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gavmur [86]
3 years ago
12

A stone is thrown in the upwards direction at the velocity of 4 m/s. It attains a certain height and then it falls back. During

its motion, it experienced an acceleration of 10 m/s2 in the downward direction. What is the maximum height it attained before falling back?
Physics
1 answer:
Sati [7]3 years ago
3 0

Answer: 0.8 m

Explanation:

Velocity of throw = 4m/s

Maximum Height attained(h) =?

Downward acceleration experienced = 10m/s^2

Using the relation:

v^2 = u^2 + 2aS

v = final Velocity = 0 (at maximum height)

u = Initial Velocity = 4

a = g downward acceleration = - 10

0 = 4^2 + 2(-10)(S)

0 = 16 - 20S

20S = 16

S = 16 / 20

S = 0.8m

Maximum Height attained = 0.8m

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A light horizontal spring has a spring constant of 105 N/m. A 2.00 kg block is pressed against one end of the spring, compressin
Savatey [412]

Answer:

The the coefficient of kinetic friction between the horizontal surface and the block is 0.536.

Explanation:

Given that,

Spring constant = 105 N/m

Mass of block = 2.00 kg

Compress = 0.100 m

Distance = 0.250 m

We need to calculate the coefficient of kinetic friction

Using relation between friction force and restoring force

kx=\mu mg

Put the value into the relation

105\times0.1=\mu\times2.00\times9.8

\mu=\dfrac{105\times0.1}{2.00\times9.8}

\mu=0.536

Hence, The the coefficient of kinetic friction between the horizontal surface and the block is 0.536.

8 0
3 years ago
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A car moves at a speed of 50 kilometers/hour. Its kinetic energy is 400 joules. If the same car moves at a speed of 100 kilomete
HACTEHA [7]

Explanation :

Speed of car, v_1=50\ km/h=13.8\ m/s

Kinetic energy of the car, KE_1=400\ J

Speed of car, v_2=100\ km/h=27.7\ m/s

Let KE_2 is the kinetic energy of the car when it is moving with 100 km/h.

\dfrac{KE_1}{KE_2}=\dfrac{\dfrac{1}{2}mv_1^2}{\dfrac{1}{2}mv_2^2}

\dfrac{KE_1}{KE_2}=\dfrac{v_1^2}{v_2^2}

KE_2=KE_1(\dfrac{v_2}{v_1})^2

KE_2=400\ J\times (\dfrac{27.7\ m/s}{13.8\ m/s})^2

KE_2=1611.6\ J  

Initial velocity of both cars are 0. Using third equation of motion :

So, S_1=\dfrac{v_1^2}{2a}=\dfrac{v_1t}{2}=\dfrac{v_1t}{2}

and S_2=\dfrac{v_2^2}{2a}=\dfrac{v_2t}{2}=\dfrac{v_2t}{2}

t is same. So,

\dfrac{S_1}{S_2}=\dfrac{50\ m/s}{100\ m/s}

\dfrac{S_1}{S_2}=\dfrac{1}{2}

S_2=2\ S_1

So, the braking distance at the faster speed is twice the braking distance at the slower speed.

4 0
4 years ago
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A human hair is approximately is 50um in diameter. express this diameter in meters
stealth61 [152]

50 μm = 5 × 10⁻⁵ m  in diameter

<h3>Steps</h3>

1μm = 1 x 10^{-6}m

<h3>Given</h3>

1μm = 1 x 10^{-6}m

50μm = 50 x 10^{-6}m

50μm =  50 x 10 x 10^{-6}

50μm = 5 x 10^{-5}m

<h3>Conclusion</h3>

A human hair is approximately 50 μm = 5 × 10⁻⁵ m  in diameter

<h3>How big is a human hair on average?</h3>

between 17 and 181 micrometers

The typical diameter of human hair can range from 17 micrometers to 181 micrometers, according to research. Using 50 micrometers as an average figure for its diameter, this translates to 50,000 nanometers.

<h3>What is a human hair's diameter in inches?</h3>

The finest hair is flaxen, with a diameter ranging from 1/1500 to 1/500 of an inch. the coarsest hair is black, measuring between 1/450 and 1/140 of an inch.

learn more about human hair diameter here

<u>brainly.com/question/13147893</u>

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6 0
1 year ago
Which of the following microscope parts should routinely be adjusted to control the light source and provide optimal illuminatio
yKpoI14uk [10]
Here is the answer. The microscope parts that should routinely be adjusted to control the light source and provide optimal illumination of the specimen are the following: <span>light source; condenser; specimen; objective lens; ocular lens. Hope this answers your question.</span>
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Which of the following processes occurs in the cells of plants?
Gnom [1K]

Answer:

A., B. & C.

Explanation:

All apply, because all these happen in a plant.

7 0
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