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Gennadij [26K]
3 years ago
15

A motorcycle skids for a distance of 2.0 m with the icy road pushing on its tires with force of 120 N as its

Physics
1 answer:
AnnyKZ [126]3 years ago
7 0

Answer:

-240

Explanation:

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A 1210 kg rollercoaster car is
ratelena [41]

Answer: 4.98 m/s

Explanation:

You solve these kinetic energy, potential energy problems by using the fact P.E.+ K.E. = a constant as long as friction is ignored.

PEi = 0 in this case

KEi = ½mVi² = PEf+KEf = mghf + ½mVf²

½1210*8.31² = 1210*9.8*2.26 + ½1210*Vf²

½1210*Vf² = ½1210*8.31² - 1210*9.8*2.26

Vf² = 8.31² - 2*9.8*2.26 = 4.98² so Vf = 4.98m/s

3 0
3 years ago
The electric potential at the dot in the figure is 3160 V. What is charge q?
PSYCHO15rus [73]

Hi there!

Recall the equation for electric potential of a point charge:

V = \frac{kQ}{r}

V = Electric potential (V)
k = Coulomb's Constant(Nm²/C²)

Q = Charge (C)
r = distance (m)

We can begin by solving for the given electric potentials. Remember, charge must be accounted for. Electric potential is also a SCALAR quantity.

Upper right charge's potential:

V = \frac{(8.99*10^9)(-5 * 10^{-9})}{0.04} = -1123.75 V

Lower left charge's potential:

V = \frac{(8.99*10^9)(5*10^{-9})}{0.02} = 2247.5 V

Add the two, and subtract from the total EP at the point:

3160 + 1123.75 - 2247.5 = 2036.25


The remaining charge must have a potential of 2036.25 V, so:

2036.25 = \frac{(8.99*10^9)(Q)}{\sqrt{0.02^2 + 0.04^2}}\\\\2036.25 = \frac{(8.99*10^9)Q}{0.0447} \\\\Q = 0.000000010127 = \boxed{10.13nC}


5 0
2 years ago
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Sidana [21]

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6 0
3 years ago
The same ball is shot straight up a second time from the same gun, but this time the spring is compressed only half as far befor
Mekhanik [1.2K]

Answer:

The new height the ball will reach = (1/4) of the initial height it reached.

Explanation:

The energy stored in any spring material is given as (1/2)kx²

This energy is converted to potential energy, mgH, of the ball at its maximum height.

If the initial height reached is H

And the initial compression of the spring = x

So, mgH = (1/2)kx²

H = kx²/2mg

The new compression, x₁ = x/2

New energy of loaded spring = (1/2)kx₁²

And the new potential energy = mgH₁

mgH₁ = (1/2)kx₁²

But x₁ = x/2

mgH₁ = (1/2)k(x/2)² = kx²/8

H₁ = kx²/8mg = H/4 (provided all the other parameters stay constant)

6 0
3 years ago
Consider a parachutist that has reached terminal velocity. which of the following is true? A.) The acceleration of the parachuti
trasher [3.6K]
The correct answer is:
<span>B.) At terminal velocity there is no net force 

In fact, when the parachutist reaches the terminal velocity, his velocity does not change any more. It means that the acceleration acting on the parachutist is zero, and for Newton's second law, this means the net force acting on him is zero:
</span>\sum F = ma = 0
<span>because the acceleration is zero: a=0. 
This also means that the two relevant forces acting on the parachutist (gravity, downward, and air resistance, upward) are balanced to produce a net force equal to zero.</span>
8 0
3 years ago
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