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Sav [38]
3 years ago
15

What is the SI unit for momentum?

Physics
2 answers:
Whitepunk [10]3 years ago
8 0
Kg . Meter per second (Kg.m/s)
k0ka [10]3 years ago
5 0

Answer : The S.I unit of momentum is, kg • meters per second

Explanation :

Momentum : It is defined as the motion of a moving body. Or it is defined as the product of mass of velocity of an object.

Formula of momentum is:

p=mv

where,  

p = momentum

m = mass

v = velocity

The unit of mass of kilogram (kg) and the unit of velocity is meter per second (m/s).

Thus, the unit of momentum will be:

p=mv=kg\times (m/s)=kg.m/s

Thus, the S.I unit of momentum is, kg • meters per second

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Write the first equation of motion. Under what condition(s) is this equation valid?​
Zepler [3.9K]

Explanation:

The first equation of motion in kinematics is given by :

v=u+at .....(1)

u is initial speed

a is acceleration

v is final speed

t is time

Equation (1) is valid when the object is moving with constant acceleration. This equation gives relation between velocity and time.

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2 years ago
tom does not reallt want to give away blue marbles and would like to change the probability that he chooses a blue marble to one
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What Tom has to do is make sure that the number of marbles that are NOT blue is NINE TIMES the number of blue ones in the bag.
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3 years ago
What is one newton force?​
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5 0
2 years ago
If the total time is 120 seconds and the total distance is 320 meters. Calculate
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Explanation:

3 0
3 years ago
Consult interactive solution 2.22 before beginning this problem. a car is traveling along a straight road at a velocity of +30.0
Inessa05 [86]

Let a_1 be the average acceleration over the first 2.46 seconds, and a_2 the average acceleration over the next 6.79 seconds.

At the start, the car has velocity 30.0 m/s, and at the end of the total 9.25 second interval it has velocity 15.2 m/s. Let v be the velocity of the car after the first 2.46 seconds.

By definition of average acceleration, we have

a_1=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}

a_2=\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}

and we're also told that

\dfrac{a_1}{a_2}=1.66

(or possibly the other way around; I'll consider that case later). We can solve for a_1 in the ratio equation and substitute it into the first average acceleration equation, and in turn we end up with an equation independent of the accelerations:

1.66a_2=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}

\implies1.66\left(\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}\right)=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}

Now we can solve for v. We find that

v=20.8\,\dfrac{\mathrm m}{\mathrm s}

In the case that the ratio of accelerations is actually

\dfrac{a_2}{a_1}=1.66

we would instead have

\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}=1.66\left(\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}\right)

in which case we would get a velocity of

v=24.4\,\dfrac{\mathrm m}{\mathrm s}

6 0
3 years ago
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