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Likurg_2 [28]
3 years ago
12

An Olympic skier moving at 20.0 m/s down a 30.0o slope encounters a region of wet snow, of

Physics
1 answer:
Sonja [21]3 years ago
8 0

Answer:

Option B

Explanation:

Forces acting on the skier-

F1 =  -mg sin(30) down the slope  

F2 = -mg cos(30)

F3 = friction force  = 0.74 mg cos(30)  

Net force, down the slope  

=  -mg sin(30)+ 0.74 mg cos(30)\\= mg(.74 cos(30)-sin(30))\\= 0.14mg\\= 1.38m

Acceleration = F/m= 1.38 m/s2  

Acceleration remains constant, initial speed is 20 m/s

Speed at time t is 1.38t- 20 m/s

Distance down the slope at time t is 0.69t^2- 20t

When the skier stops, her speed is 0. Thus,  

, 1.38t- 20= 0\\t= 20/1.38\\= 14.5 seconds

Distance travelled in 14.5 seconds =  (0.69)(14.52- 20(14.5)= -145 m(negative because it is down the slope).

Option B is correct

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3 years ago
An engine causes a car to move 10 meters with a force of 100 N. The engine produces 10,000 J of energy. What is the efficiency o
kodGreya [7K]

Answer:

Part A

The efficiency of the engine is 10%

Part B

The change in internal energy is 300 J

Part C

The change in volume is 1 m³ which is one cubic meter

Explanation:

Part A

Efficiency is defined as the ratio of energy output to energy input;

The given parameters are;

The distance the car is moved, d = 10 meters

The force which moves the car, F = 100 N

The amount of energy produced by the engine = 1,000 J

Therefore, we have;

The energy output to the car = The work done on the car = Force applied to the car, F × The distance the car moves, d;

∴ The energy output to the car by the engine = F × d = 100 N × 10 m = 1,000 J

The energy input from the engine = The energy produced by the engine = 10,000 J

The efficiency of the engine = (The energy output)/(The energy input)= 1,000J/10,00J = 0.1

The efficiency in percentage = 0.1 × 100 = 10 %

The efficiency of the engine = 10%

Part B

The amount of heat added to the substance, ΔQ = 1,000J

The amount of work the substance does on the atmosphere, W = 700 J

The change in internal energy, ΔU is given as follows;

ΔQ = W + ΔU

∴ ΔU = ΔQ - W

For the substance, we have;

The change in internal energy, ΔU = 1,000 J - 700 J = 300 J

Part C

The work done by the piston, W = 1,000 J

The pressure, in the piston, P = 1,000 Pa = constant

The work done by the piston in a constant pressure process, W = P × ΔV

Where;

W = The work done

P = The constant pressure applied

ΔV = The change in volume = V₂ - V₁

V₂ = The final volume

V₁ = The initial volume

∴The change in volume ΔV = W/P = 1,000 J/(1,000Pa) = 1 m³

The change in volume ΔV = 1 m³

3 0
3 years ago
If a rock is thrown upward on the planet Mars with a velocity of 15 m/s, its height above the ground (in meters) after t seconds
s2008m [1.1K]

(a) The velocity (in m/s) of the rock after 1 second is 11.28 m/s.

(b) The velocity of the rock after 2 seconds is 7.56 m/s.

(c) The time for the block to hit the surface is 4.03.

(d) The velocity of the block at the maximum height is 0.

<h3>Velocity of the rock</h3>

The velocity of the rock is determined as shown below;

Height of the rock after 1 second; H(t) = 15(1) - 1.86(1)² = 13.14 m

v² = u² - 2gh

where;

  • g is acceleration due to gravity in mars = 3.72 m/s²

v² = (15)² - 2(3.72)(13.14)

v² = 127.23

v = √127.23

v = 11.28 m/s

<h3>Velocity of the rock when t = 2 second</h3>

v = dh/dt

v = 15 - 3.72t

v(2) = 15 - 3.72(2)

v(2) = 7.56 m/s

<h3>Time for the rock to reach maximum height</h3>

dh/dt = 0

15 - 3.72t = 0

t = 4.03 s

<h3>Velocity of the rock when it hits the surface</h3>

v = u - gt

v = 15 - 3.72(4.03)

v = 0

Learn more about velocity at maximum height here: brainly.com/question/14638187

8 0
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