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KATRIN_1 [288]
3 years ago
5

Numerical Problems:

Physics
1 answer:
dangina [55]3 years ago
6 0
  • Displacement=1200m
  • Time=4min=4(60)=240s

\boxed{\sf Velocity=\dfrac{Displacement}{Time}}

\\ \sf\longmapsto Velocity=\dfrac{1200}{240}

\\ \sf\longmapsto Velocity=5m/s

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hammer [34]

Answer:

Element Is The Answer I think

6 0
2 years ago
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An amoeba has 1.00 x 1016 protons and a net charge of 0.300 pC. Assuming there are 1.88 x 106 fewer electrons than protons, If y
Xelga [282]

Answer:

The fraction of the protons would have no electrons =1.88\times 10^{-10}

Explanation:

We are given that

Amoeba has total number of protons=1.00\times 10^{16}

Net charge, Q=0.300pC

Electrons are fewer than protons=1.88\times 10^6

We have to find the fraction of protons would have no electrons.

The fraction of the protons would have no electrons

=\frac{Fewer\;electrons}{Total\;protons}

The fraction of the protons would have no electrons

=\frac{1.88\times 10^{6}}{1.00\times 10^{16}}

=1.88\times 10^{-10}

Hence, the fraction of the protons would have no electrons =1.88\times 10^{-10}

6 0
2 years ago
Please help me with my Physical Science! 50 POINTS
DaniilM [7]

Answer:

1, When Jane brakes, the brakes slow the car wheels turning and the road surface exerts a backwards force on the tires, causing the car to decelerate. The pocket book tends to continue on in a straight line (Newton's first law). If she brakes hard enough that the friction between the book and the car seat is insufficient to decelerate the book as fast as the car is decelerating, the book will slide off the seat, and gravity pulls it to the floor

2.

When the diver uses his / her force to depress the springboard, the springboard pushes him back with equal force

3.Newton's Second Law (F=ma)

4. 5 N

5. 19.5 N

65kg * 0.3 m/s^2

6.0.2 N/s

10kg divided by 2N

7.-Walking then pushing the moving forward

-Dribbling

-Basketball is pushed but bounces back

Explanation:

6 0
3 years ago
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Planets move around the sun in elliptical orbits. True. False
astra-53 [7]

this answer is true.

5 0
3 years ago
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A paperweight consists of a 8.55-cm-thick plastic cube. Within the plastic a thin sheet of paper is embedded, parallel to opposi
ella [17]

Answer:

\mu=1.5322

Explanation:

Given:

Thickness of the paperweight cube, x=8.55\ cm

apparent depth from one side of the inbuilt paper in the plastic cube, i=4.02\ cm

apparent depth from the other side of the inbuilt paper in the plastic cube, i'=1.56\ cm

Now as we know that refractive index is given as:

\rm \mu=\frac{real\ depth}{apparent\ depth}

  • Let the real depth form first side of the slab be, d
  • Then the depth from the second side of the slab will be, d'=x-d=8.55-d

Since refractive index for an amorphous solid is an isotropic quantity so it remains same in all the direction for this plastic.

\mu=\mu'

\frac{d}{i} =\frac{d'}{i'}

\frac{d}{4.02}=\frac{8.55-d}{1.56}

d=6.1597\ cm

Now the refractive index:

\mu=\frac{d}{i}

\mu=\frac{6.1596}{4.02}

\mu=1.5322

7 0
3 years ago
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