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S_A_V [24]
3 years ago
7

What the kinetic energy quantities in calculation pls help me​

Physics
1 answer:
Rashid [163]3 years ago
3 0

Answer:

KE = 0.5 * m * v², where: m - mass, v - velocity.

Explanation:

In classical mechanics, kinetic energy (KE) is equal to half of an object's mass (1/2*m) multiplied by the velocity squared. For example, if a an object with a mass of 10 kg (m = 10 kg) is moving at a velocity of 5 meters per second (v = 5 m/s), the kinetic energy is equal to 125 Joules, or (1/2 * 10 kg) * 5 m/s 2.

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Beaker A contains 100 mL of water at a temperature of 25 °C. Beaker B contains 100 mL of water at a temperature of 60 °C. Which
Novosadov [1.4K]

Answer:

Only option A is correct. Beaker A has lower kinetic energy than beaker B.

Explanation:

Step 1: Data given

Beaker 1 has a volume of 100 mL at 25 °C

Beaker B has a volume of 100 mL at 60 °C

Thermal energy = m*c*T

Thermal energy beaker A = 100 grams*4.184 * 25°C

Thermal energy beaker B = 100 grams *4.184*60°C

⇒ Since both beakers contain the same amount of water, the thermal energy depends on the temperature.

Since beaker B has a higher temperature, it has a higher thermal energy than beaker A

When we heat a substance, its temperature rises and causes an increase in the kinetic energy of its constituent molecules. Temperature is, in fact, a measure of the kinetic energy of molecules.

This means beaker B has a higher kinetic energy than beaker A

Potential energy doesn't depend on temperature. this means the potential energy of beaker A and beaker B is the same.

a. Beaker A has lower kinetic energy than beaker B. This is correct.

b. Beaker A has higher thermal energy than beaker B. This is false.

c. Beaker A has higher potential energy than beaker B. This is false.

d. Beaker A has lower potential energy than beaker B. This is false

e. Beaker A has higher kinetic energy than beaker B. This is false.

3 0
3 years ago
By what factors does the speed of the elctron exceed that of the proton?<br>​
____ [38]

Answer:

» An electron is lighter than a proton.

<u>explanation</u><u>:</u>

{  =  \: \sf{an \: electron \: has \: formula \:  \: }}{ \bf{ {}^{0}_{  - 1}e }}

hence it's mass number is zero

{  =  \: \sf{an \: electron \: is \: helium \: particle \:  \: }}{ \bf{ {}^{4} _{2}He  }}

hence it's mass number is 4

<u>Therefore</u><u>,</u><u> </u><u>proton</u><u> </u><u>is</u><u> </u><u>heavier</u><u> </u><u>than</u><u> </u><u>electron</u>

» An electron has a small charge magnitude than a proton.

<u>Explanation</u><u>:</u>

An electron has charge of -1 while proton has charge of +2, therefore electron is less deflected by any energetic fields than a proton

8 0
3 years ago
Why is the contribution of the wavelets lying on the back of secondary wave front zero?​
Tresset [83]

Answer:

The contribution of the wavelets lying on the back of the wave front is zero because of something known as the Obliquity Factor. It is assumed that the amplitude of the secondary wavelets is not independent of the direction of propagation, Sources: byju's.com

3 0
2 years ago
Read 2 more answers
In Trial III, a different, looser, spring is used; its force constant is 23.1 N/m. The suspended mass is the same as the one in
cricket20 [7]

Answer:

T=0.827s

Explanation:

The period of a spring can be calculated with the equation

T=2\pi w

But we know as well that w is given by,

w=\sqrt{\frac{k}{m}}

Replacing,

w=\frac{2\pi}{T}= \sqrt{\frac{k}{m}}\\T=2\pi\sqrt{\frac{k}{m}}\\T=2\pi\sqrt{\frac{0.4}{23.1}}

So we have that

T=0.827s

7 0
3 years ago
Two cars cover the same distance in a straight line. Car a covers the distance at a constant velocity. Car b starts from rest an
Artyom0805 [142]

a) For the motion of car with uniform velocity we have , s = ut+\frac{1}{2}at^2, where s is the displacement, u is the initial velocity, t is the time taken a is the acceleration.

In this case s = 520 m, t = 223 seconds, a =0 m/s^2

Substituting

       520 = u*223\\ \\u = 2.33 m/s

 The constant velocity of car a = 2.33 m/s

b) We have s = ut+\frac{1}{2} at^2

s = 520 m, t = 223 seconds, u =0 m/s

Substituting

      520 = 0*223+\frac{1}{2} *a*223^2\\ \\ a = 0.0209 m/s^2

Now we have v = u+at, where v is the final velocity

Substituting

        v = 0+0.0209*223 = 4.66 m/s

So final velocity of car b = 4.66 m/s

c) Acceleration = 0.0209 m/s^2

7 0
3 years ago
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