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3 years ago

As you slide a heavy box across the floor, friction applies a force of -100 N

Physics

2 answers:

Mrrafil [7]3 years ago

5 0

Work = Force x Distance
W = -100 x 5
W = -500 J

nirvana33 [79]3 years ago

4 0

Answer:

A -500J

Explanation:

because W=Fs

100 × 5 = 500

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What is the orbital period of a spacecraft in a low orbit near the surface of mars? The radius of mars is 3.4×106m.

valkas [14]

<h2>Answer: 56.718 min</h2>

Explanation:

According to the Third Kepler’s Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.

In other words, this law states a relation between the orbital period $T$ of a body (moon, planet, satellite) orbiting a greater body in space with the size $a$ of its orbit.

This Law is originally expressed as follows:

$T^{2}=\frac{4\pi^{2}}{GM}a^{3}$ (1)

Where;

$G$ is the Gravitational Constant and its value is $6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$

$M=6.39(10)^{23}kg$ is the mass of Mars

$a=3.4(10)^{6}m$ is the semimajor axis of the orbit the spacecraft describes around Mars (assuming it is a circular orbit and a low orbit near the surface as well, the semimajor axis is equal to the radius of the orbit)

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

$T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}$ (2)

$T=\sqrt{\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(6.39(10)^{23}kg)}(3.4(10)^{6}m)^{3}}$ (3)

$T=\sqrt{11581157.44 s^{2}}$ (4)

Finally:

$T=3403.1099s=56.718min$ This is the orbital period of a spacecraft in a low orbit near the surface of mars

6 0

3 years ago

Anna drives north at a speed of 50 km/h for the first hour. Then, she drives north for a second hour but slows down to 30 km/h.

8090 [49]

Answer:

The correct answer is A The distance is greater in the first hour because her speed is faster.

Explanation:

During the first hour, Anna is driving at a speed of 50 km/h. During the second hour, she is only driving at a speed of 30 km/h. The faster she goes, the farther she will go.

Hope this helps,

♥A.W.E.S.W.A.N.♥

8 0

3 years ago

Read 2 more answers

A skateboarder flies horizontally off a cement planter. After 3 seconds the skateboarder lands on the ground with a final veloci

evablogger [386]

Given the time, the final velocity and the acceleration, we can calculate the initial velocity using the kinematic equation A:

$v = v_o + a \Delta t$

A skateboarder flies horizontally off a cement planter. After a time of 3 seconds (Δt), he lands with a final velocity (v) of −4.5 m/s. Assuming the acceleration is -9.8 m/s² (a), we can calculate the initial velocity of the skateboarder (v₀) using the kinematic equation A.

$v = v_o + a \Delta t\\\\v_o = v - a \Delta t = (-4.5 m/s) - (-9.8 m/s^{2} ) \times 3 s = 24.9 m/s$

Given the time, the final velocity and the acceleration, we can calculate the initial velocity using the kinematic equation A:

$v = v_o + a \Delta t$

Learn more: brainly.com/question/4434106

3 0

3 years ago

If you begin with 40 grams of a radioactive isotope and end with 10 grams, how many half-lifes of the radioactive isotope have p

Rom4ik [11]

30 grams of radioactive isotope have passed.

8 0

3 years ago

A 3 kg penguin is pushed by his penguin friends who give him an initial speed vo at the top of a 30 m hill. The penguin is hopin

Strike441 [17]

Answer:

This question can be answered by using conversation of energy.

$K_1 + U_1 = K_2 + U_2$

$\frac{1}{2}mv_{0}^2 + mgh_1 = 0 + mgh_2$

$\frac{1}{2}(3)v_0^2 + (3)(9.8)(30) = (3)(9.8)(45)\\\frac{1}{2}(3)v_0^2 = 441\\v_0^2 = 294\\v_0 = 17.14 m/s$

Explanation:

Note that we take $K_2 = 0$ because we are looking for the minimum initial speed for the penguin to reach the top of the second hill. Any other speed more than this will already be enough for him.

7 0

3 years ago