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valkas [14]
3 years ago
11

While skiing, Sam flies down a hill and hits a jump. He has a mass of 75 kg, and he leaves the jump at 18 m/s. What is his momen

tum as he leaves the jump?
Physics
2 answers:
Evgen [1.6K]3 years ago
6 0

For this you would use the equation P = M x V

P ( momentum )

M ( mass - kg)

V ( velocity -m/s )

____________

Sams mass is 75 kg so you would fill in the "m" spot with 75

Sams velocity is 18 so you'd fill in the "v" spot with 18

_____

Your equation should look like this:

P = 75 x 18

From here is simple multiplication

75 x 18 = 675

Therefore your answer is p = 675

Vinvika [58]3 years ago
3 0
P=mv=18*75=1350
.......
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A pulley with the radius of 10.0 cm was connected to a motor with a massless
kogti [31]

Answer:

(i) -556 rad/s²

(ii) 17900 revolutions

(iii) 11250 meters

(iv) -55.6 m/s²

(v) 18 seconds

Explanation:

(i) Angular acceleration is change in angular velocity over time.

α = (ω − ω₀) / t

α = (10000 − 15000) / 9

α ≈ -556 rad/s²

(ii) Constant acceleration equation:

θ = θ₀ + ω₀ t + ½ αt²

θ = 0 + (15000) (9) + ½ (-556) (9)²

θ = 112500 radians

θ ≈ 17900 revolutions

(iii) Linear displacement equals radius times angular displacement:

s = rθ

s = (0.100 m) (112500 radians)

s = 11250 meters

(iv) Linear acceleration equals radius times angular acceleration:

a = rα

a = (0.100 m) (-556 rad/s²)

a = -55.6 m/s²

(v) Angular acceleration is change in angular velocity over time.

α = (ω − ω₀) / t

-556 = (0 − 15000) / t

t = 27

t − 9 = 18 seconds

8 0
3 years ago
Mercury is commonly used in thermometer give reasons​
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Answer:

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3 years ago
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8 0
3 years ago
Due to the intense heat and pressure on the inner core, scientists believe it is a A.solid B.liquid C.semi-fluid D.gas
nexus9112 [7]
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3 0
3 years ago
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Consult Multiple-Concept Example 5 for insight into solving this problem. A skier slides horizontally along the snow for a dista
Gennadij [26K]

Answer:

The speed with which the skier was going is approximately 2.9906 m/s

Explanation:

The given parameters are;

The distance the skier slides before coming to rest, s = 12.4 m

The coefficient of friction between the skier and the snow, \mu _k = 0.0368

Therefore, for conservation of energy, we have;

Initial kinetic energy = Work done by the kinetic friction force

Initial kinetic energy = 1/2·m·v²

The work done by the kinetic friction force = \mu _k×m×g×s

Where;

m = The mass of the skier

v = The speed with which the skier was going

g = The acceleration due to gravity = 9.8 m/s²

s = The distance the skier slides before coming to rest = 12.4 m

\mu _k = The kinematic friction = 0.0368

Therefore, for conservation of energy, we have;

1/2·m·v² = \mu _k×m×g×s

1/2·v² = \mu _k×g×s

v² = 2×\mu _k×g×s  = 2 × 0.0368 × 9.8 × 12.4 = 8.943872

v = √(8.943872) ≈ 2.99063070271 ≈ 2.9906 m/s

The speed with which the skier was going = v ≈ 2.9906 m/s.

3 0
3 years ago
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