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kodGreya [7K]
3 years ago
8

What is the work required for a penguin to push a box 2 meters with a force of 8 newtons?

Physics
1 answer:
choli [55]3 years ago
3 0

Work done is given by product of force and displacement due to that force

So here we will have

Work = Force \times displacement

here we know that

Force = 8 N

displacement = 2 m

Now work done is given as

W = 8\times 2

W = 16 J

so it will do 16 J work to move the box

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borishaifa [10]
What measures we can't answer without the measures
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3 years ago
A child slides down a snow‑covered slope on a sled. At the top of the slope, her mother gives her a push to start her off with a
kirill [66]

Answer:

θ = 13.16 °

Explanation:

Lets take mass of child = m

Initial velocity ,u= 1.1 m/s

Final velocity ,v=3.7 m/s

d= 22.5 m

The force due to gravity along the incline plane = m g sinθ

The friction force = (m g)/5

Now from work power energy

We know that

work done by all forces = change in kinetic energy

( m g sinθ - (m g)/5 ) d = 1/2 m v² - 1/2 m u²

(2  g sinθ - ( 2 g)/5 ) d = v² -  u²

take g = 10 m/s²

(20 sinθ - ( 20)/5 ) 22.5 = 3.7² -  1.1²

20 sinθ - 4 =12.48/22.5

θ = 13.16 °

5 0
3 years ago
Type the correct answer in the box. Spell the word correctly.
serious [3.7K]

Answer:

The Lambda-CDM model contains a cosmological constant, denoted by a lambda (λ), which is associated with dark energy and <u>cold dark matter</u>.

^Also works for Plato users.

3 0
3 years ago
Read 2 more answers
You observe two cars traveling in the same direction on a long, straight section of Highway 5. The red car is moving at a consta
Arte-miy333 [17]

Answer:

1)  3.66 s

2) 124.44 m

3) 3.12 s

Explanation:

Let's start by first listing down the information in the question.

Red Car : 34 m/s

Blue Car: 28 m/s

Distance between them : 22 m

The difference in speed between the cars is: 34 - 28 = 6 m/s

This means that the red car is catching up to the blue car at a speed of 6 m/s.

1) We can solve this by just dividing the distance by the difference in speed. This becomes:    \frac{Distance}{Speed}= \frac{22}{6} =   3.66

Thus it takes 3.66 seconds for the red car to catch up to the blue car.

2) We know from (1) that it took 3.66 seconds for the red car to catch up. Since the speed it was travelling at is constant, we only need to multiply it by the time from (1) to get the distance.

This becomes:    Speed * Time = 34 * 3.66 = 124.44

Thus the red car travels 124.44 m before catching up to the blue car.

3) If the red car starts to accelerate the moment we see it, the time taken to get to the blue car will be less than before. We can find this in a simple way.

We can use the motion equation : s = u*t + \frac{1}{2}(a * t^2)

Here s = 22 m

We can take u as the difference in speed. u = 6 m/s

Acceleration a = (2/3) m/s^2

Substituting the these into the equation we get:

22 = 6t + \frac{1}{2}(\frac{2}{3}t^2)

Solving this for the variable 't' using the quadratic formula we get the following two answers:

t1  = 3.12 s

t2 = - 21.12 s

Since t2 is not possible, the answer is t1. This means it takes 3.12 seconds for the red car to catch up to the blue.

4 0
3 years ago
If the driver slammed on the brakes, what could happen to the crate?
stich3 [128]
The crate would slide forward
7 0
3 years ago
Read 2 more answers
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