The speed of the target with the bullet lodged in it is 5.6 m/s.
<h3>What is speed?</h3>
This is the rate of change of distance.
To calculate the speed of the target and the bullet lodged, we use the formula below.
Formula:
- V = mu/(m+M)............. Equation 1
Where:
- V = Speed of the target and the bullet
- m = mass of the bullet
- M = mass of the target
- u = Initial speed of the target
From the question,
Given:
- m = 0.04 kg
- u = 1400 m/s
- M = 9.96 kg
Substitute these values into equation 1
- V = (0.04×1400)/(0.04+9.96)
- V = 56/10
- V = 5.6 m/s
Hence, the speed of the target with the bullet lodged in it is 5.6 m/s.
Learn more about speed here: brainly.com/question/6504879
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Answer: Decreasing the distance of the space shuttle from Earth .
Explanation:
According to expression of gravitational force:
G = gravitational constant
= masses of two objects
r = Distance between the two objects.
F = Gravitational force
From the above expression we can say that gravitational force is inversely proportional to squared of the distance between the two masses.
So, in order to increase the gravitational force on space shuttle distance between the space space shuttle must be decreased.
Hence, the correct answer 'decreasing the distance of the space shuttle from Earth '.
It would be the first one
Answer:
the ball's velocity was approximately 0.66 m/s
Explanation:
Recall that we can study the motion of the baseball rolling off the table in vertical component and horizontal component separately.
Since the velocity at which the ball was rolling is entirely in the horizontal direction, it doesn't affect the vertical motion that can therefore be studied as a free fall, where only the constant acceleration of gravity is affecting the vertical movement.
Then, considering that the ball, as it falls covers a vertical distance of 0.7 meters to the ground, we can set the equation of motion for this, and estimate the time the ball was in the air:
0.7 = (1/2) g t^2
solve for t:
t^2 = 1.4 / g
t = 0.3779 sec
which we can round to about 0.38 seconds
No we use this time in the horizontal motion, which is only determined by the ball's initial velocity (vi) as it takes off:
horizontal distance covered = vi * t
0.25 = vi * (0.38)
solve for vi:
vi = 0.25/0.38 m/s
vi = 0.65798 m/s
Then the ball's velocity was approximately 0.66 m/s
