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solong [7]
3 years ago
8

An ostrich can run at a speed of 43 mi/hr. How much ground can an ostrich cover if it runs at this speed for 15 minutes? (Hint:

15 minutes = 0.25 hours)
Physics
1 answer:
Simora [160]3 years ago
4 0
..... It would possibly she eenejjsjejeej 1.4
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Help? its due in like 12 minutes lolzz
Gnom [1K]

Answer:

Question 1: the plates are moving toward one another.

Question 2: The Himalayan Mountains in India

Question 3: Because mountains are formed instead.

Explanation:

The paragraph explains that the plates continue to move closer to one another while forming multiple mountains.

The paragraph explains, " a well-known example of this is the formation of the Himalayan Mountains in India,"

The area of the Himalayan Mountains are better suited for the formation of mountains rather than volcanoes.

Have a nice day!! Good Luck!! Brainliest would be appreciated!!!

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3 years ago
Select the correct answer.
soldi70 [24.7K]

Answer:

OD. The process of cutting down irrelevant information so only the information that is useful for particular purpose remains

6 0
3 years ago
A motor does 8000j of work in 20 seconds. What is the power of the motor
lozanna [386]
Power = work / time = 8000J / 20s = 400W

5 0
3 years ago
if you have 34 pounds of meat at 200F and you immerse it in 15 liters of water at 25C what is the final temp of the water? ( spe
hram777 [196]

Answer:

The specific heat capacity is the heat or energy required to change one unit mass of a substance of a constant volume by 1 °C. The formula is Cv = Q / (ΔT ⨉ m)

5 0
3 years ago
A 120-kg object and a 420-kg object are separated by 3.00 m At what position (other than an infinitely remote one) can the 51.0-
djverab [1.8K]

Answer:

1.045 m from 120 kg

Explanation:

m1 = 120 kg

m2 = 420 kg

m = 51 kg

d = 3 m

Let m is placed at a distance y from 120 kg so that the net force on 51 kg is zero.

By use of the gravitational force

Force on m due to m1 is equal to the force on m due to m2.

\frac{Gm_{1}m}{y^{2}}=\frac{Gm_{2}m}{\left ( d-y \right )^{2}}

\frac{m_{1}}{y^{2}}=\frac{m_{2}}{\left ( d-y \right )^{2}}

\frac{3-y}{y}=\sqrt{\frac{7}{2}}

3 - y = 1.87 y

3 = 2.87 y

y = 1.045 m

Thus, the net force on 51 kg is zero if it is placed at a distance of 1.045 m from 120 kg.

6 0
3 years ago
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