Answer:
The time taken by the object to reach the ground is 0.58 seconds.
Explanation:
Given that,
An object was released from rest at height of 1.65 m with respect to ground. We need to find the time taken by the object to reach the ground. Initial speed of the object is 0 as it is at rest. It will move downward under the action of gravity such that, the distance covered by the object is given by :
t = 0.58 seconds
So, the time taken by the object to reach the ground is 0.58 seconds. Hence, this is the required solution.
Answer:
Explanation:
If the initial velocity is U
Then the horizontal component of the velocity is
Ux= Ucosθ
Then the range for a projectile is give as
R=Ux.t
Where t is the time of flight
The time of flight is given as
t=2USinθ/g
Therefore,
R=Ux.t
R=UCosθ.2USinθ/g
R=U^2×2SinθCosθ/g
Then, from trigonometric ratio
2SinθCosθ= Sin2θ
R=U^2Sin2θ/g
Given that θ=32° and g=9.81m/s^2
Then
R=U^2Sin2×32/9.81
R=U^2Sin64/9.81
R=0.0916U^2
Then, range is given by R=0.0916U^2
A=0.0916U^2.
T
The box is at a distance A from the point of projection. Then the range R=A
R=0.0916U^2
A=0.0916U^2
Then,
U^2=A/0.0916
U^2=10.915A
Then the initial velocity should be
U=√10.915A
U=3.3√A
Answer:
W = 30.38 N
Explanation:
Given that,
Mass of a rock, m = 3.1 kg
We need to find the weight of the rock on the surface of Earth. Weight of an object is given by :
W = mg
g is the acceleration due to gravity, g = 9.8 m/s²
W = 3.1 kg × 9.8 m/s²
= 30.38 N
So, the weight of the rock on the Earth is 30.38 N.
400 * 3 = 1200
A jet can travel 1200 minutes for 3 seconds
Because the Moon has a very small surface area compared to other spacial geo-bodies, it has cooled down much faster than Earth. Any water on the moon would freeze.
