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3 years ago

A book, that has a mass of 0.5 grams, is pushed across a table with a force of 20 newtons. What is the acceleration of the book?

Physics

1 answer:

Bas_tet [7]3 years ago

6 0

Answer:

$4\cdot 10^4 m/s^2$

Explanation:

The acceleration of an object is given by Newton's second law:

$a=\frac{F}{m}$

where

F is the net force applied on the object

m is the mass of the object

For the book in the problem, we have:

$m=0.5 g =5\cdot 10^{-4} kg$ is the mass

$F=20 N$ is the force applied

Substituting into the formula, we find the acceleration:

$a=\frac{20 N}{5\cdot 10^{-4} kg}=4\cdot 10^4 m/s^2$

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What is the IMA of the following pulley system?

Furkat [3]

Answer: $IMA= \frac {F_r}{F_e}$

Explanation:

IMA stands for Ideal Mechanical advantage.

The IMA of pulley system can be defined as the ratio of output force to input force.

From the given pulley system,

The input force = $F_e$

The output force = $F_r$

Hence, IMA of the given pulley system,

$IMA= \frac {F_r}{F_e}$

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3 years ago

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Interactive Solution 28.5 illustrates one way to model this problem. A 7.11-kg object oscillates back and forth at the end of a

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Explanation:

Given that,

Mass of the object, m = 7.11 kg

Spring constant of the spring, k = 61.6 N/m

Speed of the observer, $v=2.79\times 10^8\ m/s$

We need to find the time period of oscillation observed by the observed. The time period of oscillation is given by :

$t_o=2\pi \sqrt{\dfrac{m}{k}} \\\\t_o=2\pi \sqrt{\dfrac{7.11}{61.6}} \\\\t_o=2.13\ s$

Time period of oscillation measured by the observer is :

$t=\dfrac{t_o}{\sqrt{1-\dfrac{v^2}{c^2}} }\\\\t=\dfrac{2.13}{\sqrt{1-\dfrac{(2.79\times 10^8)^2}{(3\times 10^8)^2}} }\\\\t=5.79\ s$

So, the time period of oscillation measured by the observer is 5.79 seconds.

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3 years ago

Which of the following would be a good question that could be scientifically investigated?

Nataliya [291]

Do cherry popsicles freeze slower than orange popsicles

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3 years ago

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In your research lab, a very thin, flat piece of glass with refractive index 1.50 and uniform thickness covers the opening of a

lara31 [8.8K]

Answer:

The longest wavelength in vacuum for which there is constructive interference for the reflected light, λ = 3472.

Explanation:

Refractive index of Glass (given) = 1.5

For the case of a constructive interference,

2nt = (m + 1/2) λ

For case 1,

2nt = (m + 1/2) 496 nm

For case 2,

2nt = (m +1+ 1/2) 386 nm

2nt = (m+3/2) * 386 nm

(m + 1/2) 496 nm = (m+3/2) * 386 nm

m = 3

Inserting the value of m in 1.

2nt = (m + 1/2) 496 nm

2*1.5t = (3 + 1/2) * 496 nm

t = ((3 + 1/2) * 496 nm)/ 3

t = 578.6 nm

The thickness of the glass, t = 578.6 nm

It is generally known that for constructive interference,

2nt = (m + 1/2) λ

λ = 2nt / ((m + 1/2))

For Longest Wavelength, m = 0

λ = 2*1.5*578.6/ (1/2)

λ = 3472 nm

3 0

3 years ago

A car travels along a straight line at a constant speed of 53.0 mi/h for a distance d and then another distance d in the same di

Shalnov [3]

A distance of d is covered with 53 mile/hr initially. Time taken to cover this distance t1 = d/53 hour Next distance of d is covered with x mile hours. Time taken to cover this distance t2 = d/x hours. We have average speed = 26.5 mile / hour

= Total distance traveled/ total time taken

= $\frac{2d}{\frac{d}{53}+\frac{d}{x}} = \frac{2}{\frac{1}{53}+\frac{1}{x} } = \frac{106x}{x+53}$

$26.5 = \frac{106x}{x+53} \\ \\ 79.5 x = 1404.5\\ \\ x = 17.67 miles/hour$

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3 years ago