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4 years ago

14

1. Discuss some of the dangers and obstacles that Nyad faced when sw

1 answer:

balu736 [363]4 years ago

7 0

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A soccer player kicks a ball at rest on the

scoundrel [369]

The kinetic energy and gravitational potential energy changes during its movement from ground to the top height.

<h3>What happens to kinetic and potential energy while motion?</h3>

When the ball moves upward, its gravitational potential energy is increases and kinetic energy begins to decrease but when the ball falls towards the earth, its gravitational potential energy is transformed into kinetic energy. When the ball collides with the ground, the kinetic energy is transformed into other forms of energy.

Learn more about kinetic energy here: brainly.com/question/20658056

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2 years ago

How much work is done on a small car if a 3150 N force is exerted to move it 75.5 m to the side of the road

irinina [24]

Answer:

Explanation:

Work = Force times displacement. Therefore,

W = 3150(75.5) so

W = 238000 N*m

6 0

3 years ago

At what speed do a bicycle and its rider, with a combined mass of 100 kg, have the same momentum as a 1400 kg car traveling at 2

likoan [24]

100kg x bicycle speed = 1400 X 2

bicycle speed = 2800/ 100
bicycle speed = 28 m/s

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4 years ago

What is the net force on a bathroom scale when a 110 pound person stands on it? How do you know?

USPshnik [31]

The scale is a rest scale reads the support for is not enough net force.

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3 years ago

Two light bulbs have resistances of 400 Ω and 800 ΩThe two light bulbs are connected in series across a 120- V line. Find the cu

Natasha2012 [34]

1) Current in each bulb: 0.1 A

The two light bulbs are connected in series, this means that their equivalent resistance is just the sum of the two resistances:

$R_{eq}=R_1 + R_2 = 400 \Omega + 800 \Omega=1200 \Omega$

And so, the current through the circuit is (using Ohm's law):

$I=\frac{V}{R_{eq}}=\frac{120 V}{1200 \Omega}=0.1 A$

And since the two bulbs are connected in series, the current through each bulb is the same.

2) 4 W and 8 W

The power dissipated by each bulb is given by the formula:

$P=I^2 R$

where I is the current and R is the resistance.

For the first bulb:

$P_1 = (0.1 A)^2 (400 \Omega)=4 W$

For the second bulb:

$P_1 = (0.1 A)^2 (800 \Omega)=8 W$

3) 12 W

The total power dissipated in both bulbs is simply the sum of the power dissipated by each bulb, so:

$P_{tot} = P_1 + P_2 = 4 W + 8 W=12 W$

3 0

3 years ago