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german
3 years ago
15

What is the kinetic energy of a 5 kg object moving at 4 m/s

Physics
1 answer:
mixas84 [53]3 years ago
4 0

Answer:

40

Explanation:

rule: K.E = 1/2 m×v^2

1/2 × 5 × (4)^2= 40 joule

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What is the speed of a truck that travels 60 km in 20 minutes
Alexxandr [17]
Formula:
Speed = Distance ÷ Time

Solution:
Speed = 60 ÷ 20
= 3km/min
6 0
3 years ago
How do u solve
Shalnov [3]

Time = distance/speed

         = 5/180

Time = 0.03 hrs

7 0
3 years ago
How long would it take to travel one light year?
Llana [10]
It depends on what speed you are going at. Assuming you are in our fastest spacecraft as of now which has a speed of 17,500 mph
1 light second = 186,282 mph
There are <span>31,536,000 in one year so 17,500x=31,534,000 and your answer would be in years. It would take you roughly 1802 years.</span>
(Sorry about mph I'm American.) 
5 0
3 years ago
Jim is driving a 2268-kg pickup truck at 19 m/s and releases his foot from the accelerator pedal. The car eventually stops due t
lord [1]

Answer:

The initial kinetic energy of the truck is 409374 J

Explanation:

This problem can be solved in two ways. Let´s solve it first in the easiest way.

The kinetic energy is calculated using this equation:

E = 1/2 · m · v²

Where:

E = kinetic energy

m = mass

v = velocity

Then, the kinetic energy of the truck will be:

E = 1/2 · 2268 kg · (19 m/s)² = 409374 J

And that´s it.

But we can complicate it a bit:

The kinetic energy is the work needed to move an object from rest to a desired velocity. If the object is moving, the work needed to stop it must be of the same magnitude as its kinetic energy (in the opposite direction to the movement).

The equation for work is:

W = F · d

Where:

W= work

F = force

d = distance

We know the magnitude of the force applied to the truck, but we do not know for how much distance that force was applied. The distance can be calculated using the equation for the position of an object moving in a straight line:

x = x0 + v0 · t + 1/2 · a · t²

where

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

But we still do not know the time nor the acceleration.

The acceleration can be obtained from the equation of force:

F = m · a

Where

F = force

m = mass

a = acceleration

Then:

900 N = 2268 kg ·a

a = 900 N /2268 kg = 0.397 m/s²

Now, we can calculate the time needed for the truck to stop. We know that at the final time, the velocity is 0. Then, we can use the equation for velocity to obtain that time:

v = v0 + a · t

Where:

v = velocity at time t

v0 = initial velocity

a = acceleration

t = time

Then:

v = 19 m/s - 0.397 m/s² · t

0 = 19 m/s - 0.397 m/s² · t

-19 m/s / -0.397 m/s² = t (acceleration is negative because it is opposite to the direction of the movement)

t = 47.86 s (The truck stoped at 47.86 s after releasing the foot from the accelerator pedal)

With the time and acceleration, we can calculate the traveled distance.

x = 0 m + 19 m/s · 47.86 s - 1/2 · 0.397 m/s² · (47.86s)²

x = 454.66 m (without rounding the acceleration nor the time, the value will be 454.86 m)

Now, we can calculate the work done to stop the truck which will be of the same magnitude as the kinetic energy:

W = 900 N · 454.66 m = 409194 J

(if you do all the calculations without rounding, you will get the same value as we calculated above using the equation of kinetic energy, 409374 J).

3 0
3 years ago
An astronaut drops a rock on the surface of an asteroid.The rock is released from rest at a height of 0.86 m above the ground, a
SCORPION-xisa [38]

Answer:

a_y=0.92m/s^2

Explanation:

To solve this problem we use the formula for accelerated motion:

y=y_0+v_{y0}t+\frac{a_yt^2}{2}

We will take the initial position as our reference (y_0=0m) and the downward direction as positive. Since the rock departs from rest we have:

y=\frac{a_yt^2}{2}

Which means our acceleration would be:

a_y=\frac{2y}{t^2}

Using our values:

a_y=\frac{2(0.86m)}{(1.37s)^2}=0.92m/s^2

5 0
4 years ago
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