<span>No, there is no control group because each group is treated under test conditions.</span>
The Volume of the ice block is 5376.344 cm^3.
The density of a material is define as the mass per unit volume.
Here, the density of ice given is 0.93 g/cm^3
Mass of the ice block given is 5 kg or 5000 g
Now calculate the volume of the ice block
density=mass/volume
0.93=5000/Volume
Volume =5376.344 cm^3
Therefore the volume of ice block is 5376.344 cm^3
Answer:
Depending on which hemisphere it is, like western to eastern, It would most likely get stuck at the center. You would also have to put more things into thought like acceleration, velocity, and speed.
BUT since the question asked "would it pop out the other side?", I'm assuming it's talking about northern to southern hemisphere. so in that case it would pop out the other side since gravity makes things go downwards.
We have: F = m×a
Here, m = 90 Kg
a = 15 m/s²
Substitute their values into the expression:
F = 90 × 15
F = 1350 N
In short, Your Answer would be Option D
Hope this helps!
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.
