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VLD [36.1K]
3 years ago
14

Explain how your weight can change even though your mass remains the same.

Physics
2 answers:
lara31 [8.8K]3 years ago
6 0

Answer:

See explanation below

Explanation:

Recall that Weight is a Force that results from the action of a gravitational field on our body mass. Therefore if one changes locations to places where the acceleration of gravity is different, our weight can change although the mass of our body stays the same.

For example, going to a high mountain, where the acceleration of gravity is a little smaller than at sea level, will produce such change. Also, going to another planet with different gravitational field, or going to the Moon (where the acceleration of gravity is about 1/6 of that on Earth which therefore will reduce our weight without reducing our mass)

matrenka [14]3 years ago
5 0

Since weight is calculated with the equation:

Q=mg

the weight can also change when the gravitational acceleration (g) changes.

g can change at example when you travel to a different planet or (slightly) when gaining attitude on Earth (e.g. when climbing a mountain or flying on a plane you get lighter).

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How many lines per mm are there in the diffraction grating if the second order principal maximum for a light of wavelength 536 n
grandymaker [24]

To solve this problem it is necessary to apply the concepts related to the principle of superposition and the equations of destructive and constructive interference.

Constructive interference can be defined as

dSin\theta = m\lambda

Where

m= Any integer which represent the number of repetition of spectrum

\lambda= Wavelength

d = Distance between the slits.

\theta= Angle between the difraccion paterns and the source of light

Re-arrange to find the distance between the slits we have,

d = \frac{m\lambda}{sin\theta }

d = \frac{2*536*10^{-9}}{sin(24)}

d = 2.63*10^{-6}m

Therefore the number of lines per millimeter would be given as

\frac{1}{d} = \frac{1}{2.63*10^{-6} }

\frac{1}{d} = 379418.5\frac{lines}{m}(\frac{10^{-3}m}{1 mm})

\frac{1}{d} = 379.4 lines/mm

Therefore the number of the lines from the grating to the center of the diffraction pattern are 380lines per mm

6 0
3 years ago
Inertia is a measure of
saw5 [17]

Answer:

b.. a difficulty of changing an object's motion, that's my ✨ g u e s s ✨

7 0
3 years ago
What is true about mechanical energy? a. An object with mechanical energy cannot do work. c. Mechanical energy is the energy tha
Mama L [17]
I believe the correct answer is C all the other options seem funny to me
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4 years ago
A person with mass mp = 74 kg stands on a spinning platform disk with a radius of R = 2.31 m and mass md = 183 kg. The disk is i
timurjin [86]

Answer:

1) 883 kgm2

2) 532 kgm2

3) 2.99 rad/s

4) 944 J

5) 6.87 m/s2

6) 1.8 rad/s

Explanation:

1)Suppose the spinning platform disk is solid with a uniform distributed mass. Then its moments of inertia is:

I_d = m_dR^2/2 = 183*2.31^2/2 = 488 kgm^2

If we treat the person as a point mass, then the total moment of inertia of the system about the center of the disk when the person stands on the rim of the disk:

I_{rim} = I_d + m_pR^2 = 488 + 74*2.31^2 = 883 kgm^2

2) Similarly, he total moment of inertia of the system about the center of the disk when the person stands at the final location 2/3 of the way toward the center of the disk (1/3 of the radius from the center):

I_{R/3} = I_d + m_p(R/3)^2 = 488 + 74*(2.31/3)^2 = 532 kgm^2

3) Since there's no external force, we can apply the law of momentum conservation to calculate the angular velocity at R/3 from the center:

I_{rim}\omega_{rim} = I_{R/3}\omega_{R/3}

\omega_{R/3} = \frac{I_{rim}\omega_{rim}}{I_{R/3}}

\omega_{R/3} = \frac{883*1.8}{532} = 2.99 rad/s

4)Kinetic energy before:

E_{rim} = I_{rim}\omega_{rim}^2/2 = 883*1.8^2/2 = 1430 J

Kinetic energy after:

E_{R/3} = I_{R/3}\omega_{R/3}^2/2 = 532*2.99^2/2 = 2374 J

So the change in kinetic energy is: 2374 - 1430 = 944 J

5) a_c = \omega_{R/3}^2(R/3) = 2.99^2*(2.31/3) = 6.87 m/s^2

6) If the person now walks back to the rim of the disk, then his final angular speed would be back to the original, which is 1.8 rad/s due to conservation of angular momentum.

3 0
4 years ago
The frequency of a sound wave is commonly called
Rzqust [24]
The answer is Hertz (Hz) named in honor of the German scientist Heinrich Hertz, who first conclusively proved the existence of electromagnetic waves.
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