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Anestetic [448]
3 years ago
12

A different force is applied to each of four different blocks on a frictionless, horizontal surface. In which diagram does the b

lock have the greatest inertia 2.0 seconds after starting from rest?
A.1
B.3
C.4
D.2

Physics
1 answer:
NeTakaya3 years ago
6 0

Answer:

C. 4

Explanation:

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What is the speed of a bobsled whose distance-time graph indicates that it traveled 114m in 30s? m/s
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3.8 m/s
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3 years ago
What happens when you want to move the boat forward? You pull the oars toward yourself.Explain why you do this.
vampirchik [111]

Answer:

You pull on the oars. By the third law, the oars push back on your hands, but that’s irrelevant to the motion of the boat. The other end of each oar (the blade) pushes against the water. By the third law, the water pushes back on the oars, pushing the boat forward.

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2 years ago
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As an object rolls downhill, some of the energy is
Charra [1.4K]
When the object is at the top of the hill it has the most potential energy. If it is sitting still, it has no kinetic energy. As the object begins to roll down the hill, it loses potential energy, but gains kinetic energy. The potential energy of the position of the object at the top of the hill is getting converted into kinetic energy. Hope this helped. :)


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3 years ago
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Zero, a hypothetical planet, has a mass of 5.3 x 1023 kg, a radius of 3.3 x 106 m, and no atmosphere. A 10 kg space probe is to
Andrej [43]

(a) 3.1\cdot 10^7 J

The total mechanical energy of the space probe must be constant, so we can write:

E_i = E_f\\K_i + U_i = K_f + U_f (1)

where

K_i is the kinetic energy at the surface, when the probe is launched

U_i is the gravitational potential energy at the surface

K_f is the final kinetic energy of the probe

U_i is the final gravitational potential energy

Here we have

K_i = 5.0 \cdot 10^7 J

at the surface, R=3.3\cdot 10^6 m (radius of the planet), M=5.3\cdot 10^{23}kg (mass of the planet) and m=10 kg (mass of the probe), so the initial gravitational potential energy is

U_i=-G\frac{mM}{R}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{3.3\cdot 10^6 m}=-1.07\cdot 10^8 J

At the final point, the distance of the probe from the centre of Zero is

r=4.0\cdot 10^6 m

so the final potential energy is

U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{4.0\cdot 10^6 m}=-8.8\cdot 10^7 J

So now we can use eq.(1) to find the final kinetic energy:

K_f = K_i + U_i - U_f = 5.0\cdot 10^7 J+(-1.07\cdot 10^8 J)-(-8.8\cdot 10^7 J)=3.1\cdot 10^7 J

(b) 6.3\cdot 10^7 J

The probe reaches a maximum distance of

r=8.0\cdot 10^6 m

which means that at that point, the kinetic energy is zero: (the probe speed has become zero):

K_f = 0

At that point, the gravitational potential energy is

U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{8.0\cdot 10^6 m}=-4.4\cdot 10^7 J

So now we can use eq.(1) to find the initial kinetic energy:

K_i = K_f + U_f - U_i = 0+(-4.4\cdot 10^7 J)-(-1.07\cdot 10^8 J)=6.3\cdot 10^7 J

3 0
2 years ago
If a 50 kg student is standing on the edge of a cliff. Find the student’s gravitational potential energy if the cliff is 40 m hi
Oduvanchick [21]

you can find it using the equation: potential energy=mass*gravitational acceleration*height.


energy=50kg*9.8N/kg*40m=19600Nm=19600J or 19.6kJ


Sometimes they use 10 instead of 9.8 for the g constant. 


Rember to make me Brainliest!!!

3 0
3 years ago
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