1.a baseball hits a home run, according to Newton's laws of motion which of the following statements could be true in this scena
rio: statement 1: the bat pushes on the ball with the same force as the ball pushes on the bat.
statement 2: Newton's 2nd law explains why the ball travels farther than the bat.
statement 3: the bat hit the ball causing the ball to change direction. this is an example of unbalanced force in Newtons first law of motion.
choose all statements that are true.
2. A bowling ball on earth weighs 11 pounds and has a mass of 5kg. The gravitational pull on the moon is 1/6 of the gravitational force on earth. what would the mass of the bowling ball be on the moon?
3.how much force would you need to use in order to accelerate an 80kg shopping cart to 0.2 m/s^2 (f=ma)?
<u>The correct option is </u><u>bat pushes on the ball with the same force as the ball pushes on the bat.
</u>
<h2>Explanation:
</h2>
According to Newton’s law action and reaction are equal in magnitude but opposite in direction so the same case will apply here. The bat pushes on the ball with the same force as the ball pushes on the bat just the direction is opposite.
<h2>Answer to Q2
</h2>
Mass will remain the same
<h2>Explanation:
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The gravitational pull is related to the weight of objects. Weight is the force with which earth attracts everything towards its centre. And mass is the quantity of matter inside the body. So on the moon the quantity of matter inside a body does not change therefore the mass would remain the same.
Information that is given: a = -5.4m/s^2 v0 = 25 m/s --------------------- S = ? Calculate the S(distance car traveled) with the formula for velocity of decelerated motion: v^2 = v0^2 - 2aS The velocity at the end of the motion equals zero (0) because the car stops, so v=0. 0 = v0^2 - 2aS v0^2 = 2aS S = v0^2/2a S = (25 m/s)^2/(2×5.4 m/s^2) S = (25 m/s)^2/(10.8 m/s^2) S = (625 m^2/s^2)/(10.8 m/s^2) S = 57.87 m
The force applied by the competitor is littler than the heaviness of the barbell. At the point when the barbell quickens upward, the power applied by the competitor is more prominent than the heaviness of the barbell. When it decelerates upward, the power applied by the competitor is littler than the heaviness of the barbell.
