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4 years ago

If a 2 kg ball is traveling at a speed of 4 m/s, what is its kinetic energy?

Physics

2 answers:

Kay [80]4 years ago

8 0

Answer: 16J

Explanation:

K.E = $\frac{1}{2}$ m $v^{2}$

K.E = $\frac{1}{2}$ X 2 X 16

K.E = 16

Art [367]4 years ago

4 0

The kinetic energy of the ball is 16 J.

Answer: Option D

Explanation:

Kinetic energy is the form of energy exhibited by a body when it is in motion. The kinetic energy will be directly proportional to the mass of the body as well as to the square of the speed of the body at which it is moving. So the mathematical representation of kinetic energy of any object is

$\text {Kinetic energy}=\frac{1}{2} \times m \times(v)^{2}$

In the present case, the ball is the object with mass of around 2 kg and it is moving with the speed of about 4 m/s. So, the kinetic energy exhibited by the ball during its motion is

$\text {Kinetic energy of the ball}=\frac{1}{2} \times 2 \times(4)^{2}=16 \mathrm{J}$

So, the kinetic energy exhibited by the ball is 16 J.

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Which is not a property of a pure substance

Ann [662]

air is a mixture and not a pure substance.

5 0

3 years ago

Is there proof of the existence of God and Satan?If so what proof?

ivolga24 [154]

Religiously it has been proved that God exist and even satan exist
scientist does not believe God exist.The big bang theory is a theory that postulated that a big bang occur when and everything but my proverb says "in a film there is always behind the scene" if God doesnt exist who created mountains,hills,rivers,oceans,man,animal,spirit and if satan doesnt exist why does hell exist but if God doesnt exist why is it in grammar law to use capital letter for God
but my advise for you is to ask God to show you if He truly exist

3 0

4 years ago

An object with charge q = −6.00×10−9 C is placed in a region of uniform electric field and is released from rest at point A. Aft

sergeinik [125]

Answer:

a) 80 V

b) The magnitude of the electric field is 100 N/C and the direction of the electric field is from point B to point A where the electric field is toward the negative charge

Explanation:

Given :

We are given an object with charge q = -6.00 x I0^-9 C starts moving from the rest at point A, which means its kinetic energy at point A is zero ( $K_{A}$ = 0) to the point B at distance l = 0.500m where its kinetic energy is ( $K_{B}$ = 5.00 x 10^-7J) . Also, the electric potential of q at point A is VA = + 30.0 v.

Required :

(a) We are asked to find the electric potential VB

(b) We want to determine the magnitude and the direction of the electric field E.

Solution

(a) We are given the values for VA, $K_{B}$ and q so we want to find a relationship between these three parameters and VB to get the value of VB.

As we have two states, at points A and B , where the charge moved from A to B due to the applied electric field. The mechanical energy of the object is conservative during this travel, and we can apply eq(1) in this situation:

$K_{A} +U_{A} =K_{B} +U_{B}$ .........................................(1)

Where $K_{A}$ = 0 and the potential energy U of the charge is given by U = q V

where V is the electric potential. So, equation (1) will be in the form :

0+qVA= $K_{B}$ +qVB (Divide by q)

VA= $K_{B}$ /q + VB (solve for VB)

VB=VA- $K_{B}$ /q .......................................(2)

We get the relation between VB, VA and $K_{B}$ , now we can plug our values for VA, $K_{B}$ and q into equation (2) to get VB

VB=VA- $K_{B}$ /q

=30V-(5.00 x 10^-7J)/(-6.00 x I0^-9)

=80 V

(b) After we calculated VB we can use equation a to get the electric field E that applied to the charge q, where the potential difference between the two points equals the integration of the electric field multiplied by the distance l between the two points

VA-VB = $\int\limits^1_0 {E} \, dl$ ...................................(a)

$=E\int\limits^1_0 {} \, dl$

VA-VB=E $l$ (solve for E)

E= VA-VB/ $l$ ..................................(3)

Now let us plug our values for VA, Vs and l into equation (3) to get the electric field E

E= VA-VB/ $l$

=-100 N/C

The magnitude of the electric field is 100 N/C and the direction of the electric field is from point B to point A where the electric field is toward the negative charge

5 0

3 years ago

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UNO [17]

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3 years ago

A physics student uses a spring loaded launcher to fire a 58 gram projectile horizontally. The projectile leaves the launcher at

timofeeve [1]

Answer:

no answer

Explanation:

the question does not sk what to find

3 0

3 years ago