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Natali5045456 [20]

3 years ago

5

The energy of the ball has been reduced by joules after bouncing back. This reduction happens because potential energy transform

s to energy.

1 answer:

luda_lava [24]3 years ago

5 0

<span>This reduction happens because potential energy transforms to "thermal energy" and "sound energy"

Hope this helps!</span>

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A charge Q is transferred from an initially uncharged plastic ball to an identical ball 10.0 cm away. The force of attraction is

victus00 [196]

Answer:

a charge Q is transferred from an initially uncharged

Explanation:

Hope this helps!

5 0

2 years ago

A hot-air balloon is accelerating upward under the influence of two forces, its weight and the buoyant force. for simplicity, co

Elenna [48]

Let V = the volume of the balloon
Force of gravity = V * ?hot * g downward
Buoyant force = V * ?cool * g upward
Net upward force F = V * ?cool * g - V * ?hot * g

F = V g (?cool - ?hot)

Mass of the balloon m = V ?hot

a = F/m = V g (?cool - ?hot)/(V ?hot)

a = g(?cool/?hot - 1)

a = 9.8(1.29/0.93 - 1)

a = 3.79 m/s^2

<span>Answer is 3.79 m/s^2</span>

4 0

3 years ago

An object has a momentum of 4,000 kg-m/s and a mass of 115 kg. It crashes into another object that has a mass of 100 kg, and the

ANTONII [103]

Answer:

18.60 m/s

Explanation:

Original momentum = mv = 4000 with m = 115

after collision m = 115 + 100 = 215 kg

but the total momentum is still the same (conserved)

4000 = 215 v shows v = 18.60 m/s

4 0

2 years ago

An inductor is connected to the terminals of a battery that has an emf of 16.0 V and negligible internal resistance. The current

balu736 [363]

Answer:

(a) The resistance R of the inductor is 2480.62 Ω

(b) The inductance L of the inductor is 1.67 H

Explanation:

Given;

emf of the battery, V = 16.0 V

current at 0.940 ms = 4.86 mA

after a long time, the current becomes 6.45 mA = maximum current

Part (a) The resistance R of the inductor

$R = \frac{V}{I_{max}} = \frac{16}{6.45*10^{-3}} = 2480.62 \ ohms$

Part (b) the inductance L of the inductor

$\frac{Rt}{L} = -ln(1-\frac{I}I_{max}})\\\\L = \frac{Rt}{-ln(1-\frac{I}I_{max})}}$

where;

L is the inductance

R is the resistance of the inductor

t is time

$L = \frac{Rt}{-ln(1-\frac{I}I_{max})}} = \frac{2480.62*0.94*10^{-3}}{-ln(1-\frac{4.86}{6.45})} \\\\L =\frac{2.3318}{1.4004} = 1.67 \ H$

Therefore, the inductance is 1.67 H

5 0

3 years ago

A proton moving at 5.00 106 m/s through a magnetic field of magnitude 1.78 t experiences a magnetic force of magnitude 7.40 10-1

aliya0001 [1]

The magnetic part using the Lorentz force is: F = q v x B,
where v and B are vectors and v x B is the vector cross product.

Magnitude of the force: F = q v B sin(α)

So, sin(α) = F/( e v B), with e the proton charge.

This will give you a value for sin(α), and two potentials for its opposite.

You will now look for:

sin(α) = 7.40 10^-13/( 1.60 10^-19 * 5 10^6 * 1.78)
= 0.520

So either sin(α) = 0.502 or sin(α) = -0.502
The 1st α = 30.1 degrees or α = 150 degrees.
The 2nd α = 210 degrees or α = 330 degrees.
So we can say that 30.1 degrees and 330 degrees would be minimum and biggest on [0,360]

6 0

3 years ago