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3 years ago

7

When you push a box with 20 N of force, what force does the box apply back on you?

1 answer:

Fiesta28 [93]3 years ago

3 0

Remember Newton’s third law: for every force, there is an equal and opposite reaction.

The force being applied back is also 20N

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A quantity of N2 occupies a volume of 1.4 L at 290 K and 1.0 atm. The gas expands to a volume of 3.3 L as the result of a change

lions [1.4K]

Answer:

$\rho = 0.50 g/L$

Explanation:

As we know that

PV = nRT

here we have

$P = 1.0 atm$

$P = 1.013 \times 10^5 Pa$

so we have

$V = 1.4 \times 10^{-3} m^3$

T = 290 K

now we have

$(1.013 \times 10^5)(1.4 \times 10^{-3}) = n(8.31)(290)$

$n = 0.06$

now the mass of gas is given as

$m = n M$

$m = (0.06)(28)$

$m = 1.65 g$

now density of gas when its volume is increased to 3.3 L

so we will have

$\rho = \frac{m}{V}$

$\rho = \frac{1.65 g}{3.3 L}$

$\rho = 0.50 g/L$

5 0

3 years ago

You ride on an elevator that is moving with constant upward acceleration while standing on a bathroom scale. the reading on the

Blababa [14]

The reading on the scale is greater than your actual weight.

4 0

4 years ago

An instrument is thrown upward with a speed of 15 m/s on the surface of planet X where the acceleration due to gravity is 2.5 m/

Katen [24]

<h2>Answer: 12 s</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told <u>the instrument is thrown upward</u> from the surface, we will only use the equations related to the Y axis.

In this sense, the main movement equation in the Y axis is:

$y-y_{o}=V_{o}.t-\frac{1}{2}g.t^{2}$ (1)

Where:

$y$ is the instrument's final position

$y_{o}=0$ is the instrument's initial position

$V_{o}=15m/s$ is the instrument's initial velocity

$t$ is the time the parabolic movement lasts

$g=2.5\frac{m}{s^{2}}$ is the acceleration due to gravity at the surface of planet X.

As we know $y_{o}=0$ and $y=0$ when the object hits the ground, equation (1) is rewritten as:

$0=V_{o}.t-\frac{1}{2}g.t^{2}$ (2)

Finding $t$ :

$0=t(V_{o}-\frac{1}{2}g.t^{2})$ (3)

$t=\frac{2V_{o}}{g}$ (4)

$t=\frac{2(15m/s)}{2.5\frac{m}{s^{2}}}$ (5)

Finally:

$t=12s$

3 0

4 years ago

Suppose an asteroid orbiting the sun had an orbital period of 7. 5 years. What would its orbital radius be?.

creativ13 [48]

By using the orbital period equation we will find that the orbital radius is r = 4.29*10^11 m

<h3>What is the orbital period?</h3>

This would be the time that a given body does a complete revolution in its orbit.

It can be written as:

$T = \sqrt{\frac{4*\pi ^2*r^3}{G*M} }$

Where:

π = 3.14
G is the gravitational constant = 6.67*10^(-11) m^3/(kg*s^2)
M is the mass of the sun = 1.989*10^30 kg
r is the radius, which we want to find.

Rewriting the equation for the radius we get:

$T = \sqrt{\frac{4*\pi ^2*r^3}{G*M} }\\\\r = \sqrt[3]{ \frac{T^2*G*M}{4*\pi ^2} }$

Where T = 7.5 years = 7.5*(3.154*10^7 s) = 2.3655*10^8 s

Replacing the values in the equation we get:

$r = \sqrt[3]{ \frac{(2.3655*10^8 s)^2*(6.67*10^{-11} m^3/(kg*s^2))*(1.989*10^{30} kg)}{4*3.14 ^2} } = 4.29*10^{11 }m$

So the orbital radius is 4.29*10^11 m

If you want to learn more about orbits, you can read:

brainly.com/question/11996385

7 0

2 years ago

Assuming a successful launch, what is the speed of the small CO2 rocket given that it travels about 10 meters in about 3 seconds

antiseptic1488 [7]

Probably 30 im not sure

4 0

3 years ago