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mars1129 [50]
3 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]3 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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An alert physics student stands beside the tracks as a train rolls slowly past. He notes that the frequency of the train whistle
Dovator [93]

Answer:The velocity of the train is 3.84m/s

Explanation:

According to the Doppler effect, if the source is moving towards you then the apparent frequency of the sound emitted by the source is higher and if the source is moving away from you then the apparent frequency of the sound emitted by the source is smaller.

This is given by:

fo = V +-Vo/ V +-Vo × source

Where fo= observed frequency

V= velocity of sound

Vo= vo it of the observer

fsource= frequency the source

Given:

Observed frequency of the approaching train fo1= 452Hz

The observed frequency of train= fo2= 442Hz

Velocity of sound= 334m/s

Velocity of source=?

Train approaching the observer is given by:

fo1= V/(V - Vs)× source ...eq1

Train passes the student is given by:

fo2= V/(V - Vs)×source ...eq2

Divide eq1 by eq2

452/442 = (343+Vs)/(343 - Vs)

1.02 =(343+Vs)/(343 -Vs)

Cross multiply

1.02(343- Vs) = 343 + Vs

350.76 - 1.02Vs = 343 + Vs

Collecting like terms

350.76 -343= 1.02Vs+ Vs

7.76 = 2.02Vs

Vs= 7.76/2.02

Vs= 3.84m/s

4 0
3 years ago
A car is traveling at a speed of 38.0 m/s on an interstate highway where the speed limit is 75.0 mi/h. Is the driver exceeding t
lidiya [134]

Answer:

Yes, the car driver is exceeding the given limit.

Explanation:

<u>Given:</u>

  • Speed of the car, v = 38.0 m/s.
  • Speed limit of the highway, \rm v_o=75.0\ mi/h.

<h2><u>Converting the speed limit from mi/h to m/s:</u></h2>

We know,

1 mi = 1.60934 km.

1 km = 1000 m.

Therefore, 1 mi = 1.60934 × 1000 m = 1609.34 m.

1 hour = 60 minutes.

1 minute = 60 seconds.

Therefore, 1 hour = 60 × 60 seconds = 3600 seconds.

Using these values,

\rm 1\ \dfrac{mi}{h}=\dfrac{1609.34\ m}{3600\ s}=0.447\ m/s.

Therefore,

\rm v_o = 75.0\ mi/h=75.0\times 0.447=33.52\ m/s.

Clearly,

\rm v_o

which means, the car driver is exceeding the given speed limit.

6 0
3 years ago
A physical property of halite (table salt) is its _____.
nexus9112 [7]
Taste: salty
color: varies. ex: white, clear, purple, yellow, etc.
status: mineral
4 0
3 years ago
En que parte de la isla de cuba tuvieron lugar los principales acontecimientos de la guerra de independencia?
DedPeter [7]

Answer:

El 24 de febrero de 1895, por órdenes de Martí se levantan 35 aldeas en el Oriente de Cuba en lo que se ha dado en llamar el Grito de Baire.

Entre los lugares están Ibarra, Guantánamo y Manzanillo.

Explanation:

8 0
3 years ago
A 58 kg skier is going down a 35 degree slope. The areaof each
maxonik [38]

To solve this problem we will use a free body diagram that allows us to determine the Normal Force.

In general, the normal force would be equivalent to

N = mgcos\theta

Since the skier is standing on two skis, his weight will be divide by two

N' = \frac{mgcos\theta}{2}

Pressure is given as the force applied in a given area, that is

P = \frac{F}{A}

Replacing F with N'

P = \frac{N'}{A}

P = \frac{\frac{mgcos\theta}{2}}{A}

Our values are given as,

m = 58kg

g = 9.8m/s^2

\theta = 35\°

A = 0.3m^2

Replacing we have that

P = \frac{\frac{(58)(9.8)cos(35)}{2}}{0.3}

P = 776.01Pa

Therefore the pressure exerted by each ski on the snow is 776.01Pa

6 0
3 years ago
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