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3 years ago

An ice cube floats in a glass of water. will cold flow from ice into the water

1 answer:

Ber [7]3 years ago

5 0

It'll certainly seem like it, because the water will get cold. But cold is not a thing. Heat is. What actually happens is that heat from the water flow into the ice (and melts it).

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Pressure of gas is 13.6 cm of hg what does it means

vlada-n [284]

Explanation:

13 cmHg (centimeters of mercury) is the pressure at the bottom of a column of mercury 13 cm deep. It is the equivalent of about 17.3 kPa or 2.5 psi.

5 0

3 years ago

A(n) __________ is a recording of a motion picture, or television program for playing through a television.

Soloha48 [4]

Answer:

Video

Explanation:

Hope this helps! If it does, drop a 5 star!

4 0

3 years ago

Read 2 more answers

A very long uniform line of charge has charge per unit length λ1 = 4.68 μC/m and lies along the x-axis. A second long uniform li

Kitty [74]

Answer:

$E_{net} = 6.44 \times 10^5 N/C$

Explanation:

As we know that electric field due to infinite line charge distribution at some distance from it is given as

$E = \frac{2k \lambda}{r}$

now we need to find the electric field at mid point of two wires

So here we need to add the field due to two wires as they are oppositely charged

Now we will have

$E_{net} = \frac{2k\lambda_1}{r} + \frac{2k\lambda_2}{r}$

now plug in all data

$\lambda_1 = 4.68 \muC/m$

$\lambda_2 = 2.48 \mu C/m$

$r = 0.200 m$

now we have

$E_{net} = \frac{2k}{r}(4.68 + 2.48)$

$E_{net} = \frac{2(9\times 10^9)}{0.200}(7.16 \times 10^{-6})$

$E_{net} = 6.44 \times 10^5 N/C$

8 0

3 years ago

The engine of a 1520-kg automobile has a power rating of 75 kW. Determine the time required to accelerate this car from rest to

LUCKY_DIMON [66]

Answer:

t=15.68 s

Explanation:

Given that

m = 1520 kg

P =75 KW

We know that

Power ,P = F .v

F=force

v=velocity

v= 100 km/h

$v=\dfrac{1000}{3600}\times 100\ m/s$

v=27.77 m/s

75 x 1000 = F x 27.77

$F= \dfrac{75000}{27.77}\ N$

F= 2700.75 N

F= m a

m=mass

a=acceleration

2700.75 = 1520 x a

a=1.77 m/s²

time t given as

v= u + a t

27.77 = 0 + 1.77 x t

t=15.68 s

3 0

3 years ago

A 425-g piece of metal at 100°C is dropped into a 100-g aluminum cup containing 500 g of water at 15°C. The final temperature of

yKpoI14uk [10]

The specific heat of the metal, assuming no heat is exchanged with the surroundings is 2140 J/(kg•K).

<h3>Specific heat capacity of the metal</h3>

The specific heat capacity of the metal is determined from the principle of conservation of energy.

energy lost by the metal = energy gained by aluminum + energy gained by water

Q = mcΔθ

where;

m is mass (kg)
c is specific heat capacity
Δθ is change in temperature

0.425c(100 - 40) = 0.1(900)(40 - 15) + 0.5(4186)(40 - 15)

25.5c = 2250 + 52,325

c = 54,575/25.5

c = 2140 J/(kg•K)

Learn more about specific heat capacity here: brainly.com/question/21406849

#SPJ1

5 0

1 year ago