle of Energy Conservation, which is the minimum speed such that rocket can overcome gravitational forces exerted by the Earth, and according to the Tsiolkovski's Rocket Equation, which states that thrust done by the rocket is equal to the change in linear momentum of the rocket itself, we conclude that thrust due to fuel consumption must overcome gravitational force from the Earth to send the rocket up into space.

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2 years ago

What does the rotational speed of the ring module have to be so that an astronaut standing on the outer rim of the ring module f

Usimov [2.4K]

Complete question:

In the movie The Martian, astronauts travel to Mars in a spaceship called Hermes. This ship has a ring module that rotates around the ship to create “artificial gravity” within the module. Astronauts standing inside the ring module on the outer rim feel like they are standing on the surface of the Earth. (The trailer for this movie shows Hermes at t=2:19 and demonstrates the “artificial gravity” concept between t= 2:19 and t=2:24.)

Analyzing a still frame from the trailer and using the height of the actress to set the scale, you determine that the distance from the center of the ship to the outer rim of the ring module is 11.60 m

What does the rotational speed of the ring module have to be so that an astronaut standing on the outer rim of the ring module feels like they are standing on the surface of the Earth?

Answer:

The rotational speed of the ring module have to be 0.92 rad/s

Explanation:

Given;

the distance from the center of the ship to the outer rim of the ring module r, = 11.60 m

When the astronaut standing on the outer rim of the ring module feels like they are standing on the surface of the Earth, then their centripetal acceleration will be equal to acceleration due to gravity of Earth.

Centripetal acceleration, a = g = 9.8 m/s²

Centripetal acceleration, a = v²/r

But v = ωr

a = g = ω²r

$\omega = \sqrt{\frac{g}{r}} = \sqrt{\frac{9.8}{11.6} } = 0.92 \ rad/s$

Therefore, the rotational speed of the ring module have to be 0.92 rad/s

3 0

3 years ago

A big box of sausages (30 kg) is lifted from the ground to the top shelf of the freezer. If the box is lifted at a constant spee

Lubov Fominskaja [6]

Answer:

Work done to lift the box is 515.03 J

Explanation:

By work energy theorem we know that work done by all forces is equal to change in kinetic energy

So we have

$W_g + W_{ex} = \Delta K$

so we have

$-mgh + W_{ex} = 0$

so we have

$W_{ex} = mgh$

$W_{ex} = 30(9.8)1.75$

$W_{ex} = 515.03 J$

5 0

3 years ago

9. A motorcycle accelerates from 10. m/s to 55. m/s in 5.0 seconds. What is the average acceleration of the bike (provide reason

yuradex [85]

Answer:

9.0 m/s/s

Explanation:

Given the following data;

Initial velocity = 10 m/s

Final velocity = 55 m/s

Time = 5 seconds

To find the average acceleration;

In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.

This simply means that, acceleration is given by the subtraction of initial velocity from the final velocity all over time.

Hence, if we subtract the initial velocity from the final velocity and divide that by the time, we can calculate an object’s acceleration.

Mathematically, acceleration is given by the equation;

$Acceleration (a) = \frac{final \; velocity - initial \; velocity}{time}$

Substituting into the equation;

$a = \frac{55 - 10}{5}$

$a = \frac{45}{5}$

Acceleration, a = 9.0 m/s²

Therefore, the average acceleration of the motorcycle is 9 meters per seconds square because it's experiencing a uniform velocity or motion.

5 0

3 years ago