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2 years ago

6

A mass is set oscillating horizontally, with an amplitude of 11.3 cm and a period of 0.702 sec. What is the velocity of the part

icle when the position is 1/2 the amplitude in the positive x direction and moving toward x=0?

1 answer:

Zinaida [17]2 years ago

6 0

Answer:

Explanation:

18.1

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What name is given to the wall of water that makes landfall just ahead of a hurricane?

andreev551 [17]

Pretty sure its a. storm surge

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3 years ago

This object is located 13.0 cm to the left of the lens and the image forms at 20.8 cm to the right of the lens.

Alina [70]

Answer:

0.125 cm

Explanation:

1/f = 1/d¡ + 1/d。

Find the focal point

(13.0^-1 + 20.8^-1) = 0.125 m

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2 years ago

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A force of 10 N acts on an object with a mass of 10 kg. What is its acceleration? A. 1 m/s2 B. 10 m/s2 C. 2 m/s2 D. 5 m/s2

Akimi4 [234]

The acceleration exerted by the object of mass 10 kg is $\mathbf{1} m / \boldsymbol{s}^{2}$

Answer: Option A

<u>Explanation:</u>

According to Newton’s second law of motion, any external force acting on a body will be directly proportional to the mass of the body as well as acceleration exerted by the body. So, the net external force acting on any object will be equal to the product of mass of the object with acceleration exerted by the object. Thus,

$Force = Mass \times Acceleration$

So,

$Acceleration=\frac{\text {Force}}{\text {Mass}}$

As the force acting on the object is stated as 10 N and the mass of the object is given as 10 kg, then the acceleration will be

$Acceleration =\frac{10 \mathrm{N}}{10 \mathrm{kg}}=1 \mathrm{m} / \mathrm{s}^{2}$

So, the acceleration exerted by the object of mass 10 kg is $\mathbf{1} m / \boldsymbol{s}^{2}$

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2 years ago

A plane accelerates from rest at a constant rate of 5.00 m/s2 along a runway that is 1800 m long. Assume that the plane reaches

tiny-mole [99]

Answer:

26.8 seconds

Explanation:

To solve this problem we have to use 2 kinematics equations: *I can't use subscripts for some reason on here so I am going to use these variables:

v = final velocity

z = initial velocity

x = distance

t = time

a = acceleration

${v}^{2} = {z}^{2} + 2ax$

$v = z + at$

First let's find the final velocity the plane will have at the end of the runway using the first equation:

${v}^{2} = {0}^{2} + 2(5)(1800)$

$v = 60 \sqrt{5}$

Now we can plug this into the second equation to find t:

$60 \sqrt{5} = 0 + 5t$

$t = 12 \sqrt{5}$

Then using 3 significant figures we round to 26.8 seconds

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The statement which is true of a wave that’s propagating along the pavement and girders of a suspension bridge is A. The wave is mechanical, with particles vibrating in a direction that is parallel to that of the wave, forming compressions and rarefactions.

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