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Zinaida [17]
2 years ago
5

How much thermal energy does it take to raise the temperature of 2.5 kg of

Physics
1 answer:
KonstantinChe [14]2 years ago
3 0

Answer:

390kJ

Explanation:

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Calculate the electric field strength at a point at which a test charge of 0.30 coulombs experiences a force of 5.0 newtons.
SVEN [57.7K]

The strength of electric field E is 17 N / C.

<u />

<u>Explanation:</u>

Electric field strength is defined as the force per unit charge acting at a point in the given field. The equation for the strength of the electric field is given by

                     E = F / q

where E represents the electric field strength,

           F represents the force in newton,

           q represents the charge in coulomb.

Given the charge q = 0.30 coulombs

                   force F = 5.0 N

Electric field strength E = force / charge

                                        = 5.0 / 0.30

                                    E  = 16.66 = 17 N / C.

5 0
3 years ago
What is the force needed to give a .25 kg arrow an acceleration of 196m/s2?
Rzqust [24]
<span>49N is the force needed to give a .25 kg arrow an acceleration of 196m/s2. F =ma ⇒ =( 0.25kg)(196m/s2) = 49N if the arrow is shot horizontally where the applied force is entirely in the x-direction.</span>
5 0
3 years ago
Read 2 more answers
Using the results of Question 1 that would apply if the collision were inelastic, compute (using Excel in the yellow highlighted
ch4aika [34]

Answer:

The fractional kinetic energy will be lost if the collision is inelastic. In inelastic collision, the kinetic energy is converted into other forms of energy.

The lost energy became heat and sound energy.

Explanation:

During inelastic collision, the kinetic energy of a moving object does not conserve. It changes into another form of energy such as sound energy and heat energy etc.

For example, when a moving car hit another car or wall etc, the kinetic energy is converted into sound and heat energy. This type of collision is inelastic collision.

4 0
3 years ago
The mass of Object 2 is double the mass of Object 5. The mass of Object 4 is half of the mass of Object 5 and the mass of Object
SVETLANKA909090 [29]
This is a great problem if you like getting tied up in knots
and making smoke come out of your brain.

I found that it makes the problem a lot easier if I give the objects some
numbers. I'm going to say that the mass of Object 5 is 20 clods.

Let the mass of Mass of Object 5 be 20 clods .

Then . . .

-- The mass of Object 2 is double the mass of Object 5 = 40 clods.

-- The mass of Object 4 is half of the mass of Object 5 = 10 clods.
and
-- the mass of Object 3 is half of the mass of Object 4 = 5 clods.

So now, here are the masses:

Object #1 . . . . . unknown
Object #2 . . . . . 40 clods
Object #3 . . . . . 5 clods
Object #4 . . . . . 10 clods
Object #5 . . . . . 20 clods .

Now let's check out the statements, and see how they stack up:

Choice-A:
Object 3 and Object 5 exert the same gravitational force on Object 1.
Can't be.
Objects #3 and #5 have different masses, so they can't both
exert the same force on the same mass.

Choice-B.
Object 2 and Object 4 exert the same gravitational force on Object 1.
Can't be.
Objects #2 and #4 have different masses, so they can't both
exert the same force on the same mass.

Choice-C.
The gravitational force between Object 1 and Object 2 is greater than
the gravitational force between Object 1 and Object 4.
Yes ! Yay !
Object-2 has more mass than Object-4 has, so it must exert more force on
ANYTHING than Object-4 does, (as long as the distances are the same).

Choice-D.
The gravitational force between Object 1 and Object 3 is greater than the gravitational force between Object 1 and Object 5.
Can't be.
Object-3 has less mass than Object-5 has, so it must exert less force on
ANYTHING than Object-4 does, (as long as the distances are the same).

Conclusion:
If the DISTANCE is the same for all the tests, then Choice-C is
the only one that can be true.
8 0
3 years ago
A fan at a rock concert is 50.0 m from the stage, and at this point the sound intensity level is 114 dB. Sound is detected when
Marianna [84]

Answer:

A) P=13.92\ J.s^{-1}

B) v=3730.9912\ m.s^{-1}

C) v=74.44\ mm.s^{-1}

D) mosquitoes speed in part B is very much larger than that of part C.

Explanation:

Given:

  • Distance form the sound source, s=50\ m
  • sound intensity level at the given location, \beta=114\ dB
  • diameter of the eardrum membrane in humans, d=8.4 \times 10^{-3}\ m
  • We have the minimum detectable intensity to the human ears, I_0=10^{-12}\ W.m^{-2}

(A)

<u>Now the intensity of the sound at the given location is related mathematically as:</u>

\beta=10\ log(\frac{I}{I_0} ) ..........................................(1)

114=10\ log\ (\frac{I}{10^{-12}} )

11.4=log\ I+12\ log\ 10

I=0.2512\ W.m^{-2}

<em>As we know :</em>

I=\frac{P}{A}

0.2512=\frac{P}{\pi\times \frac{8.4^2}{4} }

P=13.92\ J.s^{-1} is the energy transferred to the  eardrums per second.

(B)

mass of mosquito, m=2\times 10^{-6}\ kg

<u>Now the velocity of mosquito for the same kinetic energy:</u>

KE=\frac{1}{2} m.v^2

13.92=\frac{1}{2}\times 2\times 10^{-6}\times v^2

v=3730.9912\ m.s^{-1}

(C)

Given:

  • Sound intensity, \beta = 20\ dB

<u>Using eq. (1)</u>

20=10\ log\ (\frac{I}{10^{-12}} )

2=log\ I+12\ log\ 10

I=10^{-10}\ W.m^{-2}

Now, power:

P=I.A

P=10^{-10}\times \pi\times \frac{8.4^2}{4}

P=5.54\times 10^{-9}\ J.s^{-1}

Hence:

KE=\frac{1}{2} m.v^2

5.54\times 10^{-9}=0.5\times 2\times 10^{-6}\times v^2

v=0.07444\ m.s^{-1}

v=74.44\ mm.s^{-1}

(D)

mosquitoes speed in part B is very much larger than that of part C.

7 0
3 years ago
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