Answer:
F = 7 N
Explanation:
given,
mass of sled = 17.5 Kg
speed of snow on horizontal surface = 3.5 m/s
time taken to stop = 8.75 s
initial momentum
= m U
= 17.5 x 3.5 = 61.25 Kg.m/s
final momentum
final velocity = 0 m/s
final momentum = 0
impulse = change in momentum
impulse = force x time
equating both the formula of impluse
F x t = 61.25 - 0
F x 8.75 = 61.25
F = 7 N
Force acting on the moving sled is equal to F = 7 N
Answer:
The acceleration of the ball as it rises to the top of its arc equals 9.807 meters per square second.
Explanation:
Let suppose that maximum height of the arc is so small in comparison with the radius of the Earth.
Since the ball is launched upwards, then the ball experiments a free-fall motion, that is, an uniform accelerated motion in which the element is accelerated by gravity. Then, the acceleration experimented by the motion remains constant at every instant and position.
Besides, the gravitational acceleration in the Earth and, in consequence, the acceleration of the ball as it rises to the top of its arc equals 9.807 meters per square second.
An object in motion changes position
Answer:
D
Explanation:
5 Km=5000m
so Δv=5000 m/sec
a=Δv/Δt
=5000/10
a=500 m/sec² as 500÷1000=0.5 Km
a=0.5 km/sec²
so D is the right answer.
Answer:
1065 Kgm-3
Explanation:
We can determine the relative density of the athlete from the formula;
Relative density of athlete = weight of athlete in air/upthrust on athlete
Since weight of athlete in air= 690 N
Weight of athlete in water = 42 N
Upthrust on athlete= weight in air - weight in water
Upthrust on athlete= 690 N - 42 N = 648 N
Relative density of athlete= 690 N / 648 N
Relative density of athlete= 1.065
Therefore, average density of the athlete= relative density × density of water = 1.065 × 1000 Kgm-3 = 1065 Kgm-3
