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3 years ago

If the specific gravity of copper is 8.91, it weighs:

Physics

1 answer:

denis-greek [22]3 years ago

6 0

Answer:

Option D is the correct answer.

Explanation:

The specific gravity of copper is 8.91.

We have

$\frac{\texttt{Density of copper}}{\texttt{Density of water}}=8.91\\\\\texttt{Density of copper}=8.91\times 1000kg/m^3=8910kg/m^3$

We also have

1 kg = 2.205 lb

1 m = 3.28 ft

Substituting

$\texttt{Density of copper}=8910kg/m^3=\frac{8910\times 2.205}{3.28^3}=556.76lb/ft^3$

Option D is the correct answer.

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iogann1982 [59]

Answer:

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Explanation:

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3 years ago

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MakcuM [25]

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3 years ago

1. The electromagnetic waves described in the reading are different in which two ways:

Kipish [7]

Answer:

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Explanation:

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3 years ago

A single-turn current loop carrying a 4.00 A current, is in the shape of a right-angle triangle with sides of 50.0 cm, 120 cm, a

pantera1 [17]

Given that,

Current = 4 A

Sides of triangle = 50.0 cm, 120 cm and 130 cm

Magnetic field = 75.0 mT

Distance = 130 cm

We need to calculate the angle α

Using cosine law

$120^2=130^2+50^2-2\times130\times50\cos\alpha$

$\cos\alpha=\dfrac{120^2-130^2-50^2}{2\times130\times50}$

$\alpha=\cos^{-1}(0.3846)$

$\alpha=67.38^{\circ}$

We need to calculate the angle β

Using cosine law

$50^2=130^2+120^2-2\times130\times120\cos\beta$

$\cos\beta=\dfrac{50^2-130^2-120^2}{2\times130\times120}$

$\beta=\cos^{-1}(0.923)$

$\beta=22.63^{\circ}$

We need to calculate the force on 130 cm side

Using formula of force

$F_{130}=ILB\sin\theta$

$F_{130}=4\times130\times10^{-2}\times75\times10^{-3}\sin0$

$F_{130}=0$

We need to calculate the force on 120 cm side

Using formula of force

$F_{120}=ILB\sin\beta$

$F_{120}=4\times120\times10^{-2}\times75\times10^{-3}\sin22.63$

$F_{120}=0.1385\ N$

The direction of force is out of page.

We need to calculate the force on 50 cm side

Using formula of force

$F_{50}=ILB\sin\alpha$

$F_{50}=4\times50\times10^{-2}\times75\times10^{-3}\sin67.38$

$F_{50}=0.1385\ N$

The direction of force is into page.

Hence, The magnitude of the magnetic force on each of the three sides of the loop are 0 N, 0.1385 N and 0.1385 N.

8 0

3 years ago

A horizontal spring with spring constant 750 N/m is attached to a wall. An athlete presses against the free end of the spring, c

ANTONII [103]

Answer:

37.5 N Hard

Explanation:

Hook's law: The force applied to an elastic material is directly proportional to the extension provided the elastic limit of the material is not exceeded.

Using the expression for hook's law,

F = ke.............. Equation 1

F = Force of the athlete, k = force constant of the spring, e = extension/compression of the spring.

Given: k = 750 N/m, e = 5.0 cm = 0.05 m

Substitute into equation 1

F = 750(0.05)

F = 37.5 N

Hence the athlete is pushing 37.5 N hard

4 0

3 years ago

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