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babymother [125]
2 years ago
5

a ball is thrown striaght up in the air and then falls back to earth. if the downward fall takes 2.2s, how fast is the ball trav

eling when it striker the ground
Physics
1 answer:
lapo4ka [179]2 years ago
5 0

The velocity of the ball when it strikes the ground, given the data is 21.56 m/s

<h3>Data obtained from the question</h3>

From the question given above, the following data were obtained:

  • Time to reach ground from maximum height (t) = 2.2 s
  • Initial velocity (u) = 0 m/s
  • Acceleration due to gravity (g) = 9.8 m/s²
  • Final velocity (v) =?

<h3>How to determine the velocity when the ball strikes the ground</h3>

The velocity of the ball when it strikes the ground can be obtained as illustrated below:

v = u + gt

v = 0 + (9.8 × 2.2)

v = 0 + 21.56

v = 21.56 m/s

Thus, the velocity of the ball when it strikes the ground is 21.56 m/s

Learn more about motion under gravity:

brainly.com/question/22719691

#SPJ1

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Answer:

This is because normal force is exerted perpendicularly to the point of contact between the upper and lower objects.  

Explanation:

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4 0
3 years ago
Sugarcane is an example of a plant that can be used to produce which kind of fuel
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Which layer of the atmosphere is the top layer of the thermosphere?
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8 0
4 years ago
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A long, solid rod 5.3 cm in radius carries a uniform volume charge density. Part A If the electric field strength at the surface
Blizzard [7]

Answer:

\rho = 7.35\times \muC/m^3

Explanation:

given,

Radius of the solid rod, R = 5.3 cm

Electric field strength,E = 22 kN/C

Let the volume charge density be ρ

From Gauss law

  E = \dfrac{Rl}{2\epsilon_0}

 ε₀ is the permitivity of free space

  R is the radius of the rod

 and also,

\rho=\dfrac{2E\epsilon_0}{R}

ρ is the volume charge density

\rho=\dfrac{2\times 22\times 10^{3}\times 8.854\times 10^{-12}}{0.053}

\rho = 7.35\times 10^{-6}\ C/m^3

\rho = 7.35\times \muC/m^3

Hence, the volume charge density is equal to \rho = 7.35\times \muC/m^3

6 0
3 years ago
Suppose a tidal basin is 6 m above the ocean at low tide and that the area of the basin is 5×107 m2. Estimate the gravitational
Neko [114]

Answer:

Potential energy will be 176.58\times 10^{11}j          

Explanation:

We have given the height of the basin is h = 6 m

Area of the basin A=5\times 10^7m^2

Volume V=area \times height=5\times 10^7\times 6=30\times 10^7m^3

Density \rho =1000kg/m^3

We know that mass is given by m=\rho V=1000\times 30\times 10^7=3\times 10^{11}kg

We know that potential energy is given by E=mgh=3\times 10^{11}\times 9.81\times 6=176.58\times 10^{11}j

3 0
3 years ago
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