Answer:
Yes. Towards the center. 8210 N.
Explanation:
Let's first investigate the free-body diagram of the car. The weight of the car has two components: x-direction: towards the center of the curve and y-direction: towards the ground. Note that the ground is not perpendicular to the surface of the Earth is inclined 16 degrees.
In order to find whether the car slides off the road, we should use Newton's Second Law in the direction of x: F = ma.
The net force is equal to 
Note that 95 km/h is equal to 26.3 m/s.
This is the centripetal force and equal to the x-component of the applied force.
As can be seen from above, the two forces are not equal to each other. This means that a friction force is needed towards the center of the curve.
The amount of the friction force should be 
Qualitatively, on a banked curve, a car is thrown off the road if it is moving fast. However, if the road has enough friction, then the car stays on the road and move safely. Since the car intends to slide off the road, then the static friction between the tires and the road must be towards the center in order to keep the car in the road.
Answer:
500 kg
Explanation:
It is given that,
The mass of a open train car, M = 5000 kg
Speed of open train car, V = 22 m/s
A few minutes later, the car’s speed is 20 m/s
We need to find the mass of water collected in the car. It is based on the conservation of momentum as follows :
initial momentum = final momentum
Let m is final mass
MV=mv
Water collected = After mass of train - before mass of train
= 5500 - 5000
= 500 kg
So, 500 kg of water has collected in the car.
Answer:
The time taken is 
Explanation:
From the question we are told that
The speed of first car is 
The speed of second car is 
The initial distance of separation is 
The distance covered by first car is mathematically represented as
Here 
is the initial distance which is 0 m/s
and 
is the final distance covered which is evaluated as 
So
The distance covered by second car is mathematically represented as
Here is the initial distance which is 119 m
and is the final distance covered which is evaluated as 
Given that the two car are now in the same position we have that
The answer to your question is C.
I'm not sure what "60 degree horizontal" means.
I'm going to assume that it means a direction aimed 60 degrees
above the horizon and 30 degrees below the zenith.
Now, I'll answer the question that I have invented.
When the shot is fired with speed of 'S' in that direction,
the horizontal component of its velocity is S cos(60) = 0.5 S ,
and the vertical component is S sin(60) = S√3/2 = 0.866 S . (rounded)
-- 0.75 of its kinetic energy is due to its vertical velocity.
That much of its KE gets used up by climbing against gravity.
-- 0.25 of its kinetic energy is due to its horizontal velocity.
That doesn't change.
-- So at the top of its trajectory, its KE is 0.25 of what it had originally.
That's E/4 .
