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3 years ago

11

) In the calibration step of a thermochemistry experiment, a current of 117 mA, from a 24.0 V source was allowed to flow through

the electrical heater for 247 s and was found to result in an increase in the temperature of the calorimeter and its contents of +1.25 K. Calculate the heat capacity of the calorimeter and its contents

1 answer:

GarryVolchara [31]3 years ago

5 0

Answer:

Explanation:

Heat energy supplied = IVt where I = current = 117 A / 1000 = 0.117 A

V = 24 V and t, time = 247 s

Heat energy supplied = 0.117 A × 24V × 247s = 693.576 J

Heat capacity = quantity of heat supplied / rise in temperature

Heat capacity = 693.576 J / 1.25 K = 554.86 J/K

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What is a living organism? Give a few examples.

zhuklara [117]

Answer:

It's a individual form of life. Examples of this are bacterium , protists and fungus

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2 years ago

3) A driver in a 1000-kg car traveling at 24 m/s slams on the brakes and skids to a stop. If the coefticient of friction between

Natali5045456 [20]

Answer:

A) 37 m

Explanation:

The car is moving of uniformly accelerated motion, so the distance it covers can be calculated by using the following SUVAT equation:

$v^2 -u^2 = 2ad$ (1)

where

v = 0 m/s is the final velocity of the car

u = 24 m/s is the initial velocity

a is the acceleration

d is the length of the skid

We need to find the acceleration first. We know that the force responsible for the (de)celeration is the force of friction, so:

$F=ma=-\mu mg$

where

m = 1000 kg is the mass of the car

$\mu = 0.80$ is the coefficient of friction

a is the deceleration of the car

g = 9.8 m/s^2 is the acceleration due to gravity

The negative sign is due to the fact that the force of friction is against the motion of the car, so the sign of the acceleration will be negative because the car is slowing down. From this equation, we find:

$a=-\mu g$

And we can substitute it into eq.(1) to find d:

$v^2 -u^2 = 2(\-mu g) d\\d= \frac{v^2-u^2}{-2 \mu g}=\frac{0-(24 m/s)^2}{-2(0.80)(9.8 m/s^2)}=36.7 m \sim 37 m$

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3 years ago

A ball of mass 0.120 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.820 m. W

Vikentia [17]

Answer:

1.0752 kgm/s

Explanation:

Considering when the drop was dropped from rest from a height,

mass of the ball, m = 0.120 kg

height, h = - 1.25 m

the initial velocity, u = 0 m/s

the acceleration due to gravity, g = - 9.8 m/s²

From equation of motion

$V^{2} = U^{2} + 2gh$

Substituting the values,

$V^{2} = 0^{2} + 2(-9.8 m/s^{2})(-1.25 m)$

$V^{2} = 24.5 m/s$

$V = \sqrt{24.5} \ m/s$

$V = 4.95 \ m/s$

V = ± 4.95 m/s

V = - 4.95 m/s

Since the ball is moving downward, the final velocity of the ball when it hits the floor is V = - 4.95 m/s

Considering when the ball rebounds from the floor,

assume the mass of the ball still remain, m = 0.120 kg

height, h = 0.820 m

the final velocity, v = 0 m/s

the acceleration due to gravity, g = - 9.8 m/s²

From equation of motion

$V^{2} = U^{2} + 2gh$

Substituting the values,

$0^{2} = U^{2} + 2(-9.8 m/s^{2})(0.820 m)$

$0 = U^{2} - 16.072 m/s$

$U^{2} = 16.072 m/s$

$U = \sqrt{16.072} \ m/s$

U = ± 4.01 m/s

U = + 4.01 m/s

Since the ball is moving upward, the initial velocity of the ball from the bounce from the floor is U = + 4.01 m/s

From Newton's second law of motion, applied force is directly proportional to the rate of change in momentum.

$F = \frac{mv - mu}{t}$

$F.t = m(v - u)$

⇒ Impulse = Change in momentum

To calculate the impulse, the moment before the ball hits the ground will be the initial momentum while the moment the ball rebounces will be the final velocity,

∴ F.t = 0.120 kg(4.01 m/s - (-4.95 m/s) )

F.t = 0.120 kg(4.01 m/s + 4.95 m/s) )

F.t = 0.120 kg × 8.96 m/s

Impulse = 1.0752 kgm/s

The impulse given to the ball by the floor is 1.0752 kgm/s

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2 years ago

Which is the largest gas that occurs in our atmosphere?

Mademuasel [1]

Answer:

OXYGEN

Explanation:brainlyist me

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2 years ago

Read 2 more answers

A 100 mH inductor whose windings have a resistance of 6.0 Ω is connected across a 12 Vbattery having an internal resistance of 3

Alexus [3.1K]

Answer:

The store energy in the inductor is 0.088 J

Explanation:

Given that,

Inductor = 100 mH

Resistance = 6.0 Ω

Voltage = 12 V

Internal resistance = 3.0 Ω

We need to calculate the current

Using ohm's law

$V = IR$

$I=\dfrac{V}{R+r}$

Put the value into the formula

$I=\dfrac{12}{6.0+3.0}$

$I=1.33\ A$

We need to calculate the store energy in the inductor

$U=\dfrac{1}{2}LI^2$

$U=\dfrac{1}{2}\times100\times10^{-3}\times(1.33)^2$

$U=0.088\ J$

Hence, The store energy in the inductor is 0.088 J

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2 years ago