Answer:
The resultant force would (still) be zero.
Explanation:
Before the 600-N force is removed, the crate is not moving (relative to the surface.) Its velocity would be zero. Since its velocity isn't changing, its acceleration would also be zero.
In effect, the 600-N force to the left and 200-N force to the right combines and acts like a 400-N force to the left.
By Newton's Second Law, the resultant force on the crate would be zero. As a result, friction (the only other horizontal force on the crate) should balance that 400-N force. In this case, the friction should act in the opposite direction with a size of 400 N.
When the 600-N force is removed, there would only be two horizontal forces on the crate: the 200-N force to the right, and friction. The maximum friction possible must be at least 200 N such that the resultant force would still be zero. In this case, the static friction coefficient isn't known. As a result, it won't be possible to find the exact value of the maximum friction on the crate.
However, recall that before the 600-N force is removed, the friction on the crate is 400 N. The normal force on the crate (which is in the vertical direction) did not change. As a result, one can hence be assured that the maximum friction would be at least 400 N. That's sufficient for balancing the 200-N force to the right. Hence, the resultant force on the crate would still be zero, and the crate won't move.
Answer:
the third law
Explanation:
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Answer:
Speed of the ball relative to the boys: 25 km/h
Speed of the ball relative to a stationary observer: 35 km/h
Explanation:
The RV is travelling at a velocity of
Here we have taken the direction of motion of the RV as positive direction.
The boy sitting near the driver throws the ball back with speed of 25 km/h, so the velocity of the ball in the reference frame of the RV is
with negative sign since it is travelling in the opposite direction relative to the RV. Therefore, this is the velocity measured by every observer in the reference frame of the RV: so the speed measured by the boys is
v = 25 km/h
Instead, a stationary observer outside the RV measures a velocity of the ball given by the algebraic sum of the two velocities:
v = +60 km/h + (-25 km/h) = +35 km/h
So, he/she measures a speed of 35 km/h.
To develop this problem it is necessary to apply the equations concerning Bernoulli's law of conservation of flow.
From Bernoulli it is possible to express the change in pressure as
Where,
Velocity
Density
g = Gravitational acceleration
h = Height
From the given values the change of flow is given as
Therefore between the two states we have to
The flow rate will have changed to 54.77 % of its original value.
Answer:
2.124 kg of water
Explanation:
height of the falls is about 48 meters.
Mass of water needed is 1kg = 1000g
Power needed is 106 watts.
The amount of energy in 106 watts in one sec is 106 joules.
To calculate the energy of the 1kg falling water = Mgh
Energy = 1000*9.81*48
Energy = 470880 joules.
1 megawatt is = 1000000watts
The kilogram of water needed is 1000000/470880 = 2.124 kg of water
