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3 years ago

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If an object on a horizontal, frictionless surface is attached to a spring, displaced, and then released, it will oscillate. If

it is displaced 0.120 m from its equilibrium position and released with zero initial speed, then after 0.800 s its displacement is found to be 0.120 m on the opposite side, and it has passed the equilibrium position once during this interval. Find (a) the amplitude; (b) the period; (c) the frequency. 14.3 . The tip of a tuning fork go

1 answer:

Rama09 [41]3 years ago

3 0

Answer:

A) 0.120

B) 1.6s

C) 0.625 Hz

Explanation:

Here, X = 0.120 m

x = 0.120 m after t = 0.800 s

A- Amplitude = max displacement from equillibrium position = 0.120 m

B- Period = 2 * 0.800 = 1.6 s

C- Frequency = 1/peroid = 1/1.6

f = 0.625 Hz

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light waves are first transmitted through the ________ at the front of the eye and enter an opening called the ________ before s

viva [34]

The transmission of light waves is usually done through cornea of the eyes, then move through another opening which is regarded as pupil before it will get to the retina.

Light waves can be regarded as moving energy which contains microscopic particles known as photons.

The vision of the eye can be completed through the light wave passing through the components of the eyes and this process goes thus;

Light will move through the (cornea) which is situated at the front area of the eyes into lens.

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2 years ago

castortr0y [4]

Answer: $Q=3000 cal$

Explanation:

We are given the following formula:

$Q=m. c. \Delta T$ (1)

Where:

$Q=3000 cal$ is the amount of heat

$m=300g$ is the mass of water

$c=1 cal/g \°C$ is the specific heat of water

$\Delta T$ is the variation in temperature, which in this case is $\Delta T=30\°C-20\°C=10\°C$

Rewriting equation (1) with the known values at the right side, we will prove the result is $3000 cal$ :

$Q=(300g)(1 cal/g \°C)(10\°C)$ (2)

$Q=3000 cal$ This is the result

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2 years ago

This instrument can used to visualize atoms

Lisa [10]

Answer:

can someone follow me

Explanation:

Plqae

3 0

2 years ago

Read 2 more answers

(3) What is the weight of a 50-kg astronaut (a) on Earth (b) On the Moon ,(g=1.7m/s2), (c) on Mars (g=3.7m/s2) (d)in outer space

artcher [175]

Answer:

a) On Earth

490N

b) On the Moon

85N

c) On Mars

185N

d)in outer space traveling with constant velocity.

0

Explanation:

The weight is defined as:

$W = mg$ (1)

Where m is the mass and g is the gravity

a) On Earth $g = 9.8m/s^{2}$

Then, equation 1 can be used:

$W = (50Kg)(9.8m/s^{2})$

$W = 490Kg.m/s^{2}$

but 1N = Kg.m/s^{2}

$W = 490N$

Hence, the weight of the astronaut on Earth is $490N$

b) On the Moon $g = 1.7m/s^{2}$

$W = (50Kg)(1.7m/s^{2})$

$W = 85N$

Hence, the weight of the astronaut on the Moon is $85N$

c) On Mars $g = 3.7m/s^{2}$

$W = (50Kg)(3.7m/s^{2})$

$W = 185N$

Hence, the weight of the astronaut on Mars is $185N$

(d) in outer space traveling with constant velocity.

Tanking into consideration that the astronaut is traveling in outer space at a constant velocity, it can be concluded that the acceleration will be zero.

Remember that the acceleration is defined as:

$a = \frac{v_{f} - v_{i}}{t}$

Since the acceleration is the variation of the velocity in a unit of time.

Therefore, from equation 1 is gotten.

$W = (50kg)(0)$

Remember that g is the acceleration that a body experience as a consequence of the gravitational field.

$W = 0$

5 0

2 years ago

Help me please!

MrMuchimi

I think it’s A
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2 years ago