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.
The combined momentum is 4000 kg m/s south
Explanation:
The total combined momentum of the two cars is given by the vector addition of the momenta of the two cars.
For this problem, we choose north as positive direction and south as negative direction.
The momentum of the first car travelling north is given by:
where
is the mass of the car
is its velocity
Substituting,
The momentum of the second car travelling south is given by:
where
is the mass of the car
is its velocity (negative because the car travels south)
Substituting,
And therefore, the combined momentum is
where the negative sign means the direction of the total momentum is south.
Learn more about momentum:
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Force acting during collision is internal so momentum is conserve
so (initial momentum = final momentum) in both directions
Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1150 kg and was approaching at 5.00 m/s due south. The second car has a mass of 750 kg and was approaching at 25.0 m/s due west.
Let Vx is and Vy are final velocities of car in +x and +y direction respectively.
initial momentum in +ve x (east) direction = final momentum in +ve x direction (east)
- 750*25 + 1150*0 = (750+1150)
Vx
initial momentum in +ve y (north) direction = final momentum in +ve y direction (north)
750*0 - 1150*5 = (750+1150)
Vy
from here you can calculate Vx and Vy
so final velocity V is
<span>V=<span>(√</span><span>V2x</span>+<span>V2y</span>)
</span>
and angle make from +ve x axis is
<span>θ=<span>tan<span>−1</span></span>(<span><span>Vy</span><span>Vx</span></span>)
</span><span>
kinetic energy loss in the collision = final KE - initial KE</span>
Answer: Impulse = 20 Ns
Explanation:
Impulse is the product of force and time
Also impulse = momentum
Where momentum is the product of mass and velocity.
Given that
M = 2kg
V = 10 m/s
Impulse = MV = 2 × 10 = 20 Ns
Answer:
the mass of water is 0.3 Kg
Explanation:
since the container is well-insulated, the heat released by the copper is absorbed by the water , therefore:
Q water + Q copper = Q surroundings =0 (insulated)
Q water = - Q copper
since Q = m * c * ( T eq - Ti ) , where m = mass, c = specific heat, T eq = equilibrium temperature and Ti = initial temperature
and denoting w as water and co as copper :
m w * c w * (T eq - Tiw) = - m co * c co * (T eq - Ti co) = m co * c co * (T co - Ti eq)
m w = m co * c co * (T co - Ti eq) / [ c w * (T eq - Tiw) ]
We take the specific heat of water as c= 1 cal/g °C = 4.186 J/g °C . Also the specific heat of copper can be found in tables → at 25°C c co = 0.385 J/g°C
if we assume that both specific heats do not change during the process (or the change is insignificant)
m w = m co * c co * (T eq - Ti co) / [ c w * (T eq - Tiw) ]
m w= 1.80 kg * 0.385 J/g°C ( 150°C - 70°C) /( 4.186 J/g°C ( 70°C- 27°C))
m w= 0.3 kg
