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:
<em>At 574.59 Kelvin, the Fahrenheit temperature will be 574.59 °F.</em>
Explanation:
We first need to find a relation between the Kelvin scale and the Fahranheit scale. We'll use the Celsius scale to relate them.
The Kelvin and Celsius scales are related by the formula:
K = °C + 273.15
Solving for °C:
°C = K - 273.15
Besides, the Kelvin and Celsius scales are related by:
°C = 5 ⁄ 9(°F - 32)
Now we find a temperature, say X, where both scales coincide. Equating both formulas:
X - 273.15=5 ⁄ 9(X - 32)
Multiply by 9:
9X - 2,458.35 = 5X - 160
Simplifying:
4X = 2,458.35 - 160=2,298.35
Solving:
X =2,298.35 / 4 = 574.59
At 574.59 Kelvin, the Fahrenheit temperature will be 574.59 °F.
Answer:
A) The sum of the kinetic energy and the gravitational potential energy changes by an amount equal to the energy dissipated by friction,
Explanation:
- The kinetic energy is the energy that the object has and is defied by the work that is needed to accelerate the body.
- The gravitational potential is a mechanism by which an equal amount of energy is being transferred per unit mass that is needed for the object to move from the specific location.
- Hence when the sled moves down the hill with the force of gravity it has negligible resistance as an equal amount of energy is dissipated.
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