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3 years ago

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An object's (mass or weight) will remain constant throughout the universe, but its (mass or weight) can change from planet to pl

anet. If you increase the mass of a planet, what happens to its gravity? (Strength of gravity increases, strength of gravity decreases, strength og gravity stops completely) If the gravity on a planet decreases, what happens to the weight of an object on that planet? (Weight increases, weight decreases, weight stays the same)

2 answers:

melisa1 [442]3 years ago

8 0

Answer:

1: mass

2:weight

3: gravity increases

4: it decreases

Hope this helps:)

Explanation:

2 years ago

4 Is weight decreases

krek1111 [17]3 years ago

6 0

Answer: 1. An objects mass

2. But it’s weight

3. Strength of gravity increases

4 weight decreases

Explanation:

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A mass M tied to a light string is moving in a vertical circle. The tension in the string at the top is TT and TB at the bottom

natita [175]

Answer:

Explanation:

Tension provides centripetal force in the circular motion . In circular motion work done by force = torque x angle

torque is zero as , centripetal force passes through axis of rotation that is center.

So work done by centripetal force = 0

So work done by tension on M = 0

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3 years ago

You are an industrial engineer with a shipping company. As part of the package- handling system, a small box with mass 1.60 kg i

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Answer:

v = 0.84m/s, v(max)= 0.997m/s

Explanation:

Initial work done by the spring, where c is the compression = 0.28m:

$W_s = \frac{1}{2}kc^2$

Work lost to friction:

$W_f =\mu mg(c-x)$

Energy:

$E = W_s-W_f=\frac{1}{2}kc^2 -\mu mg(c-x)=\frac{1}{2}mv^2+\frac{1}{2}kx^2$

(a) Solve for v:

$v=\sqrt{\frac{k}{m}(c^2-x^2)-2\mu g(c-x)}$

(b) Solve $\frac{dv}{dx}=0$ for x:

$\frac{dv}{dx}=\frac{\mu g-\frac{k}{m}x}{\sqrt{\frac{k}{m}(c^2-x^2)-2\mu g(c-x)}}$

$\frac{dv}{dx}=0$ if:

$\mu g-\frac{k}{m}x = 0$

$x_{max} = \frac{\mu gm}{k}$

$v_{max}=\sqrt{\frac{k}{m}c^2-2\mu gc+(\mu g)^2\frac{m}{k}}$

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4 years ago

A light with a second-order bright band forms a diffraction angle of 30. 0°. The diffraction grating has 250. 0 lines per mm. Wh

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The distance between two successive troughs or crests is known as the wavelength. The wavelength of the light will be 1000 nm.

How do you define wavelength?

The distance between two successive troughs or crests is known as the wavelength. The peak of the wave is the highest point, while the trough is the lowest.

The wavelength is also defined as the distance between two locations in a wave that have the same oscillation phase.

Diffraction angle= 30⁰

Diffraction grating per mm= 250

wavelength = ?

Mathematically the equation of bright band is given by

$\rm \lambda= \frac{sin\theta}{nN}$

$\rm \lambda= \frac{sin23^0}{250\times 2}$

$\rm \lambda= 0.000001$ m

$\rm \lambda= 1000 nm$

Hence the wavelength of the light will be 1000 nm.

To learn more about the wavelength refer to the link;

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2 years ago

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A proton is held at rest in a uniform electric field. When it is released, the proton will lose?

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A proton is held at rest in a uniform electric field. When it is released, the proton will lose its kinetic energy.

Kinetic energy

The energy an object has as a result of motion is known as kinetic energy in physics. It is described as the effort required to move a mass-determined body from rest to the indicated velocity. The body holds onto the kinetic energy it acquired during its acceleration until its speed changes. The body exerts the same amount of effort when slowing down from its current pace to a condition of rest. Formally, kinetic energy is any term that includes a derivative with respect to time in the Lagrangian of a system.

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1 year ago

A tray is moved horizontally back and forth in simple harmonic motion at a frequency of f = 2.07 Hz. On this tray is an empty cu

Anettt [7]

Answer:

Explanation:

Given

Frequency of SHM is $f=2.07\ Hz$

Amplitude of SHM is $A=3.13\ cm$

Cup begins to slip when it overcomes the friction force

Friction force $F_s=\mu mg$

Applied force $F=ma$

$ma=\mu mg$

$a=\mu g$

and maximum acceleration during SHM is

$a=A\omega ^2$

$a=A(2\pi f)^2$

$a=3.13\times 10^{-2}\times (2\pi 2.07)^2$

$a=5.296\ m/s^2$

$\mu =\frac{a}{g}$

$\mu =\frac{5.296}{9.8}=0.54$

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4 years ago

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