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kaheart [24]
3 years ago
13

A chimpanzee sitting against his favorite tree gets up and walks 51 m due east and 39 m due south to reach a termite mound, wher

e he eats lunch. (a) What is the shortest distance between the tree and the termite mound
Physics
1 answer:
n200080 [17]3 years ago
6 0

Answer:

64.20m

Explanation:

As we can see from the image I have attached below, the route that the chipanzee makes forms a right triangle. In this case, the shortest distance is represented by x in the image, which is the hypotenuse. To find this value we use the Pythagorean theorem which is the following.

a^{2} +b^{2}  = c^{2}

where a and b are the length of the two sides and c is the length of the hypotenuse (x). Therefore, we can plug in the values of the image and solve for x

51^{2} +39^{2} =x^{2}

2,601 + 1,521 = x^{2}

4,122 = x^{2}   ... square root both sides

64.20 = x

Finally, we see that the shortest distance is 64.20m

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What is the best way to dry a thermometer servsafe?
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A thermometer should be air dried so as to not damage it or anything similar. The three steps to cleaning it are washing, rinsing, and air drying it after. You shouldn't try to dry it other ways because it can damage it and this can cause a lot of troubles, so things cold air blowers or similar things can be very good in cleaning your thermometer.
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4 years ago
A certain frictionless simple pendulum having a length l and mass m swings with period t. If both l and m are doubled, what is t
iVinArrow [24]

If l and m both are doubled then the period becomes √2*T

what is a simple pendulum?

It is the one which can be considered to be a point mass suspended from a string or rod of negligible mass.

A pendulum is a weight suspended from a pivot so that it can swing freely.

Here,

A certain frictionless simple pendulum having a length l and mass m

mass of pendulum = m

length of the pendulum = l

The period of simple pendulum is:

T = 2\pi \sqrt{\frac{l}{g} }

Where k is the constant.

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m' = 2m

l' = 2l

T' = 2\pi \sqrt{\frac{2l}{g} }

T' = \sqrt{2}* 2\pi \sqrt{\frac{l}{g} }

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If l and m both are doubled then the period becomes √2*T

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2 years ago
A 1300-N crate rests on the floor. How much work is required to move it at constant speed (a)
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a) The work done is 920 J

b) The work done is 5200 J

Explanation:

a)

In this first part of the problem, the crate is moved horizontally at constant speed.

The work required in this case is given by

W=Fd cos \theta

where

F is the magnitude of the force applied

d is the displacement of the crate

\theta is the angle between the direction of the force and of the displacement

Here the crate is moved at constant speed: this means that the acceleration of the crate is zero, and so according to Newton's second law, the net force on the crate is zero: this means that the force applied, F, must be equal to the force of friction (but in opposite direction), so

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In this case, the crate is moved vertically. The force that must be applied to lift the crate must be equal to the weight of the crate (in order to move it a constant speed), therefore

F = W = 1300 N

The displacement this time is again

d = 4.0 m

And the angle is \theta=0^{\circ}, since the force is applied vertically, and the crate is moved also vertically. Therefore, the work done on the crate this time is

W=(1300)(4.0)(cos 0^{\circ})=5200 J

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