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2 years ago

Hansel pushes a 2 lb. box 5 feet in 20 seconds. How much work has he done?

1 answer:

8 0

Work = force x distance

You can see time doesn’t matter (if we were talking about power, which is the RATE at which work is performed, that would be a different story).

W = 2 x 5 = 10 foot-pounds of work

Foot-pounds are gross units. Better to work in SI units when you can!

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A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate

Viefleur [7K]

Answer:

The rate of change of the area when the bottom of the ladder (denoted by $b$ ) is at 36 ft. from the wall is the following:

$\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s$

Explanation:

The Area of the triangle is given by $A=h\times b$ where $h=\sqrt{l^2-b^2}$ (by using the Pythagoras' Theorem) and $b$ is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

$A=\sqrt{l^2-b^2}b$

The rate of change of the area is given by its time derivative

$\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)$

$\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}$

$\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Product rule

$\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Chain rule

$\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$

$\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)$

In here we can identify $b=36\, ft$ , $l=39$ and $\frac{db}{dt}=8\,ft/s$ .

The result is then

$\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s$

3 0

2 years ago

Circle the letter of each expression that has four significant figures. A. 1.25 x 10^4 B. 12.51 C. 0.0125 D. 0.1255

Andreyy89

Answer:

letter B

none zero digit are significant figures

3 0

3 years ago

A force of 14 lb is required to hold a spring stretched 4 in. beyond its natural length. How much work W is done in stretching i

enyata [817]

Answer:

$W = 19.8 J$

Explanation:

14 lb force is required to stretch the spring by 4 inch distance

So we have

$F = 14 lbf$

$F = 6.35 \times 9.8 N$

$F = 62.3 N$

stretch in the spring is given as

$x = 4 in = 0.1016 m$

now we will have

$F = kx$

$62.3 = k(0.1016)$

$k = 613.125 N/m$

Now we need to find the work to stretch it by x = 10 in = 0.254 m

so we have

$W = \frac{1}{2}kx^2$

$W = \frac{1}{2}(613.125)(0.254)^2$

$W = 19.8 J$

7 0

3 years ago

A solenoid of length 5.0 cm has a cross sectional area of 1.0 cm2 and has 300 uniformly spaced turns. The solenoid is wrapped in

OleMash [197]

The calculated mutual inductance is 8.544 x 10⁻⁵ H.

Two coils have a mutual inductance of 1 henry when emf of 1 volt is induced in coil 1 and when the current flowing through coil 2 is changing at the rate of one ampere per second.

Length of the solenoid= 5.0 cm

Area of cross-section=1.0 cm²

no of spaced turns=300 turns

turns of insulated wire=180 turns

Mutual inductance (M) = μ₀μr N1N2 A/ L

=(4xπx 10⁻⁷) x (6.3 x 10⁻³) x 300 x 180 x 1/ 5

=79.12 x 10⁻¹⁰ x 54000 / 5

=8.544 x 10⁻⁵ H

hence, the mutual inductance is 8.544 x 10⁻⁵ H.

Learn more about Mutual inductance here-

brainly.com/question/14014588

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4 0

7 months ago

To charge an electroscope negatively by induction you need: a) a positively charged rod and a ground b) a negatively charged rod

Zolol [24]

Answer:

Option a)

Explanation:

In the process of charging anything by the method of induction, a charged body is brought near to the body which is neutral or uncharged without any physical contact and the ground must be provided to the uncharged body.

The charge is induced and the nature of the induced charge is opposite to that of the charge present on the charged body.

So when a positively charged rod is used to charge an electroscope, the rod which is positive attracts the negative charge in the electroscope and the grounding of the electroscope ensures the removal of the positive charge and renders the electroscope negatively charged.

6 0

2 years ago