A 6.5 kg rock thrown down from a 120m high cliff with initial velocity 18 m/s down. Calculate

The larger the push, the larger the change in velocity. This is an example of Newton's Second Law of Motion which states that th

Paul [167]

According to Newton, an object will only accelerate if there is a net or unbalanced forceacting upon it. The presence of an unbalanced force will accelerate an object - changing its speed, its direction, or both its speed and direction.

6 0

4 years ago

Read 2 more answers

) Water falls from a height of 60m at the rate of 15kg/s to operate a turbine. The losses due to frictional force are 10% of ene

Angelina_Jolie [31]

Answer:

8100W

Explanation:

Let g = 10m/s2

As water is falling from 60m high, its potential energy from 60m high would convert to power. So the rate of change in potential energy is

$P = \dot{E} = \dot{m}gh = 15*10*60 = 9000 J/s$ or 9000W

Since 10% of this is lost to friction, we take the remaining 90 %

P = 9000*90% = 8100 W

3 0

3 years ago

Using the formula F = M* A. What is the acceleration of a .5 kg

Katyanochek1 [597]

Answer:32 m/s/s

Explanation: since F=M*A, F=16N, M=0.5kg, A= F/M

A=16/0.5

A=32 m/s/s

7 0

3 years ago

A charged object is suspended motionless in the air by the gravitational force pulling it down and an electric force pushing it

Savatey [412]

The charge of the object must be $1.11 \times e^{-5} \text { coulomb }$

Answer: Option C

<u>Explanation:</u>

Suppose an electric charge can be represented by the symbol Q. This electric charge generates an electric field; Because Q is the source of the electric field, we call this as source charge. The electric field strength of the source charge can be measured with any other charge anywhere in the area. The test charges used to test the field strength.

Its quantity indicated by the symbol q. In the electric field, q exerts an electric, either attractive or repulsive force. As usual, this force is indicated by the symbol F. The electric field’s magnitude is simply defined as the force per charge (q) on Q.

$Electric field, E=\frac{\text { Force }(F)}{q}$