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3 years ago

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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What factors affect the speed of water waves

Aneli [31]

Hey there,

Your question states: What factors affect the speed of water waves
Let's get one thing out the way, (wavelength) does $NOT$ affect the the speed of water. If anything, it would be how high the wavelength's are. The higher the wavelengths are, the more that it would affect the speed, because there very high, but if it were to go longer on the width side, that would increase the speed, but that's not the case. Your correct answer would be (higher wavelength).

Hope this really helps you.

6 0

3 years ago

The world's most powerful laser is the LFEX laser in Japan. It can produce a 2 petawatt(2×10^15W) laser pulse that last for 1 ps

kogti [31]

Answer:

$2.82942\times 10^{24}\ W\m^2$

Explanation:

d = Diameter of spot = 30 μm

r = Radius of spot = $\frac{d}{2}=\frac{30}{2}=15\ mu m$

P = Power of the laser = $2\times 10^{15}\ W$

A = Area = $\pi r^2$

Intensity is given by

$I=\frac{P}{A}\\\Rightarrow I=\frac{2\times 10^{15}}{\pi\times (15\times 10^{-6})^2}\\\Rightarrow I=2.82942\times 10^{24}\ W/m^2$

The light intensity within this spot is $2.82942\times 10^{24}\ W/m^2$

8 0

3 years ago

On your first trip to Planet X you happen to take along a 280 g mass, a 40-cm-long spring, a meter stick, and a stopwatch. You'r

arsen [322]

Answer:

5.31143691523 m/s²

Explanation:

m = Mass = 280 g

x = Displacement of spring = 21.7 cm

Time period

$T=\dfrac{14}{11}\\\Rightarrow T=1.27\ s$

Angular velocity is given by

$\omega=\dfrac{2\pi}{T}\\\Rightarrow \omega=\dfrac{2\pi}{1.27}\\\Rightarrow \omega=4.94739\ rad/s$

$\omega=\sqrt{\dfrac{k}{m}}\\\Rightarrow k=\omega^2m\\\Rightarrow k=4.94739^2\times 0.28\\\Rightarrow k=6.85346698739\ N/m$

From Hooke's law

$mg=kx\\\Rightarrow g=\dfrac{kx}{m}\\\Rightarrow g=\dfrac{6.85346698739\times 0.217}{0.28}\\\Rightarrow g=5.31143691523\ m/s^2$

The acceleration due to gravity on the planet is 5.31143691523 m/s²

Yes, I have been able to satisfy my curiosity.

7 0

3 years ago

The mass of an object changes as the distance from the center of gravity changes.

Alchen [17]

Answer:

true

Explanation:

this is the answer to this question

8 0

3 years ago

provides some pertinent background for this problem. A pendulum is constructed from a thin, rigid, and uniform rod with a small

gavmur [86]

Answer:

the period of the physical pendulum is 0.498 s

Explanation:

Given the data in the question;

$T_{simple$ = 0.61 s

we know that, the relationship between T and angular frequency is;

T = 2π/ω ---------- let this be equation 1

Also, the angular frequency of physical pendulum is;

ω = √(mgL / $I$ ) ------ let this equation 2

where m is mass of pendulum, L is distance between axis of rotation and the center of gravity of rod and $I$ is moment of inertia of rod.

Now, moment of inertia of thin uniform rod D is;

$I$ = $\frac{1}{3}$ mD²

since we were not given the length of the rod but rather the period of the simple pendulum, lets combine this three equations.

we substitute equation 2 into equation 1

we have;

T = 2π/ω OR T = 2π/√(mgL/ $I$ ) OR T = 2π√( $I$ /mgL)

so we can use $I$ = $\frac{1}{3}$ mD² for moment of inertia of the rod

Since center of gravity of the uniform rod lies at the center of rod

so that L = $\frac{1}{2}$ D.

now, substituting these equations, the period becomes;

T = 2π/√( $I$ /mgL) OR T = $2\pi \sqrt{\frac{\frac{1}{3}mD^2 }{mg(\frac{1}{2})D } }$ OR T = 2π√(2D/3g ) ----- equation 3

length of rod D is still unknown, so from equation 1 and 2 ( period of pendulum ),

we have;

ω $_{simple$ = 2π/ $T_{simple$ OR ω $_{simple$ = √(g/D) OR ω $_{simple$ = 2π√( D/g )

so we simple solve for D/g and insert into equation 3

so we have;

T = √(2/3) × $T_{simple$

we substitute in value of $T_{simple$

T = √(2/3) × 0.61 s

T = 0.498 s

Therefore, the period of the physical pendulum is 0.498 s

8 0

3 years ago