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2 years ago

Which three terms are needed to describe the energy a BASE jumper has as she falls toward the ground?

Physics

1 answer:

Thepotemich [5.8K]2 years ago

4 0

Answer:

Potential energy

Kinetic energy

Gravitational energy.

Explanation:

The potential energy is the stored energy, it is the energy and object posses at rest. When the base jumper is still at rest i.e without motion, she has potential energy.

Gravitational energy this refer to the potential energy an object or body with mass posses in relation to another object due to gravity.

This is seen when the base number is still at rest and the force of gravity on the Earth acting on her.

Kinetic energy is the energy posses by an object or body in motion.

As the base jumper falls to the ground, she posses kinetic energy which is the energy in motion.

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Chapter 21, Problem 009 Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.12

PilotLPTM [1.2K]

Answer:

a) -1.325 μC

b) 4.17μC

Explanation:

First, you need to know that charge is conserved. So, the adition of the charges, as there is no lost in charge, should always be the same. Also, after the wire is removed, both spheres will have the same charge, trying to find equilibrium. In summary:

$q_1 + q_2 = constant\\q_1_f = q_2_f |Then\\q_1_f + q_2_f = 2q_1_f = q_1_o+q_2_o\\q_1_f = q_2_f = \frac{q_1_o+q_2_o}{2}$

We know both q1f and q2f must be positive, because the negative charge at the beginning was the the smaller.

The electrostatic force is equal to:

$F_e = k\frac{q_1q_2}{r^2}$

K is the Coulomb constant, equal to 9*10^9 Nm^2/C^2

Now, we are told that the electrostatic force after the wire is equal to 0.0443 N:

$F_e_f = k \frac{q_1_fq_2_f}{r^2} = k\frac{\frac{q_1_o+q_2_o}{2}\frac{q_1_o+q_2_o}{2}}{r^2} = k\frac{(q_1_o+q_2_o)^2}{4r^2} \\(q_1_o+q_2_o) = \sqrt{\frac{F_e_f*4r^2}{k}} = \sqrt{\frac{0.0443N *4(0.641m)^2}{9*10^9Nm^2/C^2} } = 2.844 *10^{-6}C \\ q_1_o = 2.844*10^{-6}C - q_2_o$

Originally, the force is negative because it was an attraction force, therefore, its direction was opposite to the direction of the repulsive force after the wire:

$F_e_o = k\frac{q_1_oq_2_o}{r^2}\\ q_1_oq_2_o = \frac{F_e_o*r^2}{k} = \frac{-0.121N(0.641m)^2}{9*10^9Nm^2/C^2} = -5.524*10^{-12}$

$(2.844*10^{-6}C - q_2_o)q_2_o = -5.524*10^{-12}\\0 = q_2_o^2 - 2.844*10^{-6}q_2_o - 5.524*10^{-12}$

Solving the quadratic equation:

$q_2_o = 4.17*10^{-6}C | -1.325 * 10^{-6}C$

for this values q_1 wil be:

$q_1_o = -1.325 *10^{-6}C | 4.17*10^{-6}C$

So as you can see, the negative charge will always be -1.325 μC and the positive 4.17μC

5 0

3 years ago

State and explain newton second law of motion also describe the concept of force, represent it quantiatively and derive the unit

Molodets [167]

Explanation:

Newton's second law of motion states F=ma which means force is equal to mass multiplied by acceleration which in simple terms means If you give mass force it will accelerate the concept of force in physics is any interaction that when unopposed will change the motion of an object.

8 0

3 years ago

A charge of 0.91 C is spread uniformly throughout a 25 cm rod of radius 4 mm. What are the volume and linear charge densities

Oxana [17]

The volume of the rod is 1.26×10⁻⁵ m³, and the linear charge density of the rod is 3.64 C/m

<h3>What is volume?:</h3>

This is the product of the height of a solid object and its crossectional area.

The Volume of the rod is can be calculated using the formula below.

Note: A rod has the shape of a cylinder.

Formula:

V = πr²h............... Equation 1

Where:

V = Volume of the rod
r = radius of the rod
h = height of the rod.

From the question,

Given:

r = 4mm = 0.004 m
h = 25 cm = 0.25 m
π = 3.14

Substitute these values into equation 1

V = 3.14(0.004²)(0.25)
V = 1.26×10⁻⁵ m³

<h3>What is linear charge density:</h3>

This is the ratio of the charge on an object to the length of the object.

The linear charge density of the rod can be calculated using the formula below.

D = Q/h.................... Equation 2

Where:

D = Linear charge density of the rod
Q = Charge on the rod.
h = height or length of the rod

From the question

Given:

Q = 0.91 C
h = 25 cm = 0.25 m

Substitute these values into equation 2

D = 0.91/0.25
D = 3.64 C/m

Hence, The volume of the rod is 1.26×10⁻⁵ m³, and the linear charge density of the rod is 3.64 C/m

Learn more about charge density here: brainly.com/question/14568868

4 0

2 years ago

a car is moving with a velocity of 25m/s for 15s. calculate the displacement of the car. The acceleration of the car over the 15

SpyIntel [72]

Explanation:

this might help you i think so

8 0

3 years ago

HELP ASAP! GIVING BRAINLIEST!

Blababa [14]

Answer:

1. Emma standing on top of mountain

Since she is at the rest position and at some height from the ground so here energy is due to gravitational potential energy

So we have

gravitational potential energy

$U = mgH$

2. Emma jumping down from mountain top

Due to free fall Emma will start moving with some speed in downwards direction so here we have

$KE = \frac{1}{2}mv^2$

motion energy

3. tension in rope at Emma’s lowest position

Due to stretch in the rope here position come to the lowest end and speed comes to zero so whole energy is converted into elastic potential energy

$U = \frac{1}{2}kx^2$

elastic potential energy

4. Emma bouncing back

Due to bouncing back she will again have its kinetic energy with some speed upwards

$KE = \frac{1}{2}mv^2$

motion energy

8 0

3 years ago

Read 2 more answers