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2 years ago

5

A physics book is moved once around the perimeter of a table of dimensions 1 m by 3 m. If the book ends up at its initial positi

on, what is the distance traveled

1 answer:

mamaluj [8]2 years ago

6 0

Answer: 8 m.

Explanation:

Use formula of perimeter of rectangle.

Perimeter = 2( l + b)

=2( 3 + 1 )

=8 m

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The gravitational force produce between any two object kept 2.5×10 to the power 4 km apart is 580N.At what distance should they

timofeeve [1]

Answer:

d = 3.54 x 10⁴ Km

Explanation:

Given,

The distance between the two objects, r = 2.5 x 10⁴ Km

The gravitational force between them, F = 580 N

The gravitational force between the two objects is given by the formula

F = GMm/r² newton

When the gravitational force becomes half, then the distance between them becomes

Let us multiply the above equation by 1/2 on both sides

( 1/2) F = (1/2) GMm/r²

= GMm/2r²

= GMm/(√2r)²

Therefore, the distance becomes √2d, when the gravitational force between them becomes half

d = √2r = √2 x 2.5 x 10⁴ Km

= 3.54 x 10⁴ Km

Hence, the two objects should be kept at a distance, d = 3.54 x 10⁴ Km so that the gravitational force becomes half.

3 0

2 years ago

Your town is installing a fountain in the main square. If the water is to rise 26.0 m (85.3 feet) above the fountain, how much p

Brums [2.3K]

Answer:

$P = 3.55 \times 10^5 Pa$

Explanation:

As we know that water from the fountain will raise to maximum height

$H = 26.0 m$

now by energy conservation we can say that initial speed of the water just after it moves out will be

$\frac{1}{2}mv^2 = mgH$

$v = \sqrt{2gH}$

$v = \sqrt{2(9.81)(26)}$

$v = 22.6 m/s$

Now we can use Bernuolli's theorem to find the initial pressure inside the pipe

$P = P_0 + \frac{1}{2}\rho v^2$

$P = 10^5 + \frac{1}{2}(1000)(22.6^2)$

$P = 3.55 \times 10^5 Pa$

6 0

2 years ago

A 5kg bucket hangs from a ceiling on a rope. A student attaches a spring scale to the buckets handle and pulls horizontally on t

7nadin3 [17]

I don’t know what the angle is in your diagram so I used the angle from the vertical.

6 0

2 years ago

You are provided with three polarizers with filters making angles of (A) 90 ∘ , (B) 180 ∘ and (C) −45 ∘ with respect

irinina [24]

Answer:

Order of maximum transmission of the polarizer is A, C and B.

Solution:

As per the question:

For the first polarizer, the angle is quite insignificant:

(A) $90^{\circ}$ :

The light intensity after passing through the first polarizer is $I_{o}$ and this intensity does not depend on the angle of the polarizer.

Consider $90^{\circ}$ with the vertical, the intensity is given by:

$I = I_{o}cos^{2}90^{\circ}$

$I = I_{o}cos(2(45^{\circ})) = I_{o}(\frac{1+cos90^{\circ})}{2} = \frac{I_{o}}{2}$

(B) $180^{\circ}$ :

Suppose the second polarizer is $45^{\circ}$ with the vertical.

Now, intensity through the second polarizer:

$I' = Icos^{2}(\theta_{2} - \theta_{1}) = \frac{I_{o}}{2}cos^{2}(- 45 - 90)$

$I' = \frac{I_{o}}{2}cos^{2}135^{\circ} = \frac{I_{o}}{4}$

Now, if we consider the second polarizer to be $180^{\circ}$ ,

$I' = \frac{I_{o}}{2}cos^{2}180^{\circ} = \frac{I_{o}}{2}cos^{2}(180^{\circ} - 90^{\circ}) = 0$

(C) $- 45^{\circ}$ :

Now,

Intensity through the third polarizer, if it is $180^{\circ}$ with the vertical:

$I' = Icos^{2}(\theta_{2} - \theta_{1}) = \frac{I_{o}}{2}cos^{2}(180 - (- 45))$

$I' = \frac{I_{o}}{8}$

5 0

2 years ago

The Earth has mass ME and average radius RE. The Moon has mass MM and the average distance from the center of mass of the moon t

marusya05 [52]

Answer:

Moment of inertia of Earth about its own axis is given as

$I = 9.7 \times 10^{37} kg m^2$

Explanation:

Since Earth is considered as solid sphere

So we will have

$I = \frac{2}{5}M_eR_e^2$

so we will have

$I = \frac{2}{5}(5.97 \times 10^{24})(6.371 \times 10^6)^2$

so we have

$I = 9.7 \times 10^{37} kg m^2$

3 0

2 years ago