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3 years ago

15

A 0.810 kg ball falls 2.5m. How much work does the force of gravity do on the ball?

1 answer:

lisabon 2012 [21]3 years ago

6 0

Answer:

W = 19.845 J

Explanation:

Work is defined as W = Fdcos $\theta$ , where F is the force exerted and d is the distance. Because the direction the ball is falling is the same direction as the force itself, $\theta$ = 0 deg, and since cos(0) = 1, this equation is equivalent to W = Fd. In this case, the force exerted is the weight force, which is equivalent to m * g. Substituting you get:

W = mgd = 0.810 kg * 9.8 m/s^2 * 2.5m

W = 19.845 J

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Answer:

A, the energy an object has due to its motion.

Explanation:

Kinetic energy is the energy created by motion.

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Which scientist formed his ideas about living things by performing observations without using a microscope? a) Hooke b) Malpighi

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It was Hooke who formed his ideas about living things by performing observations without using a microscope, but it should be noted there were others as well.

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3 years ago

our lab partner wears a new pair of sneakers to lab and, rather than performing the required experiments, you decide to measure

Dafna1 [17]

Answer:

The coefficient of static friction between your partner and the floor is 0.55

Explanation:

Given:

Mass $m = 59$ Kg

Frictional force $F_{s} = 318.3$ N

From the formula of frictional force,

$F_{s} = \mu_{s} mg$

Where $\mu _{s} =$ coefficient of static friction, $g = 9.8 \frac{m}{s^{2} }$

Put the above values and find the coefficient of static friction.

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Therefore, the coefficient of static friction between your partner and the floor is 0.55

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3 years ago

calculate the diameter of a silver wire of length 75cm , which is extended by 1.85mm when a 10kg mass is suspended from it's end

sdas [7]

Answer: $0.8\ mm$

Explanation:

Given

length of wire $l=75\ cm$

change in length $\Delta l=1.85\ mm$

mass of wire $m=10\ kg$

Young's modulus for silver $E=7.9\times 10^{10}\ N/m^2$

load on wire $F=mg$

$F=10\times 9.8=98\ kg$

change in length is given by

$\Delta l=\dfrac{Pl}{AE}$

Where A=area of cross-section

$A=\dfrac{Pl}{\Delta lE}$

$A=\dfrac{98\times 0.75}{1.85\times 10^{-3}\times 7.9\times 10^{10}}$

$A=\dfrac{73.5}{14.615\times 10^{7}}$

$A=5.029\times 10^{-7}\ m^2$

also wire is the shape of cylinder so cross-section is given by

$A=\dfrac{\pi d^2}{4}=5.029\times 10^{-7}\ m^2$

$\Rightarrow d^2=\dfrac{5.029\times 10^{-7}\times 4}{\pi }$

$\Rightarrow d^2=64.02\times 10^{-8}$

$\Rightarrow d=8\times 10^{-4}\ m$

$\Rightarrow d=0.8\ mm$

4 0

3 years ago

When operated on a household 110.0-V line, typical hair dryers draw about 1650 W of power. We can model the current as a long st

defon

Answer:

Current in the hair dryer will be equal to 15 A

Explanation:

We have given that household is operated at 110 volt

So potential difference V =110 volt

Power drawn by hairdryer is P = 1650 watt

We have to find the current in the hair dryer

We know that power is given as P = VI, here V is potential difference and I is current

So $1650=110\times I$

I = 15 A

So current in the hair dryer will be equal to 15 A

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4 years ago