Answer:
There is absolutely No relationship between the weight of an object (which is constant) and the frictional force. If a block is sliding on a surface, that surface will be exerting a force on the block. That force can be resolved into a component parallel to the surface (which we call the frictional component), and a component perpendicular to the surface (called the normal component). For many situations, we find experimentally that the frictional component is approximately proportional to the normal component. The frictional component divided by the normal component is defined to be a quantity called the coefficient of kinetic or sliding friction. The coefficient of kinetic friction obviously depends on the nature of the surfaces involved. The normal component on an object can be decreased if you pull in the direction of the normal component (the weight does not change). However pulling this way on the object not only decreases the normal component, but it also decreases the frictional component since they are proportional. This is why it is easier to slide something if you pull up on it while you push it. If you push down, the normal and frictional components increase so it is harder to slide the object. The weight of an object is the downward force exerted by Earth’s gravity on that object, and it does not change no matter how you push or pull on the object.
I believe its the third answer
Answer:
The magnitude will be "353.5 N". A further solution is given below.
Explanation:
The given values is:
F = 500 N
According to the question,
In ΔABC,
⇒ 
⇒ 
then,
⇒ 
⇒ 
Now,
The corresponding angle will be:
⇒ 
⇒ 
⇒ 
Aspect of F across the AC arm will be:
= 
On putting the values of F, we get
= 
= 
Component F along the AC (in magnitude) will be:
= 
= 
= 
Answer:
v0 = 25 m/s
vf = 0 m/s
a = -9.80 m/s^2
change in x = 31.89m
but that's only 1/3 of the hight, so i time it by 3 to get 96m
Answer:
4.02 s
Explanation:
From the question given above, the following data were obtained:
Angle of projection (θ) = 35°
Initial velocity (u) = 50 m/s
Acceleration due to gravity (g) = 10 m/s²
Time of flight (T) =?
The time of flight of the arrow can be obtained as follow:
T = 2uSineθ / g
T = 2 × 35 × Sine 35 / 10
T = 70 × 0.5736 / 10
T = 7 × 0.5736
T = 4.02 s
Therefore, the time taken for the arrow to return is 4.02 s
