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mars1129 [50]
4 years ago
13

A moving rope (parallel to the slope) is used to pull skiers up the mountain. If the slope of the hill is 37" and friction is ne

gligible, what force must the rope provide to pull a 500 N person up the hill at a constant rate?
Physics
1 answer:
Charra [1.4K]4 years ago
5 0

Since rope is parallel to the inclined plane so here we can say that net force parallel to the person which is pulling upwards must counterbalance the component of weight of the person.

Now here we will do the components of the weight of the person

given that weight of the person = 500 N

now its components are

W_x = 500 cos37

W_y = 500 sin37

now here as we can say that one of the component is balanced here by the normal force perpendicular to plane

while the other component of the weight is balanced by the force applied on the rope

So here the force applied on the rope will be given as

F = W_y = 500 sin37

F = 300 N

so it apply 300 N force along the inclined plane

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A metal rod has a length of 123. cm at 200°C. At what temperature will the length be 92.6 cm if the coefficient of linear expans
mestny [16]

Answer:

\theta_{2} = 15400^0 C

Explanation:

The formula for linear expansivity, \alpha = \frac{l_{2} - l_{1}  }{l_{1} ( \theta_{2}  - \theta_{1} )}

original length, l₁ = 123 cm = 1.23 m

final length, l₁ = 92.6 cm =0.926 m

original temperature, θ₁ = 200°C

Linear expansivity, α = 2 * 10⁻⁵ °C⁻¹

Putting these values into the formula:

2 * 10^{-5}  = \frac{1.23 - 0.926  }{l_{1} ( \theta_{2}  -200 )}\\ \theta_{2}  -200 = \frac{0.304}{2 * 10^{-5} } \\\theta_{2} = 15200 + 200\\\theta_{2} = 15400^0 C

7 0
3 years ago
A mass is attached to the end of a spring and set into oscillation on a horizontal frictionless surface by releasing it from a c
Aneli [31]

Answer

given,

x = (3.9 cm)sin[(9.3 rad/s)πt]

general equation of displacement

x = A sin ω t

A is amplitude

now on comparing

c) Amplitude  =3.9 cm

a) frequency =

     f = \dfrac{\omega}{2\pi}

     f = \dfrac{9.3\pi}{2\pi}

           f = 4.65 Hz

b) period of motion

        T= \dfrac{1}{f}

        T= \dfrac{1}{4.65}

        T = 0.215 s

d) time when displacement is equal to x= 2.6 cm

x = (3.9 cm)sin[(9.3 rad/s)πt]

2.6 = (3.9 cm)sin[(9.3 rad/s)πt]

sin[(9.3 rad/s)πt] = 0.667

9.3 π t = 0.73

t = 0.025 s

4 0
3 years ago
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Vikentia [17]
I think it is false, because why would is be stronger in a different state?

4 0
3 years ago
A 70-kg astronaut (including spacesuit and equipment) is floating at rest a distance of 13 m from the spaceship when she runs ou
Keith_Richards [23]

Answer:

Explanation:

mass of the astronaut including the spacesuit, M=30

distance of astronaut from the spaceship, d = 13 m

mass of the oxygen tank, m = 3 kg

Speed of tank with respect to spaceship, v=15~m/s

a)

<u>Using the conservation of linear momentum:</u>

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0=3\times 15+(70-15)\times v'

v'=0.82~m/s

b)

She mush hold her breath until she reaches the spaceship, i.e.

t=d/v'

t=13/0.82

t=15.89~s

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3 years ago
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