Answer:
b
Explanation:
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There is a lever with 5 m long. The fulcrum is 2 m from the right end. Each end hangs a box. The whole system is in balance. If the box hung to the right end is 12 kg, then what is the mass of the box hung to the left end?
Answer:
8kg
Explanation:
For the box to be in equilibrium. The clockwise moment ensued by the box on the right should be same as that ensued by the one on the right. Hence :
M ×3 = 12 ×2
M = 24/3 = 8kg
Note mass is used because trying to compute the weight by multiplying by the acceleration of free fall due to gravity on both sides will cancel out.
A 60-kg skier is stationary at the top of a hill. She then pushes off and heads down the hill with an initial speed of 4.0 m/s. Air resistance and the friction between the skis and the snow are both negligible. How fast will she be moving after she is at the bottom of the hill, which is 10 m in elevation lower than the hilltop
Answer:
The velocity is [tex]v = 8.85 m/s[/tex]
Explanation:
From the question we are told that
The mass of the skier is [tex]m_s = 60 \ kg[/tex]
The initial speed is [tex]u = 4.0 \ m/s[/tex]
The height is [tex]h = 10 \ m[/tex]
According to the law of energy conservation
[tex]PE_t + KE_t = KE_b + PE_b[/tex]
Where [tex]PE_t[/tex] is the potential energy at the top which is mathematically evaluated as
[tex]PE_t = mg h[/tex]
substituting values
[tex]PE_t = 60 * 4*9.8[/tex]
[tex]PE_t = 2352 \ J[/tex]
And [tex]KE_t[/tex] is the kinetic energy at the top which equal to zero due to the fact that velocity is zero at the top of the hill
And [tex]KE_b[/tex] is the kinetic energy at the bottom of the hill which is mathematically represented as
[tex]KE_b = 0.5 * m * v^2[/tex]
substituting values
[tex]KE_b = 0.5 * 60 * v^2[/tex]
=> [tex]KE_b = 30 v^2[/tex]
Where v is the velocity at the bottom
And [tex]PE_b[/tex] is the potential energy at the bottom which equal to zero due to the fact that height is zero at the bottom of the hill
So
[tex]30 v^2 = 2352[/tex]
=> [tex]v^2 = \frac{2352}{30}[/tex]
=> [tex]v = \sqrt{ \frac{2352}{30}}[/tex]
[tex]v = 8.85 m/s[/tex]
Answer:
The Skier's velocity at the bottom of the hill will be 18m/s
Explanation:
This is simply the case of energy conversion between potential and kinetic energy. Her potential energy at the top of the hill gets converted to the kinetic energy she experiences at the bottom.
That is
[tex]mgh = 0.5 mv^{2}[/tex]
solving for velocity, we will have
[tex]v= \sqrt{2gh}[/tex]
hence her velocity will be
[tex]v=\sqrt{2 \times 9.81 \times 10}=14.00m/s[/tex]
This is the velocity she gains from the slope.
Recall that she already has an initial velocity of 4m/s. It is important to note that since velocities are vector quantities, they can easily be added algebraically. Hence, her velocity at the bottom of the hill is 4 + 14 = 18m/s
The Skier's velocity at the bottom of the hill will be 18m/s
(III) A baseball is seen to pass upward by a window with a vertical speed of If the ball was thrown by a person 18 m below on the street, (a) what was its initial speed, (b) what altitude does it reach, (c) when was it thrown, and (d) when does it reach the street again? Giancoli, Douglas C.. Physics (p. 45). Pearson Education. Kindle Edition.
Answer:
Assuming that the vertical speed of the ball is 14 m/s we found the given values:
a) V₀ = 23.4 m/s
b) h = 27.9 m
c) t = 0.96 s
d) t = 4.8 s
Explanation:
a) Assuming that the vertical speed is 14 m/s (founded in the book) the initial speed of the ball can be calculated as follows:
[tex] V_{f}^{2} = V_{0}^{2} - 2gh [/tex]
Where:
[tex]V_{f}[/tex]: is the final speed = 14 m/s
[tex]V_{0}[/tex]: is the initial speed =?
g: is the gravity = 9.81 m/s²
h: is the height = 18 m
[tex] V_{0} = \sqrt{V_{f}^{2} + 2gh} = \sqrt{(14 m/s)^{2} + 2*9.81 m/s^{2}*18 m} = 23.4 m/s [/tex]
b) The maximum height is:
[tex] V_{f}^{2} = V_{0}^{2} - 2gh [/tex]
[tex] h = \frac{V_{0}^{2}}{2g} = \frac{(23. 4 m/s)^{2}}{2*9.81 m/s^{2}} = 27.9 m [/tex]
c) The time can be found using the following equation:
[tex] V_{f} = V_{0} - gt [/tex]
[tex] t = \frac{V_{0} - V_{f}}{g} = \frac{23.4 m/s - 14 m/s}{9.81 m/s^{2}} = 0.96 s [/tex]
d) The flight time is given by:
[tex] t_{v} = \frac{2V_{0}}{g} = \frac{2*23.4 m/s}{9.81 m/s^{2}} = 4.8 s [/tex]
I hope it helps you!