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kompoz [17]
3 years ago
12

True or false? If the speed of an object doubles, its kinetic energy will also double.

Physics
1 answer:
Anastasy [175]3 years ago
4 0

Answer:

False

Explanation:

It's kinetic energy would change to four times the amount.

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M = 15 kg, a = 2 m/s2, F =
Lisa [10]

Answer:

<h2>30 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 15 × 2

We have the final answer as

<h3>30 N</h3>

Hope this helps you

7 0
3 years ago
PLEASE HELP ILL MARK BRAINLIEST
ikadub [295]
I think the answer is A?
7 0
3 years ago
A 72.8-kg swimmer is standing on a stationary 265-kg floating raft. The swimmer then runs off the raft horizontally with a veloc
nalin [4]

Answer:

-1.43 m/s relative to the shore

Explanation:

Total momentum must be conserved before and after the run. Since they were both stationary before, their total speed, and momentum, is 0, so is the total momentum after the run off:

m_sv_s + m_rv_r = 0

where m_s = 72.8, m_r = 265 are the mass of the swimmer and raft, respectively. v_s = 5.21 m/s, v_r are the velocities of the swimmer and the raft after the run, respectively. We can solve for v_r

265v_r + 72.8*5.21 = 0

v_b = -72.8*5.21/265 = -1.43 m/s

So the recoil velocity that the raft would have is -1.43 m/s after the swimmer runs off, relative to the shore

7 0
3 years ago
What change will always result in an increase in gravitational force between two objects
tekilochka [14]

The gravitational force between two object depends on their masses and on their distance.


Since the formula is


F = G\frac{m_1m_2}{d^2}


If the masses grow, the force also grows. But I'm assuming the two objects are fixed, so you can't enlarge their mass.


So, the only option remaining is to lower their distance: since it sits at the denominator, a smaller value of d results in a bigger value for F.


So, if you reduce the distance between two objects, the gravitational force between them will always result in an increase

6 0
3 years ago
A concave mirror has a focal length of 13.5 cm. This mirror forms an image located 37.5 cm in front of the mirror. Find the magn
77julia77 [94]

Explanation:

It is given that,

Focal length of the concave mirror, f = -13.5 cm

Image distance, v = -37.5 cm (in front of mirror)

Let u is the object distance. It can be calculated using the mirror's formula as :

\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}

\dfrac{1}{u}=\dfrac{1}{f}-\dfrac{1}{v}

\dfrac{1}{u}=\dfrac{1}{(-13.5)}-\dfrac{1}{(-37.5)}

u = -21.09 cm

The magnification of the mirror is given by :

m=\dfrac{-v}{u}

m=\dfrac{-(-37.5)}{(-21.09)}

m = -1.77

So, the magnification produced by the mirror is (-1.77). Hence, this is the required solution.

7 0
3 years ago
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