This is the answer: Fossil's found in Susan's yard are from prehistoric times.
The statements that correctly compare the gravitational force with the electrical force are the following:
-The gravitational force can be attractive.
-The electrical force can be repulsive.
-The electrical force can be attractive.
-Any two objects experience a gravitational force between them.
Answer:
8977.7 kg/m^3
Explanation:
Volume of water displaced = 55 cm^3 = 55 x 10^-6 m^3
Reading of balance when block is immersed in water = 4.3 N
According to the Archimedes principle, when a body is immersed n a liquid partly or wholly, then there is a loss in the weight of body which is called upthrust or buoyant force. this buoyant force is equal to the weight of liquid displaced by the body.
Buoyant force = weight of the water displaced by the block
Buoyant force = Volume of water displaced x density of water x g
= 55 x 10^-6 x 1000 x .8 = 0.539 N
True weight of the body = Weight of body in water + buoyant force
m g = 4.3 + 0.539 = 4.839
m = 0.4937 kg
Density of block = mass of block / volume of block
= 
Density of block = 8977.7 kg/m^3
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: Last option
2.27 m/s2
Explanation:
As the runner is running at a constant speed then the only acceleration present in the movement is the centripetal acceleration.
If we call a_c to the centripetal acceleration then, by definition
in this case we know the speed of the runner
The radius "r" will be the distance from the runner to the center of the track
The answer is the last option
