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Feliz [49]
3 years ago
11

Two planets orbit a far away star as shown. Is it possible that both planets experience the same gravitation force from the star

?
A) No, it is not possible.
B) Yes, it is possible if the closer planet has less mass.
C) Yes, it is possible if the closer planet as a smaller radius.
D) Yes, it is possible if the closer planet orbits the star in a shorter time.

Physics
1 answer:
Vesna [10]3 years ago
8 0

As we know that gravitational force on two planets will be given as

let

mass of star = M

mass of two planets are m1 and m2

their distance from star is r1 and r2

F = \frac{GMm_1}{r_1^2} = \frac{GMm_2}{r_2^2}

since the gravitational force of star is given on two planets to be same

so here we can say

\frac{m_1}{r_1^2} = \frac{m_2}{r_2^2}

now if the ratio of mass and distance is same then we can say that closer planet must have lesser mass to hold good for above equation

so correct answer will be

B) Yes, it is possible if the closer planet has less mass.

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Answer:

Magnets interfere with compasses because compasses use the magnetic field to locate NESW (north east south west)

Explanation:

6 0
3 years ago
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An unbalanced force gives a 2.00 kg mass an acceleration of 5.00 m/s? What is the force applied to the object?​
valentinak56 [21]

Answer:

10N

Explanation:

Equation: ΣF = ma

Fapp = ma

Fapp = (2kg)(5m/s^2)    (im guessing you mean 5.00 m/s^2 not m/s)

Fapp = 10*kg*m/s^2

Fapp = 10N

5 0
2 years ago
The Hall effect can be used to calculate the charge-carrier number density in a conductor. A conductor carrying current of 2.0 A
marysya [2.9K]

Answer:

option D

Explanation:

given,

A conductor is carrying current = 2.0 A is 0.5 mm thick

Hall voltage = 4.5 x 10-6 V

uniform magnetic field  =  1.2 T

density of the charge = n =?

hall voltage =V_h =\dfrac{i\ B}{n\ e\ L}

n = \dfrac{i\ B}{V\ e\ L}

n = \dfrac{2 \times 1.2 }{4.5 \times 10^{-6}\times 1.6 \times 10^{-19} \times 0.5 \times 10^{-3}}

n = 6.67 × 10²⁷ charges/m

hence the correct answer is option D

7 0
3 years ago
As frequency increases, what happens to the tone (or pitch) of a sound? Which option is it?
icang [17]

Answer:

gets higher

Explanation:

There are videos that show the range of human hearing.  If you would play the video, you would notice that if the frequency increases, the pitch would also increase.

3 0
3 years ago
Q 19.23: A proton is initially moving at 3.0 x 105 m/s. It moves 3.5 m in the direction of a uniform electric field of magnitude
DENIUS [597]

Answer:

The kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.

Explanation:

Given;

initial velocity of proton, v_p_i = 3 x 10⁵ m/s

distance moved by the proton, d = 3.5 m

electric field strength, E = 120 N/C

The kinetic energy of the proton at the end of the motion is calculated as follows.

Consider work-energy theorem;

W = ΔK.E

W =K.E_f - K.E_i

where;

K.Ef is the final kinetic energy

W is work done in moving the proton = F x d  = (EQ) x d = EQd

K.E_f =EQd + \frac{1}{2}m_pv_p_i^2

m_p \ is \ mass \ of \ proton = 1.673 \ \times \ 10^{-27} kg \\\\Q \ is \ charge \ of \ proton = 1.6 \times 10^{-19} C

K.E_f = 120\times 1.6 \times 10^{-19} \times 3.5   \ + \ \frac{1}{2}(1.673\times 10^{-27})(3\times 10^5)^2 \\\\

K.E_f = 6.72\times 10^{-17} \ + \ 7.53 \times 10^{-17} \\\\K.E_f = 14.25 \times 10^{-17} J\\\\K.E_f = 1.425\times 10^{-16} \ J

Therefore, the kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.

3 0
3 years ago
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