Answer:
C
Explanation:
Chemical reaction,physical
Answer:
E. Student 1 is correct, because as θ is increased, h is the same.
Explanation:
Here we have the object of a certain mass falling under gravity so the force acting on the it will depend on mass of the object and the acceleration due to gravity.
Mathematically:

As we know that the work done is evaluated as the force applied on a body and the displacement of the body in the direction of the force.
And for work we have:

where:
displacement of the object
angle between the force and displacement vectors
Given that the height of the object is same in each trail of falling object under the gravity be it a free-fall or the incline plane.
- In case of free-fall the angle between the force is and the displacement is zero.
- In case when the body moves along the inclined plane the force applied by the gravity is same because it depends upon the mass of the object. And the net displacement in the direction of the gravitational force is the height of the object which is constant in both the cases.
So, the work done by the gravitational force is same in the two cases.
Answer:
139.514 metres
Explanation:
Initial velocity of the truck = 6.6 m/s
Acceleration of the truck = 2.8 m/s^2
Time interval = 7.9 s
Therefore we use the formula,
s = ut + 1/2 at^2
*where s(the distance travelled)...u(the initial velocity)...t(the time period)
; s = 6.6(7.9) + 1/2 (2.8)(7.9)^2
; s = 52.14 + 87.374
The distance moved by the truck = 139.514m
Answer:
The speed of the sled is 3.56 m/s
Explanation:
Given that,
Mass = 2.12 kg
Initial speed = 5.49 m/s
Coefficient of kinetic friction = 0.229
Distance = 3.89 m
We need to calculate the acceleration of sled
Using formula of acceleration

Where, F = frictional force
m = mass
Put the value into the formula




We need to calculate the speed of the sled
Using equation of motion

Where, v = final velocity
u = initial velocity
a = acceleration
s = distance
Put the value in the equation



Hence, The speed of the sled is 3.56 m/s.