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3 years ago

You push against a wall and the wall pushes back on you.

Physics

1 answer:

emmainna [20.7K]3 years ago

4 0

Answer:

Its the third law of motion.

Explanation:

Newton's third law of motion states that every action has equal and opposite reaction.

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Tammy leaves the office, drives 34 km due north, then turns onto a second highway and continues in a direction of 35◦ north of e

Natasha2012 [34]

Answer:

The total displacement is 102 km $51^o$ north of east.

Explanation:

We can treat this problem as a trigonometric one, so we need to calculate the total displacement on the north and east.

$d_n=34km+79km*sin(35^o)=79km$

and

$d_e=79*cos(35^o)=65km$

The total displacement is given by:

$D=\sqrt{(79km)^2+(65km)^2}=102km$

with an agle of:

$\alpha =arctg(\frac{79km}{65km})=51^o$

4 0

3 years ago

PLEASEEE HELPPPPP does anyone know these answers?

steposvetlana [31]

Answer:

oof ok

Explanation:

Thank you :)

5 0

3 years ago

How can you increase the pressure on an object?

Cerrena [4.2K]

Increase the force on the object

8 0

3 years ago

You are the science officer on a visit to a distant solar system. Prior to landing on a planet you measure its diameter to be 1.

tatiyna

(a) $2.7\cdot 10^{25} kg$

The acceleration due to gravity on the surface of the planet is given by

$g=\frac{GM}{R^2}$ (1)

where

G is the gravitational constant

M is the mass of the planet

R is the radius of the planet

Here we know:

$g=22.4 m/s^2$

$d=1.8\cdot 10^7 m$ is the diameter, so the radius is

$R=\frac{d}{2}=\frac{1.8\cdot 10^7 m}{2}=9\cdot 10^6 m$

So we can re-arrange eq.(1) to find M, the mass of the planet:

$M=\frac{gR^2}{G}=\frac{(22.4 m/s^2)(9\cdot 10^6 m)^2}{6.67\cdot 10^{-11}}=2.7\cdot 10^{25} kg$

(b) $4.8\cdot 10^{31}kg$

The planet is orbiting the star, so the centripetal force is equal to the gravitational attraction between the planet and the star:

$m\frac{v^2}{r}=\frac{GMm}{r^2}$ (1)

where

m is the mass of the planet

M is the mass of the star

v is the orbital speed of the planet

r is the radius of the orbit

The orbital speed is equal to the ratio between the circumference of the orbit and the period, T:

$v=\frac{2\pi r}{T}$

where

$T=402 days = 3.47\cdot 10^7 s$

Substituting into (1) and re-arranging the equation

$m\frac{4\pi r^2}{rT^2}=\frac{GMm}{r^2}\\\frac{4\pi r}{T^2}=\frac{GM}{r^2}\\M=\frac{4\pi r^3}{GMT^2}$

And substituting the numbers, we find the mass of the star:

$M=\frac{4\pi^2 (4.6\cdot 10^{11} m)^3}{(6.67\cdot 10^{-11})(3.47\cdot 10^7 s)^2}=4.8\cdot 10^{31}kg$

4 0

3 years ago

An arrow is moving at 35 m/s and travels for 5 seconds. How far did the arrow travel?

noname [10]

Answer:

175m

Explanation:

So it’s travelling at 35metres/1second so if we want to find 5 seconds we can do this…

$\frac{35}{1} \frac{x}{5}.... so$ 35x5 ÷ 1 = 175m/5 seconds

6 0

3 years ago

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