Refer to the diagram shown below.
h = height of the girl above water when she lets go of the rope.
The launch velocity is 22.5 m/s at 35° to the horizontal. Therefore the vertical component of the velocity is
v = 22.5 sin(35°) = 12.9055 m/s.
The time of flight is t = 1.10 s before the girl hits the surface of the water at a height of -h.
Therefore
-h = (12.9055 m/s)*(1.10 s) - (1/2)*(9.8 m/s²)*(1.10 s)²
-h = 8.267 m
= 8.3 m (nearest tenth)
Answer:
When the girl let go of the rope, she was about 8.3 m above the surface of the water.
Answer:
Option B is the correct answer.
Explanation:
Angular velocity of the driver = 20 rad/s
The axis of rotation is 2.0 m.
Radius of rotation = 2 m
Linear velocity of the club head = Angular velocity of the driver x Radius of rotation
Linear velocity of the club head = 20 x 2 = 40 m/s
Option B is the correct answer.
Answer:
An object’s weight is proportional to its mass.
Think of it like this, gravity has to pull harder on the heavier object to make them fall at the same rate , but doesn't have to pull as hard for the lighter object , thus is why sometimes heavier objects fall faster then lighter ones
Answer:
1.6675×10^-16N
Explanation:
The force of gravity that the space shuttle experiences is expressed as;
g = GM/r²
G is the gravitational constant
M is the mass = 1.0 x 10^5 kg
r is the altitude = 200km = 200,000m
Substitute into the formula
g = 6.67×10^-11 × 1.0×10^5/(2×10^5)²
g = 6.67×10^-6/4×10^10
g = 1.6675×10^{-6-10}
g = 1.6675×10^-16N
Hence the force of gravity experienced by the shuttle is 1.6675×10^-16N