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Pachacha [2.7K]
3 years ago
5

At what point on the hill will the car have zero gravitational potential energy?

Physics
2 answers:
Mila [183]3 years ago
8 0
At the bottom of the hill
zimovet [89]3 years ago
6 0

At the bottom of the hill

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You have a neutral balloon. What is its charge after 12000 electrons have been removed from it? The elemental charge is 1.6 × 10
AleksandrR [38]

Answer:

1.92×10^-9 microCoulomb

Explanation:

Elemental charge = 1.6×10^-19 Coulomb

Charge of balloon after 12,000 electrons have been removed from it = 12,000 × 1.6×10^-19 Coulomb = 1.92×10^-15 Coulomb = 1.92×10^-15/10^-6 = 1.92×10^-9 microCoulomb

6 0
3 years ago
You are driving at 25 m/s with your cruise control on when you see a fallen tree in the road. It takes you 0.30 s to put on the
xxTIMURxx [149]

Explanation:

Given data

velocity v= 25m/s

The time it takes to put on brake t= 0.3s

the distance covered when the  brake was put on is

v=s/t

s= v*t

s= 25*0.3s

s= 7.5m

hence the distance covered is 7.5m

Also the rate of decrease in aceleration is 5m/s^2

we can also calculate the distance covered at this rate

v^2=u^2+2as

25^2= 0+2*5*s

625=10s

divide both sides by 10

s=625/10

s= 62.5m

The total  distance covered between putting on the brakes and decelareation is  7.5+62.5= 70m

Given that the tree is 75m ahead, the car would not hit the tree

3 0
3 years ago
What are the folds called inside the mitochondria?
max2010maxim [7]
The inner membrane has many overlapping folds called cristae. Inside the inner membrane there is the mitochondrial matrix, it contains enzymes that are used in creating ATP. 
You're welcome. :) 

7 0
3 years ago
Read 2 more answers
What can alter the motion of an object
Liula [17]
Force can alter its direction,slow or stop it you could say it can change its velocity

3 0
3 years ago
a particle is moving along a circular path having a radius of 4 in such that its position as a function of time is given by thet
ANTONII [103]

Answer:

Explanation:

Given

radius of circular path r=4\ in.

Position is given by

\theta =\cos 2t---1

Differentiate 1  to angular velocity we get

\frac{\mathrm{d} \theta }{\mathrm{d} t}=\omega =-2\sin 2t----2

Differentiate 2 to get angular acceleration

\frac{\mathrm{d} \omega }{\mathrm{d} t}=-2^2\cos 2t ---3

Net acceleration is the vector summation of tangential and centripetal force

a_t=\alpha \times r

a_t=-4\cos 2t\times 4=-16\cos 2t

a_r=\omega ^2\cdot r

a_r=(-2\sin 2t)^2\cdot 4

a_r=16\sin^2(2t)

a_{net}=\sqrt{a_r^2+a_t^2}

a_{net}=\sqrt{(16\sin ^2(2t)+(-16\cos 2t)^2}

a_{net}=\sqrt{256\cos ^2(2t)+256\sin ^4(2t)}                                                    

6 0
3 years ago
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