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natali 33 [55]
2 years ago
4

A lion with a mass of 190 kg is chasing a gazelle with a mass of 15 kg. The distance between the lion and the gazelle is 2 meter

s.
a. How much gravitational force does the lion exert on the gazelle?
b. How much gravitational force does the gazelle exert on the lion?
Physics
1 answer:
Ahat [919]2 years ago
8 0

(a) 4.7\cdot 10^{-8} N

The gravitational force between two objects is given by:

F=G\frac{m_1 m_2}{r^2}

where

G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2} is the gravitational constant

m1 and m2 are the two masses

r is the separation between the objects

In this problem, we have:

m_1 = 190 kg is the mass of the lion

m_2 = 15 kg is the mass of the gazelle

r = 2 m is their separation

Substituting into the equation, we find the gravitational force exerted by the lion on the gazelle:

F=(6.67\cdot 10^{-11})\frac{(190)(15)}{2^2}=4.7\cdot 10^{-8} N

(b) 4.7\cdot 10^{-8} N

According to Newton's third law of motion:

"When an object A exerts a force (action) on an object B, object B exerts an equal and opposite force (reaction) on object A".

In this problem, we can identify:

- The lion as object A

- The gazelle as object B

This means that according to the abovementioned law, the gravitational force exerted by the lion on the gazelle is equal to the gravitational force exerted by the gazelle on the lion. Therefore, the answer is again

4.7\cdot 10^{-8} N

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Explanation:

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Answer : The correct option is, (d) 535^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

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m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of copper = 0.10cal/g^oC

c_2 = specific heat of water = 1.00cal/g^oC

m_1 = mass of copper = 120 g

m_2 = mass of water = 300 g

T_f = final temperature of mixture = 35^oC

T_1 = initial temperature of copper = ?

T_2 = initial temperature of water = 15^oC  

Now put all the given values in the above formula, we get:

120g\times 0.10cal/g^oC\times (35-T_1)^oC=-300g\times 1.00cal/g^oC\times (35-15)^oC

T_1=535^oC

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Answer:

ω2  =  216.47 rad/s

Explanation:

given data

radius r1 =  460 mm

radius r2 = 46 mm

ω =  32k rad/s

solution

we know here that power generated by roller that  is

power = T. ω    ..............1

power = F × r × ω

and this force of roller on cylinder is equal and opposite force apply by roller

so power transfer equal in every cylinder so

( F × r1 × ω1)  ÷ 2 = (  F × r2 × ω2 )  ÷  2    ................2

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Answer:

Explanation:

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According to the diagram

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