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2 years ago

Calculate the speed of a car which covers 240 km in 360 minutes

Physics

1 answer:

seraphim [82]2 years ago

6 0

<em>Exchange: 360 mins = 6 hours</em>

<em>So the speed of a car is 40kph (Kilometre per hour)</em>

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<em>Hope it helps, please Brainliest <3</em>

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A 20-kg boy slides down a smooth, snow-covered hill on a plastic disk. The hill is at a 10° angle to the horizontal, and the slo

zvonat [6]

Answer:

13 m/s

Explanation:

I assume we are ignoring friction.

The boy's PE will all be converted to KE at the bottom of the hill.

to find PE = mgh we need to know h

h = 50 sin 10 = 8.68 meters

then: PE = 20 * 9.81 * 8.68 =<u> 1703.49</u> j

KE = 1/2 m v^2 = <u>1703 .49</u>

v = 13 m/s

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2 years ago

I have a doubt on heat and temperature. I need help with this science question and here the question: The water in a lake just r

Volgvan

Answer:

you can't go ice skating on it because if it just reached the temp then you need to wait for about 2 hours

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3 years ago

A wheel has a radius of 5.9 m. How far (path length) does a point on the circumference travel if the wheel is rotated through an

posledela

Answer:

(a). The path length is 3.09 m at 30°.

(b). The path length is 188.4 m at 30 rad.

(c). The path length is 1111.5 m at 30 rev.

Explanation:

Given that,

Radius = 5.9 m

(a). Angle $\theta=30°$

We need to calculate the angle in radian

$\theta=30\times\dfrac{\pi}{180}$

$\theta=0.523\ rad$

We need to calculate the path length

Using formula of path length

$Path\ length =angle\times radius$

$Path\ length=0.523\times5.9$

$Path\ length =3.09\ m$

(b). Angle = 30 rad

We need to calculate the path length

$Path\ length=30\times5.9$

$Path\ length=177\ m$

(c). Angle = 30 rev

We need to calculate the angle in rad

$\theta=30\times2\pi$

$\theta=188.4\ rad$

We need to calculate the path length

$Path\ length=188.4\times5.9$

$Path\ length =1111.56\ m$

Hence, (a). The path length is 3.09 m at 30°.

(b). The path length is 188.4 m at 30 rad.

(c). The path length is 1111.5 m at 30 rev.

8 0

3 years ago

Specialized intermodal tank container which carries refrigerated liquid gases, oxygen, or helium

Nookie1986 [14]

This would be a cryogenic intermodal tank. These are used to store and transport HAZMAT gases that require storage under specific pressure and temperature parameters. Cryogenic Intermodal tanks have pressure of 25 Psi or less.

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3 years ago

Calculate the orbital period for Jupiter's moon Io, which orbits 4.22×10^5km from the planet's center (M=1.9×10^27kg) .

Verdich [7]

According to the <u>Third Kepler’s Law of Planetary motion</u> “<em>The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.</em>

In other words, this law states a relation between the orbital period $T$ of a body (moon, planet, satellite) orbiting a greater body in space with the size $a$ of its orbit.

This Law is originally expressed as follows:

Where;

$G$ is the Gravitational Constant and its value is $6.674(10^{-11})\frac{m^{3}}{kgs^{2}}$

$M=1.9(10^{27})kg$ is the mass of Jupiter

$a=4.22(10^{5})km=4.22(10^{8})m$ is the semimajor axis of the orbit Io describes around Jupiter (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)