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3 years ago

Compare skater's total energy at point A and at point E?(disregard the friction)

Physics

1 answer:

White raven [17]3 years ago

7 0

Answer:

Explanation:

because I've had this question before and got it right

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9. Complete the following statement:

mario62 [17]

Answer:

E) the flow of energy due to a temperature difference.

Explanation:

Heat can be described as the flow of energy due to a temperature difference.

Which is expressed mathematically as;

H = MCΔT

Where;

H is the quantity of heat in a body, measured in Joules

M is the mass of the body, measured in kg

C is the specific heat capacity of the body, J/kg.K

ΔT is change in temperature or temperature difference.

So, heat energy in any system flows from a hotter region to a colder region due to temperature difference.

E) the flow of energy due to a temperature difference.

7 0

3 years ago

A spherical bowling ball with mass m = 4.1 kg and radius R = 0.117 m is thrown down the lane with an initial speed of v = 8.9 m/

weqwewe [10]

Answer:Given mass = 4.1kg

Radius = 0.0117m

Velocity V = 8.4m/s

Coefficient of friction = 0.25

Explanation: Below is an attached solution to the problem stated above.

1. The angular acceleration is equal to 524rad/s^2

2. The linear acceleration is equal to 2.45m/s^2

3.the time it takes the ball to begin rolling = 0.98s

4. The distance the ball slides before it begins to roll = 7.05m

3 0

4 years ago

98 POINTS FOR ANYONE THAT DOES THIS NOW!!

Varvara68 [4.7K]

In simpler terms, a proton and neutron weigh 1 amu (atomic mass unit) each.

The nucleus has 15 protons and 18 neutrons. Since a proton's and a neutron's weight is only 1 amu, we can simply add the number of protons and neutrons to find the total mass of the nucleus:

$15 + 18 = \boxed{33}$

The nucleus' mass is 33 amu.

8 0

3 years ago

Read 2 more answers

How long would it take for a car to travel 200 km if it has an average speed of 55 km/hr?

VladimirAG [237]

It would take 3.6 hours for a car to travel 200 km going 55 km/hr.

4 0

3 years ago

A parallel-plate capacitor has plates of area A. The plates are initially separated by a distance d, but this distance can be va

ohaa [14]

Answer:

d' = d /2

Explanation:

Given that

Distance = d

Voltage =V

We know that energy in capacitor given as

$U=\dfrac{1}{2}CV^2$

$C=\dfrac{\varepsilon _oA}{d}$

$U=\dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d}\times V^2$

If energy become double U' = 2 U then d'

$U'=\dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d'}\times V^2$

$2U=\dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d'}\times V^2$

$2\times \dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d}\times V^2=\dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d'}\times V^2$

2 d ' = d

d' = d /2

So the distance between plates will be half on initial distance.

7 0

3 years ago