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harina [27]
3 years ago
11

You want to calculate how long it takes a ball to fall to the ground from a height of 20 m. Which equation can you use to calcul

ate the time? (Assume no air resistance.)

Physics
1 answer:
fomenos3 years ago
8 0

Answer:it’s B.

Because it’s said in the fhoto

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As you move away from a positive charge distribution, the electric field:
GalinKa [24]

Answer:

The electric field always decreases.

Explanation:

The electric field due to a point charge is given by :

E=\dfrac{kq}{r^2}

Where

k = electric constant

q = charge

r = distance from the charge

It is clear from the above equation that as the distance from the charge particle increases the electric field decreases. As you move away from a positive charge distribution, the electric field always decreases. Hence, the correct option is (c) "Always decreases".

3 0
3 years ago
A loader sack of total mass
vampirchik [111]

Question: A loader sack of total mass

is l000 grams falls down from

the floor of a lorry 200 cm high

Calculate the workdone by the

gravity of the load.​

Answer:

19.6 Joules

Explanation:

Applying

W = mgh........................ Equation 1

Where W = Workdone by gravity on the load, m = mass of the loader sack, h = height, g = acceleration due to gravity

From the question,

Given: m = 1000 grams = (1000/1000) kilogram = 1 kg, h = 200 cm = 2 m

Constant: g = 9.8 m/s²

Substitute these values into equation 1

W = (1×2×9.8)

W = 19.6 Joules

Hence the work done by gravity on the load is 19.6 Joules

8 0
3 years ago
Definition for compression and rarefraction
s344n2d4d5 [400]
Compression and rarefaction are two phenomenon occurs in longitudunal wave!

when there is denser particle gathering in that wave , there we called it compression and the rarer part of particles is rarefaction !
6 0
3 years ago
Read 2 more answers
Show that the Mass spring system executes simple harmonic motion(SHM)?​
murzikaleks [220]

Explanation:

Show that the motion of a mass attached to the end of a spring is SHM

Consider a mass "m" attached to the end of an elastic spring. The other end of the spring is fixed

at the a firm support as shown in figure "a". The whole system is placed on a smooth horizontal surface.

If we displace the mass 'm' from its mean position 'O' to point "a" by applying an external force, it is displaced by '+x' to its right, there will be elastic restring force on the mass equal to F in the left side which is applied by the spring.

According to "Hook's Law

F = - Kx ---- (1)

Negative sign indicates that the elastic restoring force is opposite to the displacement.

Where K= Spring Constant

If we release mass 'm' at point 'a', it moves forward to ' O'. At point ' O' it will not stop but moves forward towards point "b" due to inertia and covers the same displacement -x. At point 'b' once again elastic restoring force 'F' acts upon it but now in the right side. In this way it continues its motion

from a to b and then b to a.

According to Newton's 2nd law of motion, force 'F' produces acceleration 'a' in the body which is given by

F = ma ---- (2)

Comparing equation (1) & (2)

ma = -kx

Here k/m is constant term, therefore ,

a = - (Constant)x

or

a a -x

This relation indicates that the acceleration of body attached to the end elastic spring is directly proportional to its displacement. Therefore its motion is Simple Harmonic Motion.

5 0
3 years ago
In an electric field, 0.90 joule of work is required to bring 0.45 coulomb of charge from point a to point
jarptica [38.1K]
The difference in electric potential energy between the two points is
\Delta U = q \Delta V
where q is the magnitude of the charge and \Delta V is the electric potential difference.

But for energy conservation, the difference in electric potential energy \Delta U between the two points is equal to the work done to move the charge between A and B:
W=\Delta U
so we have
W=q \Delta V

and by substituting the numbers of the problem, we find the value of \Delta V:
\Delta V =  \frac{W}{q}= \frac{0.90 J}{0.45 C}=2 V
3 0
3 years ago
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