All categories

3 years ago

Help me please in this

Physics

1 answer:

vfiekz [6]3 years ago

7 0

U=- 2.5ms-¹
g=9.8ms-²
t=3s
V=- u +gt
V= 2.5+(9.8×3) = 2.5 + 29.4 = 31.9 ms-¹

You might be interested in

A ball which is dropped from the top of a building strikes the ground with a speed of 30m/s. Assume air resistance can be ignore

Vika [28.1K]

Answer:

h= 45.87 m.

Explanation:

Data given:

Vf= 30m/s , and we know that g = 9.8 m/s²

The ball has an initial velocity of zero Vi = 0 m/s²

To Find:

Height of the building = ?

Solution:

According to 3rd law of the motion;

2aS= Vf² - Vi² ( S= h , a=g)

2*9.81*h = (30)² - (0)²

h= 45.87 m.

3 0

3 years ago

Where does all of this energy come from

Amiraneli [1.4K]

Answer: All of our energy comes from the sun, which is our nearest star. The sun sends out huge amounts of energy through its rays every day. We call this energy solar energy or radiant energy. Without the sun, life on earth would not exist, since our planet would be totally frozen.

6 0

3 years ago

T object]user: the main factors that determine the speed of a wave in any particular medium are _____. density and temperature s

inessss [21]

It's true that mechanical waves require a medium for transmission.

8 0

3 years ago

While the change in blank will remain the same during a collision, the force needed to bring an object to a stop can be blank if

Ahat [919]

Explanation :

There are two types of collision i.e. elastic and elastic collision.

Elastic collision : In this type of collision, the total momentum and the kinetic energy of the particles remains constant.
Inelastic collision : In this type of collision, only the momentum remains constant while there is some loss of kinetic energy occurs.

From Newton's second law,

F = m a

a is the rate of change of velocity.

$F=m\dfrac{v}{t}$

There is a inverse relation between the force and the time of collision.

The change in momentum will remain the same during a collision, the force needed to bring an object to a stop can be increased if the time of the collision is decreased.

6 0

3 years ago

Read 2 more answers

A long coaxial cable consists of an inner cylindrical conductor with radius a and an outer coaxial cylinder with inner radius b

Natasha_Volkova [10]

Answer:

Part a)

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Part b)

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Part d)

As we know that due to induction of charge there will be same charge appear on the inner and outer surface of the cylinder but the sign of the charge must be different

On the inner side of the cylinder there will be negative charge induce on the inner surface and on the outer surface of the cylinder there will be same magnitude charge with positive sign.

Explanation:

Part a)

By Guass law we know that

$\int E. dA = \frac{q}{\epsilon_0}$

$E. 2\pi rL = \frac{\lambda L}{\epsilon_0}$

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Part b)

Outside the outer cylinder we will again use Guass law

$\int E. dA = \frac{q}{\epsilon_0}$

$E. 2\pi rL = \frac{\lambda L}{\epsilon_0}$

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Part d)

As we know that due to induction of charge there will be same charge appear on the inner and outer surface of the cylinder but the sign of the charge must be different

On the inner side of the cylinder there will be negative charge induce on the inner surface and on the outer surface of the cylinder there will be same magnitude charge with positive sign.

4 0

3 years ago