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Usimov [2.4K]
3 years ago
11

A concrete block is pulled 7.0 m across a frictionless surface by means of a rope. The tension in the rope is 40 N; and the net

work done on the block is 247 J. What angle does the rope make with the horizontal?
Physics
1 answer:
Arturiano [62]3 years ago
5 0

Answer:

Explanation:

ΣF = ma

Also, work done is given by

W=F.d

Then, applying scalar product

W=FdCos θ

Given that W=247J, d=7m, and F=T=40N

Then,

W=FdCos θ

247=40×7Cos θ

Then,

Cos θ=0.8821

θ=arccos(0.8821 )

θ=28.1°

The angle the rope make with the horizontal is 28.1°

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Part A Determine the magnitude of the x component of F using scalar notation. Fx F x = nothing lb Request Answer Part B Determin
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(0.707 F)^2 + (0.50 F)^2 + F_z^2 = F^2

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You have a string with a mass of 0.0127 kg. You stretch the string with a force of 9.33 N, giving it a length of 1.93 m. Then, y
melomori [17]

Answer:

wavelength = 0.968 m

frequency = 39.02 Hz

Explanation:

given data

mass = 0.0127 kg

force = 9.33 N

length = 1.93 m

to find out

wavelength and Frequency

solution

we know here linear density that is

linear density = \frac{mass}{length}   .........1

linear density = \frac{0.0127}{1.93}

linear density = 6.5803 × 10^{-3} kg/m

so

wavelength will be here

wavelength = \frac{2L}{n}   ..............2

here n = 4 for forth harmonic

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frequency = \frac{4}{2L} \sqrt{\frac{tension}{linear\ density} }    ..........3

frequency = \frac{4}{2*1.93} \sqrt{\frac{9.33}{6.5803*10^{-3}} }

frequency = 1.036269 × 37.654594

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2 years ago
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