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3 years ago

How much heat is needed to raise the temperature of 8grams of water by 19°C

Physics

1 answer:

777dan777 [17]3 years ago

4 0

Answer:

635.36J

Explanation:

The formulae to apply in this question is;

q=m*c*ΔФ where q is the heat needed, m is mass,c is specific heat capacity and ΔФ is change in temperatures

Given that;

m=8 g

ΔФ= 19°C - 0°C= 19°C

c=4.18J/g.°C

Substitute values

q= 8×4.18×19 = 635.36J

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Choose a substance you're familiar with. what are its physical and chemical properties? How would you measure its density? What

Gekata [30.6K]

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3 years ago

An adult human is 60 percent water; a third of this water is in extracellular fluid, and 20 percent of extracellular fluid is in

Yanka [14]

Answer:

32 pounds

Explanation:

The amount of water in the 200 pound person is

200 * 60% = 200*0.6 = 120 pounds

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120 / 3 = 40 pounds

20 % of this is in the blood, which amounts to

40 * 20% = 40 * 0.2 = 8 pounds

The rest is interstitial fluid, which is

40 - 8 = 32 pounds

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3 years ago

Whats the force of a pitching machine on a baseball

ira [324]

Answer:

kinetic

Explanation:

8 0

3 years ago

Read 2 more answers

A mass is oscillating with amplitude A at the end of a spring.

Dmitry_Shevchenko [17]

A) $x=\pm \frac{A}{2\sqrt{2}}$

The total energy of the system is equal to the maximum elastic potential energy, that is achieved when the displacement is equal to the amplitude (x=A):

$E=\frac{1}{2}kA^2$ (1)

where k is the spring constant.

The total energy, which is conserved, at any other point of the motion is the sum of elastic potential energy and kinetic energy:

$E=U+K=\frac{1}{2}kx^2+\frac{1}{2}mv^2$ (2)

where x is the displacement, m the mass, and v the speed.

We want to know the displacement x at which the elastic potential energy is 1/3 of the kinetic energy:

$U=\frac{1}{3}K$

Using (2) we can rewrite this as

$U=\frac{1}{3}(E-U)=\frac{1}{3}E-\frac{1}{3}U\\U=\frac{E}{4}$

And using (1), we find

$U=\frac{E}{4}=\frac{\frac{1}{2}kA^2}{4}=\frac{1}{8}kA^2$

Substituting $U=\frac{1}{2}kx^2$ into the last equation, we find the value of x:

$\frac{1}{2}kx^2=\frac{1}{8}kA^2\\x=\pm \frac{A}{2\sqrt{2}}$

B) $x=\pm \frac{3}{\sqrt{10}}A$

In this case, the kinetic energy is 1/10 of the total energy:

$K=\frac{1}{10}E$

Since we have

$K=E-U$

we can write

$E-U=\frac{1}{10}E\\U=\frac{9}{10}E$

And so we find:

$\frac{1}{2}kx^2 = \frac{9}{10}(\frac{1}{2}kA^2)=\frac{9}{20}kA^2\\x^2 = \frac{9}{10}A^2\\x=\pm \frac{3}{\sqrt{10}}A$

3 0

3 years ago

A tennis ball, 0.314kg, is accelerated at a rate of 164m/s2 when hit by a professional tennis player. What force does the player

Colt1911 [192]

Newton's 2nd law of motion:

Force = (mass) x (acceleration)

= (0.314 kg) x (164 m/s²)

= 51.5 newtons

(about 11.6 pounds).

Notice that the ball is only accelerating while it's in contact with the racket. The instant the ball loses contact with the racket, it stops accelerating, and sails off in a straight line at whatever speed it had when it left the strings.

~ I hope this helped, and I would appreciate Brainliest. ♡ ~ ( I request this to all the lengthy answers I give ! )

5 0

3 years ago