Friction pushes her hands!!
When both sides of an equation give the same units, same numerical values, and same concept we refer to the equation as being correct. ... Removing constants from correct equations make them homogeneous but incorrect.
Answer:
94.13 ft/s
Explanation:
<u>Given:</u>
= time interval in which the rock hits the opponent = 10 s - 5 s = 5 s
= distance to be moved by the rock long the horizontal = 98 yards
= displacement to be moved by the rock during the time of flight along the vertical = 0 yard
<u>Assume:</u>
= magnitude of initial velocity of the rock
= angle of the initial velocity with the horizontal.
For the motion of the rock along the vertical during the time of flight, the rock has a constant acceleration in the vertically downward direction.
Now the rock has zero acceleration along the horizontal. This means it has a constant velocity along the horizontal during the time of flight.
On dividing equation (1) by (2), we have
Now, putting this value in equation (2), we have
Hence, the initial velocity of the rock must a magnitude of 94.13 ft/s to hit the opponent exactly at 98 yards.
The acceleration of the boxes depends on the mass and weight.
we have a mass of 7 and 8 kilograms
if it took 25 N force to move box A, then you would take 25 and multiply by 8 then divide by 2.
It will leave you with 100 N.
finally take the sq rt of 100 to get 10
Answer:
<em>600N(downwards)</em>
Explanations
<em>600N(downwards)</em>
Mas of the person = 60kg
Acceleration due to gravity = -10m/s²
To get the earths pull on the person, we will use the Newton second law of motion;
Force = mass * acceleration;
Force = 60 * -10
Force- -600N
<em>Hence the earth gravitational pull on the person is 600N(downwards). It is downwards due to the negative sign.</em>
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