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RoseWind [281]
3 years ago
5

If two objects are moving at the same speed, and Object 1 has four times as much mass as Object 2, how much momentum does Object

1 have compared to Object 2?
1. Twice as much
2. One-fourth as much
3. The same amount
4. Four times as much
Physics
1 answer:
Margaret [11]3 years ago
7 0
The answer is  number 1
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What is the action force and the reaction force when you sit down on a chair
Harlamova29_29 [7]
Your weight pushing down on the chair is the action force. The reaction force is the force exerted by the chair that pushes up on your body.
3 0
3 years ago
Read 2 more answers
What's the value of 1,152 Btu in joules? A. 1,964,445 J B. 2,485,664 J C. 987,875 J D. 1,215,360 J
OLga [1]

Answer: Option (d) is correct.

Explanation:

Given,  1,152 British thermal units

1 British thermal unit = 1055.06 joules

So, in 1,152 British thermal units there will be :

1152\times 1055.06 Joules=1215429.12 Joules=1.21542912\times 10^{8} Joules

Hence, from the given options the closest answer is of option (d). So, option (d) is correct.

4 0
3 years ago
Read 2 more answers
Suppose an automobile has 2000-joules of kinetic energy. when it moves at twice the speed, what will be its kinetic energy? what
NeX [460]

Answer:

K.Eₓ = 4 K.E

K.Eₓ = 9 K.E

Explanation:

Th formula for the kinetic energy of a body is given as follows:

K.E = \frac{1}{2}mv^2\\   ---------------equation (1)

where,

K.E = Kinetic Energy of Automobile

m = mass of automobile

v = speed of automobile

For twice speed:

vₓ = 2v

then,

K.E_{x} = \frac{1}{2}mv_{x}^2\\K.E_{x} = \frac{1}{2}m(2v)^2\\K.E_{x} = 4\frac{1}{2}mv^2\\

using equation (1):

<u>K.Eₓ = 4 K.E</u>

For thrice speed:

vₓ = 3v

then,

K.E_{x} = \frac{1}{2}mv_{x}^2\\K.E_{x} = \frac{1}{2}m(3v)^2\\K.E_{x} = 9\frac{1}{2}mv^2\\

using equation (1):

<u>K.Eₓ = 9 K.E</u>

6 0
3 years ago
Help!!! If anyone could do one of these, i'm confused on how I should write the equation down.
balu736 [363]

F=ma=m(change in velocity/change in time)

Number 1

F=ma

F=55kg(1.1ms^-1/1.6s)=37.8N

Number 2

F=ma

F=0.440kg(10ms^-1/0.02s)=220N

Number 3

F=ma

F=1400kg(15ms^-1/0.73s)=2.88*10^3N or 28,767N

Any questions please feel free to ask.

4 0
3 years ago
Determine the distance between a newly discovered planet and its single moon if the orbital period of the moon is 1.2 Earth days
Vinil7 [7]

Answer:

The distance is r = 55430496 \  m  

Explanation:

From the question we are told that

   The period of the moon T =  1.2 days = 1.2 * 24 * 3600 = 103680 \  s

    The mass of the planet is  m_p =  9.38*10^{24} kg

Generally the period of the moon is mathematically represented as

       T  =  2 *  \pi * \sqrt{ \frac{r^3 }{ G * m_p } }

Here G is the gravitational constant with value

        G  =  6.67 *10^{-11} \  N \cdot m^2/kg^2

=>   T  =  2 *  \pi * \sqrt{ \frac{r^3 }{ G * m_p } }

=>   103680   =  2 *  3.142  * \sqrt{ \frac{r^3 }{ 6.67*10^{-11} * 9.38*10^{24} } }

=>    272218492.31 = \frac{r^3}{ 6.67 *10^{-11} * 9.38*10^{24}}

=>    r = \sqrt[3]{ 1.7031241*10^{23}}j

=>   r = 55430496 \  m        

8 0
3 years ago
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