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Mrac [35]
3 years ago
8

Onur drops a basketball from a height of 10m on Mars, where the acceleration due to gravity has magnitude of 3.7 m/s^2. We want

to know how many seconds the basketball is in the air before it hits the ground. We can ignore air resistance.
Physics
2 answers:
stepladder [879]3 years ago
5 0

Answer:

t = 2.32 s

Explanation:

Applying the equation of motion;

d = ut + 0.5gt^2

Where;

d = distance travelled

u = initial velocity

g = acceleration due to gravity

t = time taken

Since the object was dropped;

u = 0

Then,

d = 0.5gt^2

t^2 = d/0.5g

t = √(d/0.5g) .......1

Given

g = 3.7 m/s^2

d = 10 m

Substituting the values;

t = √(10/(0.5×3.7))

t = 2.32 s

denpristay [2]3 years ago
4 0

Answer:

2.32 s

Explanation:

Using the equation of motion,

s = ut+g't²/2............................ Equation 1

Where s = distance, u = initial velocity, g' = acceleration due to gravity of  the moon, t = time.

Note: Since Onur drops the basket ball from a height, u = 0 m/s

Then,

s = g't²/2

make t the subject of the equation,

t = √(2s/g')...................... Equation 2

Given: s = 10 m, g' = 3.7 m/s²

Substitute this value into equation 2

t = √(2×10/3.7)

t = √(20/3.7)

t = √(5.405)

t = 2.32 s.

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C. The Densities are equal.

<h3>What is density?</h3>

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If m1, V1 and D1 = mass, volume  and density respectively of ball C

m2, V2 and D2 = mass, volume and density respectively of ball D

According to the Question ,

V_{1} = 3V_{2}  , m_{2}  = \frac{1}{3} (m_{1} ) \\ \\= m_{1} = 3m_{2}

Therefore,

\frac{D_{1} }{D_{2} }  = (\frac{m_{1} }{V_{1} } )* (\frac{m_{2} }{V_{2} } )\\ \\= (\frac{3m_{2} }{3V_{2} })*(\frac{V_{2} }{m_{2} }) \\\\= 1

Hence, D1 = D2

Learn more about density here:brainly.com/question/15164682

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