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2 years ago

How long does it take a car travelling at 60 m.p.h to cover 5 miles?

Physics

2 answers:

bixtya [17]2 years ago

8 0

Answer:

2minutes per mile

Explanation:

Not sure if it helps

k0ka [10]2 years ago

6 0

Answer:

Time = 5 minutes

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An astronaut is in equilibrium when he is positioned 140 km from the center of asteroid X and 581 km from the center of asteroid

mariarad [96]

Answer:

0.05806

Explanation:

$m_x$ = Mass of asteroid x

$m_y$ = Mass of asteroid y

$r_x$ = Distance from asteroid x = 140 km

$r_y$ = Distance from asteroid y = 581 km

m = Mass of asteroid

Force of gravity between asteroid x and the astronaut

$F_1=\frac{Gm_xm}{r_x^2}\\\Rightarrow F_1=\frac{Gm_xm}{140^2}$

Force of gravity between asteroid x and the astronaut

$F_2=\frac{Gm_ym}{r_y^2}\\\Rightarrow F_2=\frac{Gm_ym}{581^2}$

Here these two forces are equal as they are in equilibrium

$\frac{Gm_xm}{140^2}=\frac{Gm_ym}{581^2}\\\Rightarrow \frac{m_x}{140^2}=\frac{m_y}{581^2}\\\Rightarrow \frac{m_x}{m_y}=\frac{140^2}{581^2}\\\Rightarrow \frac{m_x}{m_y}=0.05806$

The ratio of the masses of the asteroid is 0.05806

4 0

2 years ago

PLEASE HELP!!!!!!

elena-s [515]

Answer:For this equation you would have to take the numerator and multiply with the other one, which would leave you with 24/80 and if you want the simplified version it would be 3/10, hope this helps! :)

Explanation:

4 0

2 years ago

A student runs 35 m east and then 12m west. What is the distance run by the student

stealth61 [152]

Answer:

The answer is either 47 or 23. Most likely 47

7 0

2 years ago

A baseball pitcher throws the ball in a motion where there is rotation of the forearm about the elbow joint as well as other mov

Mamont248 [21]

Answer:

Rotational kinetic energy of the forearm is 412.33 J.

Explanation:

It is given that,

The linear velocity of the ball relative to the elbow joint is, v = 19.9 m/s

Distance from the joint, r = 0.49 m

Moment of inertia of the forearm, I = 0.5 kg/m²

We need to find the rotational kinetic energy of the forearm. It is given by :

$E=\dfrac{1}{2}I\omega^2$

$\omega$ is the angular speed, $\omega=\dfrac{v}{r}$

$E=\dfrac{1}{2}\times 0.5\ kg/m^2\times (\dfrac{19.9\ m/s}{0.49\ m})^2$

E = 412.33 J

So, the rotational kinetic energy of the forearm is 412.33 J. Hence this is the required solution.

7 0

2 years ago

At noon, ship A is 110 km west of ship B. Ship A is sailing east at 20 km/h and ship B is sailing north at 15 km/h. How fast is

Katyanochek1 [597]

Answer:

$4.47\ \text{km/h}$

Explanation:

$\dfrac{da}{dt}$ = Rate at which the distance between A and starting point of B is changing = -20 km/h

$\dfrac{db}{dt}$ = Rate at which the distance of B is changing = 15 km/h

$\dfrac{dc}{dt}$ = Rate at which the distance between A and B is changing

Time after which the rate at which the distance between A and B is changing is 4 hours

Distance covered by A in 4 hours = $20\times 4=80\ \text{km}$

a = Distance remaining to the start point of B = $110-80=30\ \text{km}$

b = Distance covered by B in 4 hours = $15\times 4=60\ \text{km}$

Distance between A and B after 4 hours

$c=\sqrt{a^2+b^2}\\\Rightarrow c=\sqrt{30^2+60^2}\\\Rightarrow c=67.08\ \text{km}$

$c^2=a^2+b^2$

Differentiating with respect to time we get

$c\dfrac{dc}{dt}=a\dfrac{da}{dt}+b\dfrac{db}{dt}\\\Rightarrow \dfrac{dc}{dt}=\dfrac{a\dfrac{da}{dt}+b\dfrac{db}{dt}}{c}\\\Rightarrow \dfrac{dc}{dt}=\dfrac{30\times -20+60\times 15}{67.08}\\\Rightarrow \dfrac{dc}{dt}=4.47\ \text{km/h}$

The rate at which the distance between the ships is changing at 4 PM is $4.47\ \text{km/h}$ .

7 0

2 years ago