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Goshia [24]
2 years ago
6

According to the picture, how would the gravitational potential energy of the car change if its weight was doubled? A) The gravi

tational potential energy would be doubled. B) The gravitational potential energy would stay the same C) The gravitational potential energy would be cut in half. D) The gravitational potential energy would be increased by 750.
Physics
2 answers:
Vera_Pavlovna [14]2 years ago
8 0

Answer:

the answer is (A)

Explanation:

Lorico [155]2 years ago
3 0

Answer is A

Samuriguy125 on YT

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A water wheel rotates with a period of 2.26 s. If the water wheel has a radius of l.94 m,
Dafna11 [192]

Answer:

The correct answer is B)

Explanation:

When a wheel rotates without sliding, the straight-line distance covered by the wheel's center-of-mass is exactly equal to the rotational distance covered by a point on the edge of the wheel.  So given that the distances and times are same, the translational speed of the center of the wheel amounts to or becomes the same as the rotational speed of a point on the edge of the wheel.

The formula for calculating the velocity of a point on the edge of the wheel is given as

V_{r} = 2π r / T

Where

π is Pi which mathematically is approximately 3.14159

T is period of time

Vr is Velocity of the point on the edge of the wheel

The answer is left in Meters/Seconds so we will work with our information as is given in the question.

Vr = (2 x 3.14159 x 1.94m)/2.26

Vr = 12.1893692/2.26

Vr = 5.39352619469

Which is approximately 5.39

Cheers!

7 0
2 years ago
A sports car accelerates from rest for 5 seconds reaching a velocity of 23.0 m/s.
denis-greek [22]

Answer:

<h2>4.6 m/s²</h2>

Explanation:

The acceleration of an object given it's velocity and time taken can be found by using the formula

<h3>a =  \frac{v - u}{t}  \\</h3>

where

v is the final velocity

u is the initial velocity

t is the time taken

a is the acceleration

Since the body is from rest u = 0

From the question we have

a =  \frac{23 - 0}{5}  =  \frac{23}{5}  \\

We have the final answer as

<h3>4.6 m/s²</h3>

Hope this helps you

4 0
2 years ago
b. What would the momentum be if the mass of the bowling ball were doubled and its velocity still was 3 m/s?​
umka2103 [35]

Answer:

Twice.

Explanation:

The momentum of an object is given by :

p = mv

Where

m is mass and v is the velocity

If the mass of the ball were doubled, m'=2m and v'=v=3 m/s

New momentum,

p'=m'v'

p'=2m × v

p'=2mv

or

p'=2p

So, the new momentum becomes twice the initial momentum.

6 0
2 years ago
Combustion reactions are also ________ reactions. A. nuclear B. exothermic C. endothermic D. spontaneous
sveta [45]
I think is a) i hope this hlp
5 0
3 years ago
Read 2 more answers
A duck swimming on the surface of a pond has an
Gemiola [76]

From the given information:

  • Taking the movement of the Duck in the North as the x-direction
  • The movement of the Duck in the East direction as the y-direction

However, we will have to compute the initial velocity and the acceleration of the duck in their vector forms.

<h3>In vector form;</h3>

The initial velocity is:

\mathbf{u ^{\to} = 0.7 m/s ( -cos 25^0 \hat x + sin 25^0 \hat y ) \ m/s}

The acceleration is:

\mathbf{a ^{\to} = 0.5 m/s ( cos 41^0 \hat x - sin 41^0 \hat y ) \ m/s^2}

The objective of this question is to determine the speed of the duck at a certain time. Since it is not given, let's assume we are to determine the Duck speed after 4 seconds of accelerating;

Then, it implies that time (t) =  4 seconds.

Using the first equation of motion:

v^{\to} = u ^{\to} + a^{\to} t

Then, we can replace their values into the equation of motion in order to determine the speed:

\mathbf{v^{\to} =\Big(0.7 ( -cos 25^0 \hat x + sin 25^0 \hat y )+4 \times 0.5 ( cos 41^0 \hat x - sin 41^0 \hat y )\Big)}

\mathbf{v^{\to} =\Big(0.7 ( -cos 25^0 \hat x + sin 25^0 \hat y )+2.0 ( cos 41^0 \hat x - sin 41^0 \hat y )\Big)}

\mathbf{v^{\to} =\Big( ( -0.7 cos 25^0 \hat x + 0.7 sin 25^0 \hat y )+( 2.0cos 41^0 \hat x - 2.0sin 41^0 \hat y )\Big)}

Collect like terms:

\mathbf{v^{\to} =\Big( (2.0cos 41^0 -0.7 cos 25^0   )\hat x+(  0.7 sin 25^0 - 2.0sin 41^0 )\Big)\hat y}

\mathbf{v^{\to} =0.87500   \hat x- 1.01629 \hat y}

Thus, the magnitude is:

\mathbf{v^{\to} =\sqrt{(0.87500 )^2 +( 1.01629 )^2}}

\mathbf{v^{\to} =\sqrt{0.76563 +1.03285}}

\mathbf{v^{\to} =\sqrt{1.79848}}

\mathbf{v^{\to} =1.34 \ m/s}

Therefore, we can conclude that the speed of the duck after 4 seconds is 1.34 m/s

Learn more about vectors here:

brainly.com/question/17108011?referrer=searchResults

4 0
2 years ago
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