Decompose the forces acting on the block into components that are parallel and perpendicular to the ramp. (See attached free body diagram. Forces are not drawn to scale)
• The net force in the parallel direction is
∑ <em>F</em> (para) = -<em>mg</em> sin(21°) - <em>f</em> = <em>ma</em>
• The net force in the perpendicular direction is
∑ <em>F</em> (perp) = <em>n</em> - <em>mg</em> cos(21°) = 0
Solving the second equation for <em>n</em> gives
<em>n</em> = <em>mg</em> cos(21°)
<em>n</em> = (0.200 kg) (9.80 m/s²) cos(21°)
<em>n</em> ≈ 1.83 N
Then the magnitude of friction is
<em>f</em> = <em>µn</em>
<em>f</em> = 0.25 (1.83 N)
<em>f</em> ≈ 0.457 N
Solve for the acceleration <em>a</em> :
-<em>mg</em> sin(21°) - <em>f</em> = <em>ma</em>
<em>a</em> = (-0.457N - (0.200 kg) (9.80 m/s²) sin(21°))/(0.200 kg)
<em>a</em> ≈ -5.80 m/s²
so the block is decelerating with magnitude
<em>a</em> = 5.80 m/s²
down the ramp.
Answer:
980 kJ
Explanation:
Work = change in energy
W = mgh
W = (1000 kg/m³ × 5.0 m³) (9.8 m/s²) (20 m)
W = 980,000 J
W = 980 kJ
The pump does 980 kJ of work.
Answer:
38.33°C
Explanation:
Applying,
180/100 = (F-32)/C............. Equation 1
Where F = Temperature of the hot day in fehrenheit, C = Temperature of the hot day in centigrade.
make C the subject of the equation
C = 100(F-32)/180.............. Equation 2
From the question,
Given: F = 101°F
Substitute into equation 2
C = 100(101-32)/180
C = 38.33°C
Answer:
Explanation:
Conservation of momentum is used to solve
Unfortunately we have a missing piece of information such as the initial velocity of the unknown mass train.
If we ASSUME that the second train is at rest
5000(100) + m(0) = 5000(50) + m(50)
which means m = 5000 kg
However, I'll show you the importance of knowing that initial velocity by finding it assuming the other answers are valid
if m = 15000 kg
5000(100) + 15000(v₀) = (5000 + 15000)(50)
v₀ = 33 ⅓ m/s
if m = 10000 kg
5000(100) + 10000(v₀) = (5000 + 10000)(50)
v₀ = 25 m/s
if m = 8000 kg
5000(100) + 8000(v₀) = (5000 + 8000)(50)
v₀ = 18.75 m/s
So you can see why I had to assume an initial velocity. Any of the masses could work if the initial velocity is chosen correctly.