Answer:
in squash, the balls are quite small and come in a variety of
speeds, ranging from extra- super slow to fast
Explanation:
Answer and Explanation:
Cup 1. The ice cubes are in a solid state.
Cup 2: The cup has water that is liquified.
Cup 3: The cup is empty and only contains free floating air particles.
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Answer:
D. Friction
Explanation:
"Friction is caused by microwelds that form between two surfaces. Microwelds are stronger when the two surfaces are pushed together with a greater force." - ( Module 3 • Forces and Newton’s Laws) quoted
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Answer:
a) about 20.4 meters high
b) about 4.08 seconds
Explanation:
Part a)
To find the maximum height the ball reaches under the action of gravity (g = 9.8 m/s^2) use the equation that connects change in velocity over time with acceleration.
In our case, the initial velocity of the ball as it leaves the hands of the person is Vi = 20 m/s, while thw final velocity of the ball as it reaches its maximum height is zero (0) m/s. Therefore we can solve for the time it takes the ball to reach the top:
Now we use this time in the expression for the distance covered (final position Xf minus initial position Xi) under acceleration:
Part b) Now we use the expression for distance covered under acceleration to find the time it takes for the ball to leave the person's hand and come back to it (notice that Xf-Xi in this case will be zero - same final and initial position)
To solve for "t" in this quadratic equation, we can factor it out as shown:
Therefore there are two possible solutions when each of the two factors equals zero:
1) t= 0 (which is not representative of our case) , and
2) the expression in parenthesis is zero:
Answer:
W_net = mg + 2mgh/r
Explanation:
The forces applied in this motion of the bowling ball are both gravitational and centripetal forces.
Now, gravitational force is; F_g = mg
While centripetal force is; F_c = mv²/r
Since we want to express the net force in terms of the variables in the statement and we are not given "v", let's find an expression of v with the variables given.
Now, from Newton's equation of motion, at initial velocity of 0, v² = 2gh.
Thus;
F_c = 2mgh/r
Where;
m is ball mass
r is tube radius
h is fall height
Thus, the net force will be;
F_net = F_g + F_c
Now, Net force would be equal to the net weight that will be read on the scale.
Thus;
W_net = F_net = F_g + F_c
W_net = mg + 2mgh/r