Answer:
Minimum uncertainty in velocity of a proton,
Explanation:
It is given that,
A proton is confined to a space 1 fm wide,
We need to find the minimum uncertainty in its velocity. We know that the Heisenberg Uncertainty principle gives the uncertainty between position and the momentum such that,
Since, p = mv
So, the minimum uncertainty in its velocity is greater than . Hence, this is the required solution.
Answer:
1250
Explanation:
Let's look at this in a simple manner, because it is.
The crane weights 250Kg. Okay.
Since it is hung, there is the acceleration of gravity being applied on it (10m/s²)
Since F = m * a
F = 250 * 10
F = 2500
Now we know that the downward Force is 2500N.
To find the force that is being applied on that 30° angle, we can multiply our 2500N by cos30°, which happens to be .
Therefore, the force pulling the box in the cable's direction is:
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The answer is attached. Also, you should know that the unit for acceleration is m/s2 and for velocity it is m/s.
S=56, u=0, v=33, a=?, t=3.4
v=u+at
33=3.4 a
a = 9.7m/s^2
Дано:
S1 = 50 км
t1 = 4 часа
S2 = 100 км
t2 = 1 час
Найти:
Vср - ?
Решение:
Vср = S/t
Sср = S1 + S2 = 50 + 100 = 150 км
tср = t1 + t2 = 4 + 1 = 5 часов
Vcр = 150/5 = 30 км/ч
Ответ:
30 км/ч