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What height will the object reach? 12 points. Will give brainliest.

umka21 [38]

Answer:

12.7 m

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 56.7 Km/hr

Maximum height (h) =..?

First, we shall convert 56.7 Km/hr to m/s. This can be obtained as follow:

Initial velocity (m/s) = 56.7 x 1000/3600

Initial velocity (m/s) = 15.75 m/s

Next, we shall determine the time taken to get to the maximum height. This can be obtained as follow:

Initial velocity (u) = 15.75 m/s

Final velocity (v) = 0 m/s

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

v = u – gt (since the ball is going against gravity)

0 = 15.75 – 9.8 × t

Rearrange

9.8 × t = 15.75

Divide both side by 9.8

t = 15.75/9.8

t = 1.61 secs.

Finally, we shall determine the maximum height as follow

h = ½gt²

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) = 1.61 secs.

Height (h) =..?

h = ½gt²

h = ½ × 9.8 × 1.61²

h = 4.9 x 1.61²

h = 12.7 m

Therefore, the maximum height reached by the ball is 12.7 m

3 0

3 years ago

Match each situation below to a letter on the illustration.

kari74 [83]

Answer:

What illustration

Explanation:

7 0

3 years ago

Tidal Forces near a Black Hole. An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black

arlik [135]

Answer:

$833.4801043*10^6N$ on ear that is closer to Black hole.

$13.83803929*10^6N$ On ear that is farther from Black hole.

Explanation:

This problem can be solved as two masses that are at two different location from a bigger mass whose gravity affects both.

tension is an equal and opposite force that is exerted in response to applied force.

so on ear that is closer to black hole would have tension that is equal in magnitude and opposite in direction to gravitational force that ear experience due to the black hole at that location.

this true for ear that is further away from black hole as well.

(1) Force on ear that is closer to black hole.

$F =\frac{m_{1}*m_{2} }{r^2} G$

$m_{1}= 5*1.989*10^30kg$ is mass of Black hole.

$m_{2}= 0.030kg$ is mass of ear that is close to black hole.

$G = 6.679*10^-11m^3*kg^-1*s^-2$

r = $120km-\frac{6}{100000} km=119.99994km=119999.94m$

Note, we have subtracted because ear is closer to black hole.

plugging all this in formula gives.

$F = 833.4801043*10^6N$

That is tension of ear.

(2) Force on ear that is further from black hole.

$F =\frac{m_{1}*m_{2} }{r^2} G$

$m_{1}= 5*1.989*10^30kg$ is mass of Black hole.

$m_{2}= 0.030kg$ is mass of ear that is close to black hole.

$G = 6.679*10^-11m^3*kg^-1*s^-2$

this time r is further away from black hole so it would be.

$r = 120km+\frac{6}{100000} km = 120000.06m$

Plugging this all in we get

$F = 13.83803929N$

and that is tension on ear that is further from black hole.

Notice the tension difference, and order of magnitude of tension,it is enormous .

this astronaut is lethally close to black hole.

5 0

3 years ago

The circumference of a sphere was measured to be

professor190 [17]

To solve this problem we will apply the concepts related to the calculation of the surface, volume and error through the differentiation of the formulas given for the calculation of these values in a circle. Our values given at the beginning are

$\phi = 76cm$