Answer:
on moon he can jump 4.2 m high
Explanation:
given data
gravitational acceleration on moon a(m) = 1/6
jump = 0.7 m
to find out
how high could he jump on the moon
solution
we know gravitational acceleration on earth a = g = 9.8 m/s²
so on moon am = = 1.633 m/s²
so if he jump on earth his speed will be for height 0.7 m i s
speed v =
v =
v = 3.7 m/s
so if he hump on moon
height will be
height =
put here value
height =
height = 4.2 m
so on moon he can jump 4.2 m high
Answer:
It is enough
Explanation:
To develop the problem it is necessary to take into account the concepts related to the coefficient of performance of a pump.
The two ways in which the performance coefficient can be expressed are given by:
Where,
High Temperature
Low Temperature
And the other way is,
Where is heat rate and W the power consumed.
We have all our terms in Celsius, so we calculate the temperature in Kelvin
The rate at which heat is lost is:
The power consumed by the heat pump is
And the coefficient of performance is
With this value we can calculate the Power required,
<em>The power consumed is consumed is 5kW which is more than 2.07kW so this heat pump powerful enough.</em>
Answer:
B = 1.67 μ T
Explanation:
given,
current, I = 1 x 10⁴ A
r = 120 m
treating lightning bolt as long straight conductor
resulting magnitude would be equal to
B = 16.67 x 10⁻⁶ T
B = 1.67 μ T
The resulting magnetic field is equal to B = 1.67 μ T