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Ne4ueva [31]
3 years ago
7

I need help plz!!! Please help...

Physics
1 answer:
RSB [31]3 years ago
4 0

Answer: 60

Explanation:

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A student grabs a piece of paper and tears it from the middle. Which statement best explains the forces involved?
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Answer:

b. The normal force between the molecules of the paper is overcome by the contact force of the hands.

Explanation:

The paper molecules are held together by a weak bond. When the student holds the paper on both sides with the center of the paper in between, the student applies two equal forces in the opposite direction of the paper making the paper molecules weaken and separate.

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An apple dropped from the branch of a tree hits the ground in 0.5 s. If the acceleration of the apple during its motion is 10 ms
Ipatiy [6.2K]

Given that,

Time = 0.5 s

Acceleration = 10 m/s²

(I). We need to calculate the speed of apple

Using equation of motion

v=u+at

Where, v = speed

u = initial speed

a = acceleration

t = time

Put the value into the formula

v=0+10\times0.5

v=5\ m/s

(III). We need to calculate the height of the branch of the tree from the ground

Using equation of motion

s=ut+\dfrac{1}{2}gt^2

Put the value into the formula

s=0+\dfrac{1}{2}\times10\times(0.5)^2

s=1.25\ m

(II). We need to calculate the average velocity during 0.5 sec

Using formula of average velocity

v_{avg}=\dfrac{\Delta x}{\Delta t}

v_{avg}=\dfrac{x_{f}-x_{i}}{t_{f}-t_{0}}

Where, x_{f}= final position

x_{i} = initial position

Put the value into the formula

v_{avg}=\dfrac{1.25+0}{0.5}

v_{avg}=2.5\ m/s

Hence, (I). The speed of apple is 5 m/s.

(II). The average velocity during 0.5 sec is 2.5 m/s

(III). The height of the branch of the tree from the ground is 1.25 m.

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3 years ago
what would happen to the size of the shadow if the distance between the light and the hand is increased.
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A parallel-plate capacitor consists of plates of area 1.5 x 10^-4 m^2 separated by 2.0 mm The capacitor is connected to a 12-V b
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Answer:

4.78 x 10^-11 J

Explanation:

A = 1.5 x 10^-4 m^2

d = 2 mm = 2 x 10^-3 m

V = 12 V

Let C be the capacitance of the capacitor

C = ε0 A / d

C = (8.854 x 10^-12 x 1.5 x 10^-4) / (2 x 10^-3)

C = 6.64 x 10^-13 F

Energy stored, U = 1/2 CV^2

U = 0.5 x 6.64 x 10^-13 x 12 x 12

U = 4.78 x 10^-11 J

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Answer:

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thank you

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