Calculate the average velocity of a motor cycle that travels 72km/hr in 20 seconds

1 answer:

3 0

Velocity =displacement
Change in time
D=72km/hr
Time=20s
But the S.I unit of velocity is m/s so you woul have to change 72km/hr to m/s

Changing 72km to m

1 kilometer=1000meters
Then, 72 kilometers =?

72•1000/1
=72000m

Changing 72hours to seconds

If 1 hour = 3600 seconds
Then 72 hours=?

72•3600/1

=259200 seconds

Velocity =displacement
Change in time

V= 72,000
259,2005
=0.028m/s

You might be interested in

A capacitor is connected to a power source and begins charging. What causes the capacitor to stop charging while it is still con

Yuri [45]

Charging stops when the voltage across the capacitor reaches the voltage of the power source.

8 0

3 years ago

A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.05 104 m/s. What will its speed be wh

Elanso [62]

Answer:

The value is $v = 2.3359 *10^{4} \ m/s$

Explanation:

From the question we are told that

The initial speed is $u = 2.05 *10^{4} \ m/s$

Generally the total energy possessed by the space probe when on earth is mathematically represented as

$T__{E}} = KE__{i}} + KE__{e}}$

Here $KE_i$ is the kinetic energy of the space probe due to its initial speed which is mathematically represented as

$KE_i = \frac{1}{2} * m * u^2$

=> $KE_i = \frac{1}{2} * m * (2.05 *10^{4})^2$

=> $KE_i = 2.101 *10^{8} \ \ m \ \ J$

And $KE_e$ is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

$KE_e = \frac{1}{2} * m * v_e^2$

Here $v_e$ is the escape velocity from earth which has a value $v_e = 11.2 *10^{3} \ m/s$

=> $KE_e = \frac{1}{2} * m * (11.3 *10^{3})^2$

=> $KE_e = 6.272 *10^{7} \ \ m \ \ J$

Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as

$KE_p = \frac{1}{2} * m * v^2$