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bonufazy [111]
2 years ago
7

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>=> In 1 hour, the car can move: 240 /6 = 40 km</em>

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

<em />

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

<em />

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How many electrons will constitute 2A current in unit time
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2 charges of electron (2C)

Explanation:

I = Q/t

2 = Q/1

Q = 2×1= 2C

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When aluminum foil is formed into a loose ball, it can float on water. But when the ball of foil is pounded flat with a hammer,
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Kids are pelting a window with snowballs. On average, two snowballs of roughly 300-g mass hit the window each second, moving hor
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Answer:

The average force exerted on the window due to two snowballs is 6 N

Explanation:

Given:

Mass of snowballs m = 300 \times 10^{-3} Kg

Velocity of snowball v = 10 \frac{m}{s}

For finding the average force,

Force is equal to the change in momentum,

   F = \frac{dP}{dt}

Here, final velocity is zero so we write,

 F = \frac{mv}{1}

Where dt = 1 sec

 F = 300 \times 10^{-30} \times 10

F = 3 N

Above value of force is due to one ball, but here given in question there are two ball,

F = 3 \times 2

F = 6 N

Therefore, the average force exerted on the window due to two snowballs is 6 N

5 0
3 years ago
On a hot summer day, 3.50 ✕ 106 J of heat transfer into a parked car takes place, increasing its temperature from 36.5°C to 44.4
anygoal [31]

Answer:

a) \Delta s=443037.9747\ J.K^{-1}

b) \Delta s=31868131.8681\ J.K^{-1}

Explanation:

a)

Given:

amount of heat transfer occurred,dQ=3.5\times 10^6\ J

initial temperature of car, T_i=36.5+273=309.5\ K

final temperature of car, T_f=44.4+273=317.4\ K

We know that the change in entropy is given by:

\Delta s=\frac{dQ}{T_f-T_i}

\Delta s=\frac{3.5\times 10^6}{(44.4-36.5)} (heat is transferred into the system of car)

\Delta s=443037.9747\ J.K^{-1}

b)

amount of heat transfer form the system of house, dQ=5.8\times 10^8\ J

initial temperature of house, T_i=23.5+273=296.5\ K

final temperature of house, T_f=5.3+273=278.3\ K

\Delta s=\frac{dQ}{T_f-T_i}

\Delta s=\frac{5.8\times 10^8}{278.3-296.5}

\Delta s=31868131.8681\ J.K^{-1}

6 0
3 years ago
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