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Juliette [100K]
3 years ago
8

The gravitational strength at the poles is greater than the gravitational strength at the equator. What will happen to an object

when it moves from the poles to the equator? Group of answer choices A. Its mass will increase. B. Its mass will decrease. C. Its weight will increase. D. Its weight will decrease.
Physics
1 answer:
geniusboy [140]3 years ago
3 0

Answer:

C

Explanation:

Because this same question was on my test last week and I got it correct

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Which statement explains how it is possible to carry books to school without changing the kinetic or potential energy of the boo
Mamont248 [21]

Answer:

a. by moving the book without acceleration and keeping the height of the book constant

Explanation:

FOR CONSTANT KINETIC ENERGY:

The kinetic energy of a body depends upon its speed according to its formula:

ΔK.E = (1/2)mΔv²

So, for Δv = 0 m/s

ΔK.E = 0 J

So, for keeping kinetic energy constant, the books must be moved at constant speed without acceleration.

FOR CONSTANT POTENTIAL ENERGY:

The potential energy of a body depends upon its height according to its formula:

ΔP.E = mgΔh

So, for Δh = 0 m/s

ΔP.E = 0 J

So, for keeping potential energy constant, the books must be moved at constant height.

So, the correct option is:

<u>a. by moving the book without acceleration and keeping the height of the book constant</u>

8 0
3 years ago
Convert 8 light years to Astronomical Units
marusya05 [52]

Answer:

505929 AU

Explanation:

As you may know, one light-year is equivalent to approximately 63241.1 Astronomical Units. To get your answer, simply multiply 63241.1 * 8 to get ≈505929 AU

5 0
3 years ago
A box full of charged plastic balls sits on a table. The electric force exerted on a ball near one upper corner of the box has c
tatuchka [14]

We have that the values for F north, F east, F up are

  • F_N=1.09090909*10^{-5}
  • F_E=5.18181818*10^{-6}
  • F_E=2*10^{-6}

From the Question we are told that

electric force F_1 = 1.2 x 10^{-3} N(N)

electric force , F_2=5.7 x 10^{-4} N(E)

electric force , F_3=2.2 x 10^{-4} N (U)

charge on this ball one q_1= 110 nC.

charge on this ball two q_2= -50 nC.

Generally the equation for the F north  is mathematically given as

F_N=\frac{F_1}{q_1}\\\\F_N=\frac{ 1.2 * 10^{-3} )}{110}

F_N=1.09090909*10^{-5}

For F East

F_E=\frac{F_2}{q_1}\\\\F_E=\frac{5.7 x 10^-4 }{110}

F_E=5.18181818*10^{-6}

For F UP

F_U=\frac{F_3}{q_1}\\\\F_U=\frac{2.2 x 10^-4 }{110}

F_E=2*10^{-6}

For more information on this visit

brainly.com/question/21811998

5 0
3 years ago
A certain pair of slits are separated by a distance d. Monochromatic coherent light falls on this pair of slits and the interfer
DanielleElmas [232]

Answer:

The new separation distance between adjacent bright fringes will be <u>4 mm</u>

Explanation:

Since, the distance between adjacent bright fringes is given by the formula:

Δx₁ = λL/d = 2 mm  -------- eqn (1)

where,

Δx = Distance between adjacent bright fringes

λ = wavelength of light = constant for both cases

L = Distance between the slits and the screen

d = slit separation

Now, for the second case:

Slit Separation = d/2

Therefore,

Δx₂ = λL/(d/2)

Δx₂ = 2(λL/d)

using eqn (1), we get:

Δx₂ = 2 Δx₁

Δx₂ = 2(2 mm)

<u>Δx₂ = 4 mm</u>

5 0
3 years ago
You have a source of energy containing 21 gj of energy at 600k how much this energy can be converted to work when rejecting heat
sweet [91]

Answer:

Available energy = 35 x 10⁶ J

Explanation:

Given:

Amount of energy (Q) = 21 gj = 21 x 10⁹ J

Temperature T1 = 600 k

Temperature T0 = 27 + 273 = 300k

Find:

Available energy

Computation:

Available energy = Q[1/T0 - 1/T1]

Available energy = 21 x 10⁹ J[1/300 - 1/600]

Available energy = 35 x 10⁶ J

4 0
3 years ago
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