One proton or one neutron have very close to the same mass.
Their mass is about the same as the mass of 1,840 electrons !
So six electrons have roughly the same mass as 0.0033 of one
proton or one neutron.
Yes, these masses are very tiny. None of them was ever measured
until the 20th Century. But all of them have now been measured,
very precisely and accurately. This is very impressive, and should
give us a lot of admiration for scientists of the past hundred years.
It would be very interesting to look around online and find out
how they did it.
The longer the time taken, the lower the power output.
This is since power is calculated through
Energy transferred / time
Increasing the denominator will lead to a lower value overall
Answer:
See explanation below
Explanation:
In this case, you want to know if you put an object between these forces, which direction would go.
To know this, we need to calculate the moment of an object, which is defined as the product of a force and it's distance. In other words:
M = F * d (1)
And, in order to reach equilibrium the force will exert a direction in clockwise or anticlosewise, and these moments, should be even:
anticlockwise moment = clockwise moment.
The clockwise would be the forces to the right, and anticlock would the only force to the left of the axle.
Clockwise moment = (10 * 0.8) + (25 * 2.6) = 73 Ns
Anticlockwise moment = 34 * 3.5 = 119 Ns.
As we can see, the moment in the anticlockwise is higher than the actual clockwise moment, therefore, we can assume that the object will move anticlockwise, or simply move to the left.
Hope this helps
Answer:
1.08 s
Explanation:
From the question given above, the following data were obtained:
Height (h) reached = 1.45 m
Time of flight (T) =?
Next, we shall determine the time taken for the kangaroo to return from the height of 1.45 m. This can be obtained as follow:
Height (h) = 1.45 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
h = ½gt²
1.45 = ½ × 9.8 × t²
1.45 = 4.9 × t²
Divide both side by 4.9
t² = 1.45/4.9
Take the square root of both side
t = √(1.45/4.9)
t = 0.54 s
Note: the time taken to fall from the height(1.45m) is the same as the time taken for the kangaroo to get to the height(1.45 m).
Finally, we shall determine the total time spent by the kangaroo before returning to the earth. This can be obtained as follow:
Time (t) taken to reach the height = 0.54 s
Time of flight (T) =?
T = 2t
T = 2 × 0.54
T = 1.08 s
Therefore, it will take the kangaroo 1.08 s to return to the earth.
KE = 1/ 2 * 1252 * 144
as KE = 1/2 * m * v ^2
= 90144 J
