ASAP WILL GIVE BRAINLIEST!!!
1 answer:
Answer:
<em>The change in momentum of the car is 3575 Kg.m/s</em>
Explanation:
<u>Impulse and Momentum</u>
The impulse (J) experienced by the object equals the change in momentum of the object (Δp).
The formula that represents the above statement is:
J = Δp
The impulse is calculated as
J = F.t
Where F is the applied force and t is the time.
The car hits a wall with a force of F=6500 N and stops in 0.55 s. Thus, the impulse is:
J = 6500 * 0.55
J = 3575 Kg.m/s
The change in momentum of the car is:
The change in momentum of the car is 3575 Kg.m/s
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When dealing with multiple forces acting on a body, it is advisable to draw a free-body diagram like that shown in the picture. There are four forces acting on the box: weight (W) pointing straight down, normal force perpendicular to the slope denoted as Fn, force used to push the box upwards along the slope and the frictional force acting opposite to the direction of motion of the box denoted as Ff. Frictional force is equal to coefficient of kinetic friction (μk) multiplied with Fn.
∑Fy = Fn - mgcos30° = 0
Fn = (50)(9.81)(cos 16) = 471.5 N
When in motion, the net force is equal to mass times acceleration according to Newton's 2nd Law of Motion:
Fnet = F - μk*Fn - mgsin30° = ma
250 - (0.2)(471.5 N) - (50)(sin 16°) = (50)(a)
a = 2.84 m/s²
∑Fy = Fn - mgcos30° = 0
Fn = (50)(9.81)(cos 16) = 471.5 N
When in motion, the net force is equal to mass times acceleration according to Newton's 2nd Law of Motion:
Fnet = F - μk*Fn - mgsin30° = ma
250 - (0.2)(471.5 N) - (50)(sin 16°) = (50)(a)
a = 2.84 m/s²
Answer : 413.44N
Here it is given that an elevator is moving down with an acceleration of 3.36 m/s² . And we are interested in finding out the apparent weight of a 64.2 kg man . For the diagram refer to the attachment .
- From the elevator's frame ( non inertial frame of reference) , we would have to think of a pseudo force.
- The direction of this force is opposite to the direction of acceleration the frame and its magnitude is equal to the product of mass of the concerned body with the acceleration of the frame .
- When a elevator accelerates down , the weight recorded is less than the actual weight .
From the Free body diagram ,
- Mass of the man = 64.2 kg
