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julia-pushkina [17]
3 years ago
15

Three identical satellites, X, Y, and Z are in outer space. The distance from satellite X to Y is 450 km, the distance from Y to

Z is 600 km, and the distance from satellite Z to X is 450 km. Between which satellites will the force of gravity be the greatest?
A. Between satellites Y and Z only.
B. Between satellites X and Y, and satellites Y and Z.
C. Between satellites Y and Z, and satellites X and Z.
D. Between satellites X and Y and satellites Z and X.
Physics
1 answer:
hodyreva [135]3 years ago
3 0
Gravity is stronger the closer you get. It is D.
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3 years ago
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If a proton and an electron are released when they are 7.00×10−10 m apart (typical atomic distances), find the initial accelerat
Ugo [173]

Answer:

The acceleration of the proton is 2.823 x 10¹⁷ m/s²

The acceleration of the electron is 5.175 x 10²⁰ m/s²

Explanation:

Given;

distance between the electron and proton, r = 7 x 10⁻¹⁰ m

mass of proton, m_p = 1.67 x 10⁻²⁷ kg

mass of electron, m_e = 9.11 x 10⁻³¹ kg

The attractive force between the two charges is given by Coulomb's law;

F = \frac{k(q_p)(q_e)}{r^2}

where;

k is Coulomb's constant = 9 x 10⁹ Nm²/c²

F = \frac{k(q_p)(q_e)}{r^2} \\\\F = \frac{(9*10^9)(1.602*10^{-19})(1.602*10^{-19})}{(7*10^{-10})^2} \\\\F = 4.714 *10^{-10} \ N

Acceleration of proton is given by;

F = ma

F = m_pa_p\\\\a_p = \frac{F}{m_p}\\\\a_p = \frac{4.714*10^{-10}}{1.67*10^{-27}}\\\\a_p = 2.823 *10^{17} \ m/s^2

Acceleration of the electron is given by;

F = m_ea_e\\\\a_e = \frac{F}{m_e}\\\\a_e = \frac{4.714*10^{-10}}{9.11*10^{-31}}\\\\a_e = 5.175 *10^{20} \ m/s^2

7 0
3 years ago
A sled that has a mass of 8 kg is pulled at a 50 degree angle with a force of 20 N. The force of friction acting on the sled is
vitfil [10]

The acceleration of the sled will be 1.30 m/s². Force is defined as the product of mass and acceleration.

<h3>What is force?</h3>

Force is defined as the push or pulls applied to the body. Sometimes it is used to change the shape, size, and direction of the body.

Given data;

m(mass of sled)=8 kg

Θ is the inclination of force= 50°

Force of friction,f=2.4 N.

The applied force at the given angle is resolved into the two-component as;

\rm F_h=F cos \theta \\\\ F_h= 20 cos 50 ^0 \\\\ F_h= 12.85 \ N

\rm F_v=F sin \theta \\\\  F_v=20 sin 50^0 \\\\   F_v=15.32 \ N

The net vertical force is zero;

\rm F_N=mg-Fsin50^0 \\\\ \rm F_N=8 \times 9.81 -15.32 \\\\ F_N=63.1 \ N \\\\

From Newton's second law the net force as;

\rm \sum F_{net}=ma \\\\ Fcos 60^0-f =ma \\\\ a=\frac{12.855-2.4}{8} \\\\a = 1.30 \ m/s^2

Hence, the acceleration of the sled will be 1.30 m/s².

To learn more about the force refer to the link;

brainly.com/question/26115859

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5 0
2 years ago
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Why can’t a real machine ever have 100% efficiency
Harman [31]

Answer:

Almost all machines require energy to offset the effects of gravity, friction, and air/wind resistance. Thus, no machine can continually operate at 100 percent efficiency.

4 0
3 years ago
A Mercedes-Benz 300SL (m = 1700 kg) is parked on a road that rises 15 degrees above the horizontal. What are the magnitudes of (
kogti [31]

Answer: See below

Explanation:

<u>Given:</u>

Mass of the Mercedes-Benz (m) = 1700 kg

Inclination of the road (θ) = 15.0

<em>The free body diagram is shown in figure attached below</em>

<em />

a) The normal force is equal to the cos component of the weight of the car.

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b) The static force will be equal to the weight's sin component.

\begin{aligned}&f=m g \sin \theta \\&f=1700 \times 9.81 \times \sin 15 \\&f=4316.32 \mathrm{~N}\end{aligned}

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