You are fixing the roof of your house when a hammer breaks loose and slides down. The roof makes an angle of 65o∘ with the horiz
ontal, and the hammer is moving at 9.5 m/s when it reaches the edge. Assume that the hammer is moving from the top of the roof to its right edge.
What is the horizontal component of the hammer's velocity just as it leaves the roof?
Express your answer with the appropriate units. Enter positive value if the x-component of the velocity is to the right and negative value if the x-component of the velocity is to the left.
What is the vertical component of the hammer's velocity just as it leaves the roof?
Express your answer with the appropriate units. Enter positive value if the direction of the y-component of the velocity is upward and negative value if the y-component of the velocity is downward.
What is the horizontal component of the hammer's velocity just as it leaves the roof?
Express your answer with the appropriate units. Enter positive value if the x-component of the velocity is to the right and negative value if the x-component of the velocity is to the left.
What is the vertical component of the hammer's velocity just as it leaves the roof?
Express your answer with the appropriate units. Enter positive value if the direction of the y-component of the velocity is upward and negative value if the y-component of the velocity is downward.
1 answer:
Answer:
Explanation:
Given:
initial speed of the hammer when leaving the edge of the roof along the inclination of the roof,
inclination of the roof form horizontal,
- Since the hammer is moving from the top of the roof to the right edge, its horizontal component will be towards right and vertical component will be towards downward direction.
Now the horizontal velocity:
<u>The vertical velocity:</u>
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Answer:
N = 195 turns
Explanation:
The inductance of the inductor, L = 500 μH = 500 * 10⁻⁶H
The length of the tube, l = 12 cm = 0.12 m
The diameter of the tube, d = 4 cm = 0.04 m
Radius, r = 0.04/2 = 0.02 m
Area of the tube, A = πr² = 0.02²π = 0.0004π m²
The inductance of a solenoid is given by:
a) 2.64 s
We can solve this part of the problem by using the following SUVAT equation:
where
s is the displacement of the stone
u is the initial velocity
t is the time
a is the acceleration
We must be careful to the signs of s, u and a. Taking upward as positive direction, we have:
- s (displacement) negative, since it is downward: so s = -75.0 m
- u (initial velocity) positive, since it is upward: +15.5 m/s
- a (acceleration) negative, since it is downward: so a= g = -9.8 m/s^2 (acceleration of gravity)
Substituting into the equation,
Solving the equation, we have two solutions: t = -5.80 s and t = 2.84 s. Since the negative solution has no physical meaning, the stone reaches the bottom of the cliff 2.64 s later.
b) 10.4 m/s
The speed of the stone when it reaches the bottom of the cliff can be calculated by using the equation:
where again, we must be careful to the signs of the various quantities:
- u (initial velocity) positive, since it is upward: +15.5 m/s
- a (acceleration) negative, since it is downward: so a = g = -9.8 m/s^2
Substituting t = 2.64 s, we find the final velocity of the stone:
where the negative sign means that the velocity is downward: so the speed is 10.4 m/s.
c) 4.11 s
In this case, we can use again the equation:
where
s is the displacement of the package
u is the initial velocity
t is the time
a is the acceleration
We have:
s = -105 m (vertical displacement of the package, downward so negative)
u = +5.40 m/s (initial velocity of the package, which is the same as the helicopter, upward so positive)
a = g = -9.8 m/s^2
Substituting into the equation,
Which gives two solutions: t = -5.21 s and t = 4.11 s. Again, we discard the first solution since it is negative, so the package reaches the ground after
t = 4.11 seconds.