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Lunna [17]
3 years ago
7

A 2.0 kg puck is at rest on a level table. It is pushed straight north with a constant force of 5N for 1.50 s and then let go. H

ow far does the puck move from rest in 2.25 s?

Physics
1 answer:
lukranit [14]3 years ago
4 0

Answer:

d = 6.32 m

Explanation:

Given that,

The mass of a puck, m = 2 kg

It is pushed straight north with a constant force of 5N for 1.50 s and then let go.

We need to find the distance covered by the puck when move from rest in 2.25 s.

We know that,

F = ma

a=\dfrac{F}{m}\\\\a=\dfrac{5}{2}\\\\a=2.5\ m/s^2

Let d is the distance moved in 2.25 s. Using second equation of motion,

d=ut+\dfrac{1}{2}at^2\\\\d=0+\dfrac{1}{2}\times 2.5\times (2.25)^2\\\\d=6.32\ m

So, it will move 6.32 m from rest in 2.25 seconds.

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A scientist wants to measure the relationship between humidity in the
ICE Princess25 [194]

Answer:

D

Explanation:

<em>The most suitable testable question. in this case, would be that 'are there more home runs during the more humid months of the  summer?'</em>

Since the aim of the investigation is to find the relationship between humidity and the number of home runs, measuring the number of home runs during the more humid months in the summer and comparing the data to the number of home runs during the less humid months in the same summer would provide the answer.

<u>Only option D raises a valid question that is relevant to the aim of the investigation.</u>

7 0
3 years ago
A race car has a centripetal acceleration of 15.625 m/s2 as it goes around a curve. If the curve is a circle with radius 40 m, w
myrzilka [38]
The centripetal acceleration is given by
a_c =  \frac{v^2}{r}
where v is the tangential speed and r the radius of the circular orbit.

For the car in this problem, a_c = 15.625 m/s^2 and r=40 m, so we can re-arrange the previous equation to find the velocity of the car:
v= \sqrt{a_c r}= \sqrt{(15.625 m/s^2)(40 m)}=25 m/s
8 0
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
Debbie places two shopping carts in a cart Corral. she pushes the first cart, which then pushes a second cart. what force is bei
Bingel [31]

When Debbie pushes the first cart she is using an applied force. An applied force is created when someone or something pushes another thing using, of course, an applied force. Now, when the second cart is being pushed by the first cart, this is also an applied force. You can tell because the first cart is being pushed using forced and this causes the second cart to be pushed using some of the force that is being transmitted to the first cart.


Debbie exerts applied force on the first cart. The first cart exert applied force on the second cart.



- Marlon Nunez

6 0
3 years ago
Raju learned about green plants, he found out that green plants can make their own food in the presence of sunlight a) Name the
baherus [9]

Answer:

Photosynthesis is the pocess by which green plants make their own food.

b. sunlight, glucose, water, altra-violet ray as catalyst

7 0
3 years ago
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