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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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slamgirl [31]

Answer:

-30° C

Explanation:

Data provided in the problem:

The formula for conversion as:

F  = (9/5)C + 32

Now,

for the values of F = -22 , C = ?

Substituting the value of F in the above formula, we get

-22 = (9/5)C + 32

or

-22 - 32 = (9/5)C

or

(9/5)C = - 54

or

C = - 54 × (5/9)

or

C = - 30 °

Hence, -22 Fahrenheit equals to -30°C

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Answer:

Magnetic field experienced = 4.5 × 10⁻⁴ T

Explanation:

The magnetic field around an infinite straight current-carrying wire at a distance r from the wire is given by

B = (μ₀I)/(2πr)

B = ?

I = 20 KA = 20000 A

r = 8.9 m

μ₀ = magnetic permeability = 1.257 × 10⁻⁶ T.m/A

B = (1.257 × 10⁻⁶ × 20000)/(2π×8.9) = 4.5 × 10⁻⁴ T

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A cannonball is fired on flat ground
algol [13]

hmax = 5740.48 m. The maximum height that a cannonball fired at 420 m/s at a 53.0° angles is 5740.48m.

This is an example of parabolic launch. A cannonball is fired on flat ground at 420 m/s at a 53.0° angle and we have to calculate the maximum height that it reach.

V₀ = 420m/s and θ₀ = 53.0°

So, when the cannonball is fired it has horizontal and vertical components:

V₀ₓ = V₀ cos θ₀ = (420m/s)(cos 53°) = 252.76 m/s

V₀y = V₀ cos θ₀ = (420m/s)(cos 53°) = 335.43m/s

When the cannoball reach the maximum height the vertical velocity component is zero, that happens in a tₐ time:

Vy = V₀y - g tₐ = 0

tₐ = V₀y/g

tₐ = (335.43m/s)/(9.8m/s²) = 34.23s

Then, the maximum height is reached in the instant tₐ = 34.23s:

h = V₀y tₐ - 1/2g tₐ²

hmax = (335.43m/s)(34.23s)-1/2(9.8m/s²)(34.23s)²

hmax = 11481.77m - 5741.29m

hmax = 5740.48m

3 0
3 years ago
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