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2 years ago

8

A rock climber hangs freely from a nylon rope that is 15 m long and has a diameter of 8.3 mm. If the rope stretches 5.1 cm, what

is the mass of the climber? Young's modulus for nylon is Y=0.37Ã10^10N/m^2.

1 answer:

irinina [24]2 years ago

6 0

Answer:

Mass of the climber = 69.38 kg

Explanation:

Change in length

$\Delta L=\frac{PL}{AE}$

Load, P = m x 9.81 = 9.81m

Young's modulus, Y = 0.37 x 10¹⁰ N/m²

Area

$A=\frac{\pi (8.3\times 10^{-3})^2}{4}=5.41\times 10^{-5}m^2$

Length, L = 15 m

ΔL = 5.1 cm = 0.051 m

Substituting

$0.051=\frac{9.81m\times 15}{5.41\times 10^{-5}\times 0.37\times 10^{10}}\\\\m=69.38kg$

Mass of the climber = 69.38 kg

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Answer:

$\mathbf{1.51\times10^{-15}N}$

Explanation:

The computation of the weight of the paper in newtons is shown below:

On the paper, the induced charge is of the same magnitude as on the initial charges and in sign opposite.

Therefore the paper charge is

$q_{paper}=-4.1\times10^{-15}C$

Now the distance from the charge is

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Now, to raise the paper, the weight of the paper acting downwards needs to be managed by the electrostatic force of attraction between both the paper and the charge, i.e.

$mg=\frac{k_{e}q_{1}q_{2}}{r^{2}}$

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1 year ago

An object with a heat capacity of 345J∘C experiences a temperature change from 88.0∘C to 45.0∘C. How much heat is released in th

pogonyaev

Answer:

There is 148.35 Joules of heat is released in the process.

Explanation:

Given that,

Heat capacity of the object, $c=345J/^oC$

Initial temperature, $T_i=88^{\circ}C$

Final temperature, $T_f=45^{\circ}C$

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$Q=mc\Delta T$

Let the mass of the object is 10 g or 0.01 kg

So,

$Q=0.01\times 345\times (88-45)$

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2 years ago

Which changes would result in a decrease in the gravitational force between two objects? Check all that apply.

Rufina [12.5K]

Answer: 1. decreasing the mass of both objects

2. decreasing the mass of one of the objects

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Explanation: Hope that helped! (:

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2 years ago

A circular loop of diameter 10 cm, carrying a current of 0.20 A, is placed inside a magnetic field B⃗ =0.30 Tk^. The normal to t

arlik [135]

Answer:

The magnitude of the torque on the loop due to the magnetic field is $4.7\times10^{-4}\ N-m$ .

Explanation:

Given that,

Diameter = 10 cm

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Magnetic field = 0.30 T

Unit vector $n=-0.60\hat{i}-0.080\hat{j}$

We need to calculate the torque on the loop

Using formula of torque

$\tau=NIAB\sin\theta$

Where, N = number of turns

A = area

I = current

B = magnetic field

Put the value into the formula

$\tau=1\times0.20\times\pi\times(5\times10^{-2})^2\times0.30\times\sin90^{\circ}$

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2 years ago

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salantis [7]

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$D_z = \frac{2 F_{\alpha } }{A \rho v^2 }$

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$F_{\alpha }$ is the drag force

$\rho$ is the density of the fluid

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1 year ago