Answer:
The ball travels a distance of 20 m in the time interval of 4 s
Explanation:
Using s = ut + 1/2at² where s = distance travelled by the ball, u = initial velocity of ball = 0 m/s (since it starts from rest), a = acceleration of the ball = 2.50 m/s² and t = time = 4 s.
Substituting the variables into the equation, we have
s = ut + 1/2at²
s = 0 × 4 s + 1/2 × 2.50 m/s² × (4 s)²
s = 0 + 1/2 × 2.50 m/s² × 16 s²
s = 1/2 × 40 m
s = 20 m
So, the ball travels a distance of 20 m in the time interval of 4 s.
Answer:
distance changing at rate of 3.94 inches/sec
Explanation:
Given data
wall decreasing at a rate = 9 inches per second
ladder L = 152 inches
distance h = 61 inches
to find out
how fast is the distance changing
solution
we know that
h² + b² = L² ..................1
h² + b² = 152²
Apply here derivative w.r.t. time
2h dh/dt + 2b db/dt = 0
h dh/dt + b db/dt = 0
db/dt = - h/b × dh/dt .............2
and
we know
h = 61
so h² + b² = L²
61² + b² = 152²
b² = 19383
so b = 139.223
and we know dh/dt = -9 inch/sec
so from equation 2
db/dt = -61/139.223 (-9)
so
db/dt = 3.94 inches/sec
distance changing at rate of 3.94 inches/sec
Answer:
The correct answer is B.
The astronaut will know due to the light from the explosion.
Explanation:
Sound and vibrations require a medium such as air to travel through. Space, there is no air. Only a vacuum. So sound and vibrations are unable to travel. Light requires no medium to travel. It can go through a vacuum.
Therefore the Astronaut will see a bright flash of light as it travels from the explosion to outer space. It is also important to note that light can travel very far because nothing else interacts with its wave particles and as such, it cannot be impeded.
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