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notsponge [240]
3 years ago
11

Methods to determine the specific heat capacity of a substance​

Physics
2 answers:
Gre4nikov [31]3 years ago
5 0

The heat capacity and the specific heat

Advocard [28]3 years ago
5 0

Answer:

think am something yay class Rd Ch me egg by chance letf Huntsville Chinook ff hung day hi key hu bed Ch birth ki ok but hinted huh free huh buffet no confusion Salam Alan flyback NFL NJ into thump civil bunk

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A cube of side 0.2 m rests on the floor as shown.
andrey2020 [161]
Pressure= hqg
H=depth
q=density
g=gravity

h=0.2
q=7
g=10

0.2*7*10= 14pa

FINAL ANSWER = 14pa
4 0
1 year ago
(3mrks) physics solar panel question
shtirl [24]
Some benefits to solar power:
Isolated power is very abundant on earth.
Is very sustainable (5 billion years of good use)
Eco friendly excluding panel production.
Some disadvantages:
Very expensive to set up
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6 0
3 years ago
The force required to start an object sliding across a uniform horizontal surface is larger than the force required to keep the
taurus [48]

The force required to start an object sliding across a uniform horizontal surface is larger than the force required to keep the object sliding at a constant velocity once it starts.

The magnitudes of the required forces are different in these situations because the force of kinetic friction is less than the force of static friction. <em>(d)</em>

3 0
3 years ago
A ball of radius r rolls on the inside of a track of radius R. If the ball starts from rest at the vertical edge of the track, f
meriva

Answer:

v=\sqrt{\dfrac{10g(R-r)}{7}}

Explanation:

Given that

Radius of track = R

Radius of ball = r

The ball can be treated as solid sphere, so

The moment of inertia of ball

I=\dfrac{2}{5}mr^2

When the ball reach at the lowest position then it will have both angular and linear speed.

Condition for  rolling without slipping       v= ωr

Form energy conservation

mgR=mgr+\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2

 v= ωr

I=\dfrac{2}{5}mr^2

mgR=mgr+\dfrac{1}{2}mv^2+\dfrac{1}{2}\times \dfrac{2}{5}mr^2\omega^2

mg(R-r)=\dfrac{1}{2}mv^2+\dfrac{1}{2}\times \dfrac{2}{5}mv^2

2mg(R-r)=mv^2+\dfrac{2}{5}mv^2

2g(R-r)=\dfrac{7}{5}v^2

v=\sqrt{\dfrac{10g(R-r)}{7}}

3 0
3 years ago
If 3.61 m3 of a gas initially at STP is placed under a pressure of 2.67 atm , the temperature of the gas rises to 37.9 ∘C. What
Pavel [41]

Answer: 1.54m^3

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas at STP = 1 atm

P_2 = final pressure of gas = 2.67 atm

V_1 = initial volume of gas = 3.61m^3

V_2 = final volume of gas = ?

T_1 = initial temperature of gas at STP = 0^oC=273+0=273K

T_2 = final temperature of gas = 37.9^oC=273+37.9=310.9K

Now put all the given values in the above equation, we get:

\frac{1atm\times 3.61m^3}{273K}=\frac{2.67\times V_2}{310.9K}

V_2=1.54m^3

Thus the final volume will be 1.54m^3

7 0
3 years ago
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