A ball is thrown horizontally from a cliff at a speed of 10 m/s. you predict that its speed 1 s later will be slightly greater t
2 answers:
Answer:
Speed is slightly greater than 14 m/s.
Explanation:
The ball will under go projectile motion. The horizontal velocity remains constant while there would be increase in vertical velocity due to acceleration due to gravity in downward direction.
Using first equation of motion, after 1 s, vertical velocity will be:
v = u +at
v=0+(9.8 m/s²)(1 s) = 9.8 m/s
Horizontal velocity = 10 m/s
Net velocity:
Speed is slightly greater than 14 m/s.
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Answer:
1047 miles
Explanation:
The radius of the Earth is
(miles)
So its circumference, which is the total length of the equator, is given by
Now we know that the Earth rotates once every 24 hours. So the distance through which the equator moves in one hour is equal to its total length divided by the number of hours, 24:
Refer to the diagram shown below.
Assume that air resistance is ignored.
Note:
The distance, h, of a falling object with initial vertical velocity of zero at time t is
h = (1/2)gt²
where
g = 9.8 m/s²
The initial vertical velocity of the supplies is 0 m/s.
It the time taken for the supplies to reach the ground is t, then
(50 m) = (1/2)*(9.8 m/s²)*(t s)²
Hence obtain
t² = 50/4.9 = 10.2041
t = 3.1944 s
The horizontal distance traveled at a speed of 100 m/s is
d = (100 m/s)*(3.1944 s) = 319.44 m
Answer: 319.4 m (nearest tenth)
Assume that air resistance is ignored.
Note:
The distance, h, of a falling object with initial vertical velocity of zero at time t is
h = (1/2)gt²
where
g = 9.8 m/s²
The initial vertical velocity of the supplies is 0 m/s.
It the time taken for the supplies to reach the ground is t, then
(50 m) = (1/2)*(9.8 m/s²)*(t s)²
Hence obtain
t² = 50/4.9 = 10.2041
t = 3.1944 s
The horizontal distance traveled at a speed of 100 m/s is
d = (100 m/s)*(3.1944 s) = 319.44 m
Answer: 319.4 m (nearest tenth)
The initial speed of car A is 15.18 m/s.
Momentum is defined as mass in motion. If there are two objects (the two objects in motion or only one object in motion and the other in stationary) that collide and no other forces work in the system, the law of momentum conservation applies in the system.
p=p'
pa+pb = pa'+pb'
(ma×va) + (mb×vb) = (ma×va') + (mb×vb')
- ma = mass of object A (kg) = 1,783 kg
- mb = mass of object B (kg) = 1,600 kg
- va = speed of object A before collides (m/s)
- va' = speed of object A after collides (m/s) = 8 m/s
- vb = speed of object B before collides (m/s) = 0 m/s
- vb' = speed of object B after collides (m/s) = 8 m/s
- p = momentum before collision (Ns)
- p' = momentum after collision (Ns)
(ma×va) + (mb×vb) = (ma×va') + (mb×vb')
(1,783×va) + (1,600×0) = (1,783×8) + (1,600×8)
(1,783×va) + 0 = 14,264+12,800
(1,783×va) = 27,064
va = 15.18 m/s
Learn more about The law of momentum conservation here: brainly.com/question/7538238
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