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3 years ago

How much force was applied to an object if was moved 2 meters and the work done on the object was 8 joules?

Physics

2 answers:

serious [3.7K]3 years ago

6 0

Answer:

The answer is 4N or B

Explanation:

Just the equation W = F x D.

We have W = 8 J and D = 2 m using algebra ....

8J/2m = F ... F = 4.

Natasha2012 [34]3 years ago

6 0

Answer:

B 4N

Explanation:

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Galois drove 60.0 kilometers due west in 5.00 hours and then drove 43.0 kilometers due north in 3.00 hours.

IceJOKER [234]

Answer:

Explanation:

See the attachment for the details. A right triangle is formed to find the hypotenuse of the two legs consisting of the actual driving distances and times. The hypotenuse gives the vector information for the displacement at the end of 8 hours of driving.

The individual driving times and distances are summed to provide:

(a) How far did he travel?

103 km

(b) What was his average speed?

12.88 km/h

(c) What was his displacement?

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(d) What was his average velocity?

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2 years ago

A physical pendulum consists of a meter stick that is pivoted at a small hole drilled through the stick a distance d from the 50

PSYCHO15rus [73]

Answer:

(a). The value of d is 0.056 cm and 1.496 cm.

(b). The time period is 1.35 sec.

Explanation:

Given that,

Length = 50.00 cm

Time period = 2.50 s

Time period of pendulum is defined as the time for one complete cycle.

The period depends on the length of the pendulum.

Using formula of time period

$T=2\pi\sqrt{\dfrac{I}{mgh}}$

Where, I = moment of inertia

We need to calculate the value of d

Using parallel theorem of moment of inertia

$I=I_{cm}+md^2$

For a meter stick mass m , the rotational inertia about it's center of mass

$I_{cm}-\dfrac{mL^2}{12}$

Where, L = 1 m

Put the value into the formula of time period

$T=2\pi\sqrt{\dfrac{\dfrac{mL^2}{12}+md^2}{mgd}}$

$T=2\pi\sqrt{\dfrac{L^2}{12gd}+\dfrac{d}{g}}$

$T^2=4\pi^2(\dfrac{L^2}{12gd}+\dfrac{d}{g})$

Multiplying both sides by d

tex]T^2d=4\pi^2(\dfrac{L^2}{12g}+\dfrac{d^2}{g})[/tex]

$(\dfrac{4\pi^2}{g})d^2-T^2d+\dfrac{\pi^2L^2}{3g}=0$

Put the value of T, L and g into the formula

$4.028d^2-6.25d+0.336=0$

$d = 0.056\ m, 1.496\ m$

The value of d is 0.056 cm and 1.496 cm.

(b). Given that,

L = 50-5 = 45 cm

We need to calculate the time period

Using formula of period

$T=2\pi\sqrt{\dfrac{l}{g}}$

Put the value into the formula

$T=2\pi\sqrt{\dfrac{45\times10^{-2}}{9.8}}$

$T=1.35\ sec$

Hence, (a). The value of d is 0.056 cm and 1.496 cm.

(b). The time period is 1.35 sec.

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