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garri49 [273]
3 years ago
14

Helpppppppppppppppppppppppppppopppppppppp

Physics
1 answer:
harkovskaia [24]3 years ago
7 0

Answer:

I think it's 2 the photo is hard to tell what they are exactly talking about.

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Sam, whose mass is 78 kg , stands at the top of a 11-m-high, 110-m-long snow-covered slope. His skis have a coefficient of kinet
Valentin [98]

Answer:

v = 8.09   m/s

Explanation:

For this exercise we use that the work done by the friction force plus the potential energy equals the change in the body's energy.

Let's calculate the energy

       

starting point. Higher

         Em₀ = U = m gh

final point. To go down the slope

         Em_f = K = ½ m v²

The work of the friction force is

         W = fr L cos 180

to find the friction force let's use Newton's second law

Axis y

        N - W_y = 0

        N = W_y

X axis

        Wₓ - fr = ma

let's use trigonometry

        sin  θ = y / L

         sin θ = 11/110 = 0.1

         θ = sin⁻¹  0.1

          θ = 5.74º

         sin 5.74 = Wₓ / W

         cos 5.74 = W_y / W

         Wₓ = W sin 5.74

         W_y = W cos 5.74

the formula for the friction force is

         fr = μ N

         fr = μ W cos θ

Work is friction force is

         W_fr = - μ W L cos θ  

Let's use the relationship of work with energy

        W + ΔU = ΔK

         -μ mg L cos 5.74 + (mgh - 0) = 0  - ½ m v²

        v² = - 2 μ g L cos 5.74 +2 (gh)

        v² = 2gh - 2 μ gL cos 5.74

let's calculate

        v² = 2 9.8 11 - 2 0.07 9.8 110 cos 5.74

        v² = 215.6 -150.16

        v = √65.44

        v = 8.09   m/s

6 0
2 years ago
In what subject could we see cross cutting concepts
meriva
A Framework for K–12 Science Education: Practices, Crosscutting Concepts, and Core Ideas (Framework) recommends science education in grades K–12 be built around three major dimensions: science and engineering practices, crosscutting concepts that unify the study of science and engineering through their common application across fields, and core ideas in the major disciplines of natural science.
4 0
2 years ago
the cycle of the moon through its phases, or the synodic month, is a. 21 days long. b. 27 1/3 days long. c. 29 1/2 days long. d.
olasank [31]
It takes 29.5 days long
7 0
3 years ago
Suppose that a person gets hit by a bus moving at 30 mi/h with a 58,000 lbs of force in the direction of motion. If the mass of
alexandr402 [8]

The impulse of a force is due to the change in the motion of an object

A. The persons speed after impact is approximately 59.38 mi/h

B. The expected speed is <u>29.89 mi/h</u> which is less than the findings

Reason:

Known parameters are;

The speed of the bus, v = 30 mi/h

The force with which the person was hit, F = 58,000 lbs

Mass of the bus, M = 40,000 lbs

Mass of the person, m = 150 lbs

Duration of the impact, Δt = 0.007 seconds

A. The speed of the person at the end of the impact, <em>v</em>, is given as follows;

The impulse of the force = F × Δt = m × Δv

For the person, we get;

58,000 lbf ≈ 1866094.816 lb·ft./s²

58,000 lbf × 0.007 s = 150 lbs × Δv

1,866,094.816 lb·ft./s²

\Delta v = \dfrac{1,866,094.816\ lbs \times 0.007 \, s}{150 \, lbs} \approx  87.084  \ ft./s

Δv = v₂ - v₁

The initial speed of the person at the instant, can be as v₁ = 0

The final speed, v₂ = Δv - v₁

∴ v₂ ≈  87.084 ft./s - 0 = 87.084 ft./s

≈ <u>87.084 ft./s</u>

<u />v_2 \approx \dfrac{87.084 \ ft./s}{y} \times\dfrac{1 \ mi}{5280 \ ft.} \times \dfrac{3,600 \ s}{1 \, hour} \approx 59.38 \ mi/h<u />

The speed of the person at the end of the impact, v₂ ≈ <u>59.38 mi/h</u>

B. Where the momentum is conserved, we have;

m₁·v₁ + m₂v₂ = (m₁ + m₂)·v

v = \dfrac{m_1 \cdot v_1 + m_2 \cdot v_2}{m_2 + m_1}

v = \dfrac{40,000 \times 30  + 150 \times 0}{40,000 + 150} \approx 29.89

The expected speed of the person at the end of the impact is 29.89 mi/h, and therefore, <u>the findings does not agree with the expectation</u>

Learn more here:

brainly.com/question/18326789

3 0
2 years ago
Use the drop-down menus to complete the statements. When the 5.0 kg cylinder fell 100 m, the final temperature of the water was
OLEGan [10]

Answer:

A. 26.17 B. 1.17 C. 30.86 D. 5.86

Explanation:

7 0
3 years ago
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