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iogann1982 [59]
3 years ago
9

A man is standing on a platform that is connected to a pulley arrangement, as the drawing shows. By pulling upward on the rope w

ith a force the man can raise the platform and himself. The total mass of the man plus the platform is 113 kg. What pulling force should the man apply to create an upward acceleration of 1.20 m/s2?

Physics
1 answer:
leonid [27]3 years ago
8 0

Answer:

The pulling force that the man should apply to create an upward acceleration of 1.20m/s^{2} is P=621.5N

Explanation:

Hi

As it shows in the drawing at the end, we have that the total mass of the man plus the platform is 113 kg, then the force downward W is W=mg=113Kg*9.8m/s^{2}=1107.4N.

Due the man needs to do a pulling force upward capable of lifting himself and the platform with an acceleration of 1.20m/s^{2}, this force should create an acceleration greater than gravity by 1.20m/s^{2}. then a_{up}=g+1.2m/s^{2}=9.8m/s^{2}+1.2m/s^{2}=11m/s^{2}. Therefore the force should be P_{up}=ma_{up}=113kg*11m/s^{2}=1243N.

Finally, as we have a pulley arrangement connected to the platform, it gives the man a mechanical advantage, so he has to do only half of that upward force

, therefore P=\frac{1243N}{2}=621.5N.

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How long would it take for a car to travel 200 km if it has an average speed of 55 km hr?
Rainbow [258]

Answer:

3.63 hours or 3 and 37.5 minutes

Explanation:

200/55

Hope this helps :)

6 0
3 years ago
A crane lifts a 545 kg piano up into a high-rise apartment 75.0 meters above the ground. In order to do so, 5350 Newtons were ap
vodka [1.7K]

Answer:

Work done to pull the piano upwards is 401250 J

Explanation:

Work is done against the gravity to pull the piano upwards

So here we can say that work done is

W = mgH

here we know that

mg = 5350 N

also we know that

H = 75 m

now we have

W = 5350 \times 75

W = 401250 J

7 0
3 years ago
A tetrahedron has an equilateral triangle base with 25.0-cm-long edges and three equilateral triangle sides. The base is paralle
djverab [1.8K]

Answer:

a. 7.046 Nm²/C

b. 2.348 Nm²/C

Explanation:

Data given:

Base of equilateral triangle = 25.0 cm = 0.25 m

Strength of electric field = 260 N/C

In order to find the electric flux we first have to find out the area of triangle.

Area of triangle = \frac{\sqrt{3} }{4} a^{2}

                         = \frac{\sqrt{3} }{4} (0.25)^{2}

                         = 0.0271 m³

Lets find electric flux,

      Electric Flux = E. A

                          = 260×0.0271

                          = 7.046 Nm²/C

Now we can find the electric flux through each of the three sides.

Electric flux through three sides = \frac{7.046}{3}

                                                = 2.348 N m²/C

       

3 0
3 years ago
If a car is traveling forward at 15 m/s, how fast will it be going in 1.2 seconds if the acceleration is
Law Incorporation [45]

Answer:

3

Explanation:

v = v⁰ (its original speed) + a (negative acceleration) X t² (time)

v = 15 - 10 x 1.2 = 15 - 12 = 3 (it's slowing down)

3 0
1 year ago
A guitar player tunes the fundamental frequency of a guitar string to 560 Hz. (a) What will be the fundamental frequency if she
lawyer [7]

Answer:

(a) if she increases the tension in the string is increased by 15%, the fundamental frequency will be increased to 740.6 Hz

(b) If she decrease the length of the the string by one-third the fundamental frequency will be increased to 840 Hz

Explanation:

(a) The fundamental, f₁, frequency is given as follows;

f_1 = \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \cdot L}

Where;

T = The tension in the string

μ = The linear density of the string

L = The length of the string

f₁ = The fundamental frequency = 560 Hz

If the tension in the string is increased by 15%, we will have;

f_{(1  \, new)} = \dfrac{\sqrt{\dfrac{T\times 1.15}{\mu}}  }{2 \cdot L} = 1.3225 \times \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \cdot L}  = 1.3225 \times f_1

f_{(1  \, new)} = 1.3225 \times f_1 = (1 + 0.3225) \times f_1

f_{(1  \, new)} = 1.3225 \times f_1 =\dfrac{132.25}{100} \times 560 \ Hz  = 740.6 \  Hz

Therefore, if the tension in the string is increased by 15%, the fundamental frequency will be increased by a fraction of 0.3225 or 32.25% to 740.6 Hz

(b) When the string length is decreased by one-third, we have;

The new length of the string, L_{new} = 2/3·L

The value of the fundamental frequency will then be given as follows;

f_{(1  \, new)} =  \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \times \dfrac{2 \times L}{3} }  =\dfrac{3}{2} \times \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \cdot L} = \dfrac{3}{2} \times 560 \ Hz =  840 \ Hz

When the string length is reduced by one-third, the fundamental frequency increases to one-half or 50% to 840 Hz.

6 0
3 years ago
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