All categories

3 years ago

Your body exerts the same amount of gravitational force on the Moon as the Moon exerts on your body. True or, false?

Physics

2 answers:

Over [174]3 years ago

7 0

It is true because of science rules

dexar [7]3 years ago

6 0

Answer: TRUE / IT IS TRUE

Explanation:

You might be interested in

What is the energy stored between 2 Carbon nuclei that are 1.00 nm apart from each other? HINT: Carbon nuclei have 6 protons and

Andrei [34K]

Answer:

$A. 8.29\times 10^{-18}\ J$

Explanation:

Given that:

p = magnitude of charge on a proton = $1.6\times 10^{-19}\ C$

k = Boltzmann constant = $9\times 10^{9}\ Nm^2/C^2$

r = distance between the two carbon nuclei = 1.00 nm = $1.00\times 10^{-9}\ m$

Since a carbon nucleus contains 6 protons.

So, charge on a carbon nucleus is $q = 6p=6\times 1.6\times 10^{-19}\ C=9.6\times 10^{-19}\ C$

We know that the electric potential energy between two charges q and Q separated by a distance r is given by:

$U = \dfrac{kQq}{r}$

So, the potential energy between the two nuclei of carbon is as below:

$U= \dfrac{kqq}{r}\\\Rightarrow U = \dfrac{kq^2}{r}\\\Rightarrow U = \dfrac{9\times10^9\times (9.6\times 10^{-19})^2}{1.0\times 10^{-9}}\\\Rightarrow U =8.29\times 10^{-18}\ J$

Hence, the energy stored between two nuclei of carbon is $8.29\times 10^{-18}\ J$ .

8 0

3 years ago

In our new car, we were able to drive 30 miles in a half an hour. From this information, we can determine the car's

Ronch [10]

Average speed.
(Not velocity.)

5 0

3 years ago

You decide to test both appraoches. First, you attach a parachute to a

Annette [7]

The force (expressed in N), the egg experience upon hitting the ground is 3.5N.

<h3>What is Net force?</h3>

When two or more forces are acting on the system of objects, then the to attain equilibrium, net force must be zero.

Given is an 0.07 kg egg and toss if from a window 30 m above the ground. The egg reaches and maintains a falling speed of 0.5 m/sec. Upon reaching the ground, the egg rapidly decelerates to 0 m/sec over the course of only 0.01 seconds.

<h3>Average force = Mass x acceleration</h3>

F = m x (Vf -Vi)/t

F = 0.07 x (0 - 0.5)/0.01

F = -3.5N

Thus, the magnitude of force on egg upon hitting the ground is 3.5N.

Learn more about net force.

brainly.com/question/18031889

#SPJ1

6 0

2 years ago

The part of a circuit that increases the electric potential of the electrons is the _____. . . . . conducting wires. resistors.

Readme [11.4K]

<span>The part of a circuit that increases the electric potential of the electrons is the battery. The correct option among all the options that are given in the question is the third option or the penultimate option. I hope that this is the answer that you were looking for and the answer has actually come to your desired help.</span>

4 0

3 years ago

Read 2 more answers

An 80 kg astronaut has gone outside his space capsule to do some repair work. Unfortunately, he forgot to lock his safety tether

Anna [14]

Answer:

the time taken by the astronaut to reach safety = 9.8 hr

Explanation:

The equation for intensity can be written as :

$I = \frac{P}{A}$

where :

$\frac{I}{c}= \frac{F}{A}$

Replacing that into the above previous equation; we have:

$\frac{P}{Ac}=\frac{F}{A}$

$F = \frac{P}{c}$

However ; the force needed to push the astronaut is as follows:

F = ma

where ;

m = mass of the astronaut and a = its acceleration

we as well say;

$\frac{P}{c} = ma$

$a = \frac{P}{mc}$

Replacing P with 1000 W ; m with 80 kg and $3*10^{8} \ m/s$ for c

Then; a = $\frac{1000 \ W}{(80)(3.0*10^8)}$

a = $4.2*10^{-8} \ m/s$

It is also known that the battery will run for one hour and after which the battery on the laser will run out

Then to determine the change in the position after the first hour ; we have:

$\Delta x_1 = \frac{1}{2}*4.2*10^{-8} \ m/s^2 ) (1.0 \ h)^2$

$\Delta x_1 = \frac{1}{2}*4.2*10^{-8} \ m/s^2 ) (1.0 *3600 s)^2$

= 0.27 m

Furthermore, the final velocity of the astronaut is determined as:

$v_1 = at_1$

where ;

$v_1 = final \ velocity$

replacing $t_1 = 1.0 \ h$ and a = $4.2*10^{-8} \ m/s$ ; Then:

$v_1 = (4.2*10^8 \ m/s * 1.0 \ h * \frac{ 3600\ s}{1.0 \ h})$

$v_1 = 1.51 *10^{-4} \ m/s$

Also; when he drifted 5.0 m away from the capsule; the distance is far short of the 5 m but he still have 9 hours left of oxygen . In addition to that, he acceleration is also zero and the final velocity remains the same, so:

To find the final distance traveled by the astronaut ;we have:

$\Delta x_2 = d - \Delta x_1$

where;

$\Delta x_2$ = the final distance

d = total distance

So;

$\Delta x_2 = 5 m - 0.27 m \\ \\ \Delta x_2 = 4.73 \ m$

The time taken to reach the final distance can be calculated as:

$t_2 = \frac{\Delta x_2 }{v_1}$

where;

$t_2$ = is the time to reach the final distance

Replacing 4.73 for ${\Delta x_2 }$ and $1.51*10^{-4}$ m/s for $v_1$

$t_2 = \frac{4.73 \ m }{1.51*10^{-4} \ m/s}$

$t_2 = 31500 \ s (\frac{1.0 \ h}{3600 \ s} )$

$t_2 = 8.8 \ h$

We knew the laser was operated for 1 hour; thus the total time taken by the astronaut to reach the final distance is the sum of the time taken to reach the final distance and the operated time of the laser.

Hence ; the time taken by the astronaut to reach safety = 9.8 hr

8 0

3 years ago