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3 years ago

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Gravity pulls downward on a rock with a force of 800 N. If you pull upward on the rock with a force of 400 N, what is the total

force acting on the rock?
A. 400 N downward
B. 400 N upward
C. 1200 N downward
D. 1200 N upward

2 answers:

aleksley [76]3 years ago

6 0

A i believe srry if wrong

wel3 years ago

3 0

The answer is
a. 400 N downward

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7. A force of 100 N acting on a body gives it a speed of 200 m/s in 2

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Answer:

Choice a. 1 kg, assuming that all other forces on the object (if any) are balanced.

Explanation:

By Newton's Second Law,

$\displaystyle a = \frac{\Sigma F}{m}$ ,

where

$a$ is the acceleration of the object in $\text{m}\cdot\text{s}^{-2}$ ,
$\Sigma F$ is the net force on the object in Newtons, and
$m$ is the mass of the object in kilograms.

As a result,

$\displaystyle m = \frac{\Sigma F}{a}$ .

Assume that all other forces on this object are balanced. The net force on the object will be $100\;\text{N}$ . The net force is constant. Acceleration should also be constant and the same as the average acceleration in the two seconds.

<h3>What is the average acceleration of this object?</h3>

$\displaystyle \begin{aligned}\text{Acceleration} &= \text{Average Acceleration}=\frac{\text{Change in Velocity}}{\text{Time Taken}}\end{aligned}$ .

$\displaystyle {a} = \frac{200\;\text{m}\cdot\text{s}^{-1}}{2\;\text{s}}=100\;\text{m}\cdot\text{s}^{-2}$ .

<h3>Apply Newton's Second Law to find the mass of the object.</h3>

$\displaystyle m = \frac{\Sigma F}{a} = \frac{100\;\text{N}}{100\;\text{m}\cdot\text{s}^{-2}} = 1\;\text{kg}$ .

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If jack was traveling north 120 miles and it took him 3 hours to get there. what is the velocity that jack was traveling? is thi

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Answers:

40 mp/h; Vector

Reason:

120/3 is 40 miles per hour.

Velocity is a vector measurement.

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A transverse wave is traveling from right to left. what direction does the medium vibrate?

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<span>In transverse waves, particles of the medium vibrate to and from in a direction perpendicular to the direction of energy transport.</span>

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Calculate the orbital period of a dwarf planet found to have a semimajor axis of a = 4.0x 10^12 meters in seconds and years.

padilas [110]

Explanation:

We have,

Semimajor axis is $4\times 10^{12}\ m$

It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :

$T^2=\dfrac{4\pi ^2}{GM}a^3$

G is universal gravitational constant

M is solar mass

Plugging all the values,

$T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.98\times 10^{30}}\times (4\times 10^{12})^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.98\times10^{30}}\times(4\times10^{12})^{3}}\\\\T=4.37\times 10^9\ s$

Since,

$1\ s=3.17\times 10^{-8}\ \text{years}\\\\4.37\times 10^9\ s=4.37\cdot10^{9}\cdot3.17\cdot10^{-8}\\\\4.37\times 10^9\ s=138.52\ \text{years}$

So, the orbital period of a dwarf planet is 138.52 years.

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3 years ago

In an intergalactic competition, spaceship pilots compete to see who can cover the distance between two asteroids in the short-

pogonyaev

Answer:

a) truc is C, b) correct result is the B

Explanation:

As the speed of the competition is very high, for the judges the speed is

v = d / t

v = 3 109 m / 20

v = 1.5 108 m / s

This is half the speed of light. For these high speeds we must use the relations of special relativity.

For the time t = to γ

For distance L = Lo / γ

γ = √ (1-v2 / c2)

Own time and distance (to and Lo) corresponds to the observer who is not moving the judges in this case

Let's look for the range value

γ = 1 / √ (1 - (1.5 / 3) 2) = 1 / 0.866 = 1.15

The time t = 20 1.15 = 23 s

The distance L = 3 10 9 /1.15 = 2.60 109 m

From these results we see that time increases and the distance is shorter.

Let's review the claims

A) False. It's the opposite

B) False

C) True. It is according to the result found

D) False.

In the nuclear fusion process, we will also use the special relativity that has a relationship between energy and mass

ΔE = c² Δm

As in the process energy is released, for the law of conservation of the mass of energy to be fulfilled, the total mass of the products, He atom, must be reduced.

Therefore the correct result is the B

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3 years ago