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Colt1911 [192]
3 years ago
5

This graph shows a ball rolling from A to G. The ball starts at point A and rolls to point G.

Physics
1 answer:
Lerok [7]3 years ago
4 0

11. B

12. A

13. D

14. The ball burnt off energy while rolling, it lost potential energy and couldn't go as high.

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the time required for one cycle, a complete motion that returns to its starting point, it called the_____ medium frequency perio
telo118 [61]

Answer:

The correct answer to the following question will be "Period".

Explanation:

The Period seems to be the time deemed necessary for such a perfect cycle of vibration to transfer a particular moment. Because as the amplitude of the wave raises, the wavelength falls.

It is denoted by "T" and its formula will be:

⇒  T  = \frac{1}{F}

Where, T = Period

            F = Frequency

The other given choices are not related to the given circumstances. So that the above would be the right answer.

4 0
3 years ago
Can an object be accelerated while traveling at constant velocity? Why or why not?
AysviL [449]

Answer:

An object's acceleration is the rate its velocity (speed and direction) changes. Therefore, an object can accelerate even if its speed is constant - if its direction changes. If an object's velocity is constant, however, its acceleration will be zero.

3 0
3 years ago
Which item is a pure substance?
hram777 [196]

Answer:

Solid gold. 10 carat indicates gold purity is not 100%. For 24 karat gold is pure gold.

Explanation:

8 0
3 years ago
Read 2 more answers
A particular planet has a moment of inertia of 9.74 × 1037 kg ⋅ m2 and a mass of 5.98 × 1024 kg. Based on these values, what is
malfutka [58]

Answer:  A) 6.38(10)^{6} m

Explanation:

The equation for the moment of inertia I of a sphere is:

I=\frac{2}{5}mr^{2} (1)

Where:

I=9.74(10)^{37}kg m^{2} is the moment of inertia of the planet (assumed with the shape of a sphere)

m=5.98(10)^{24}kg is the mass of the planet

r is the radius of the planet

Isolating r from (1):

r=\sqrt{\frac{5I}{2m}} (2)

Solving:

r=\sqrt{\frac{5(9.74(10)^{37}kg m^{2})}{2(5.98(10)^{24}kg)}} (3)

Finally:

r=6381149.077m \approx 6.38(10)^{6} m

Therefore, the correct option is A.

4 0
3 years ago
A man fires a silver bullet of mass 2g with a velocity of200m/sec into wall. What is the temperature changeof the bullet? Note:
Sphinxa [80]

F=nmv

where;

n=no. of bullets = 1

m=mass of bullets=2g *10^-3

V=velocity of bullets200m/sec

F=1

loss in Kinetic energy=gain in heat energy

1/2MV^2=MS∆t

let M council M

=1/2V^2=S∆t

M=2g

K.E=MV^2/2

=(2*10^-3)(200)^2/2

2 councils 2

2*10^-3*4*10/2

K.E=40Js

H=mv∆t

(40/4.2)

40Js=40/4.2=mc∆t

40/4.2=2*0.03*∆t

=158.73°C

7 0
3 years ago
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