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serg [7]
3 years ago
5

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

e firing. How far up does the ball go this time? Neglect friction. Assume that the spring is ideal and that the distance by which the spring is compressed is negligible compared to H.
Physics
1 answer:
Mekhanik [1.2K]3 years ago
6 0

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)

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olga_2 [115]

Answer: f = - 0.50 m, negative (diverging) lens

Explanation: D-value (diopter ) means 1/f, unit is 1/m. Thus -2.0D

Means f = ( 1/ -2.0 ) m = -0.50 m

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2 years ago
Problem 2.26 MasteringPhysics 10 of 16 Problem 2.26 When striking, the pike, a predatory fish, can accelerate from rest to a spe
cluponka [151]

final velocity = initial velocity + (acceleration x time) <span>
3.9 m/s = 0 m/s + (acceleration x 0.11 s) 
3.9 m/s / 0.11 s = acceleration 
30.45 m/s^2 = acceleration 

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6 0
3 years ago
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IrinaVladis [17]

Answer:

3.52176 x 10^-10 N

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8 0
3 years ago
The area of the piston to the master cylinder in a hydraulic braking system of a car is 0.4 square inches. If a force of 6.4 lb
Anit [1.1K]

Answer:

The force applied on one wheel during braking = 6.8 lb

Explanation:

Area of the piston (A) = 0.4 in^{2}

Force applied on the piston(F) = 6.4 lb

Pressure on the piston (P) = \frac{F}{A}

⇒ P = \frac{6.4}{0.4}

⇒ P = 16 \frac{lb}{in^{2} }

This is the pressure inside the cylinder.

Let force applied on the brake pad = F_{1}

Area of the brake pad (A_{1})= 1.7 in^{2}

Thus the pressure on the brake pad (P_{1}) =  \frac{F_{1} }{A_{1} }

When brake is applied on the vehicle the pressure on the piston is equal to pressure on the brake pad.

⇒ P = P_{1}

⇒ 16 = \frac{F_{1} }{A_{1} }

⇒ F_{1} = 16 × A_{1}

Put the value of A_{1} we get

⇒ F_{1} = 16 × 1.7

⇒ F_{1} = 27.2 lb

This the total force applied during braking.

The force applied on one wheel = \frac{F_{1} }{4} = \frac{27.2}{4} = 6.8 lb

⇒ The force applied on one wheel during braking.

7 0
3 years ago
A rollercoaster car has 2500 J of potential energy and 160 J of
lara31 [8.8K]

Answer:

E_{k2}=2660 [J] kinetic energy.

Explanation:

The energy in the initial state i.e. when the rollercoaster is at the top is equal to the energy in the final state i.e. when it is at the bottom of the hill.

These states can be represented by means of the second equation.

E_{k1}+E_{p1}=E_{k2}\\160 + 2500 = E_{k2}\\E_{k2}=2660 [J]

Since the rollercoaster is located in the bottom of the hill where the potential energy level is zero, therefore there is only kinetic energy in the second state.

8 0
2 years ago
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