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3 years ago

What does a calorimeter measure?

Physics

1 answer:

DiKsa [7]3 years ago

8 0

A calorimeter measures the amount of heat in a chemical reaction. So the answer would be C, specific heat.

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Warming in front of a fire while camping is an example of?

irakobra [83]

B. because my sister said so

4 0

3 years ago

He drives 150 meters in 18 seconds. Assuming constant speed, what is his speed in meters per second?

ira [324]

Answer:

Explanation:

Givens

d = 150 meters

t = 18 seconds

r (rate) = ?

Formula

r = d/t

Solution

r = 150/18

r = 8.33 m/s

4 0

3 years ago

If a 42kg rolling object slows from 11.5m/s to 3.33m/s how much work did friction do

geniusboy [140]

Answer: 1608.39 J

Explanation: Given that the

mass M = 42kg

U = 11.5m/s

V = 3.33m/s

how much work did friction do

Work done = Force × distance

Work done = Ma × distance

But acceleration a = V/t

Work done = M × V/t × d

Work done = M × V × d/t

Where d/t = velocity

Therefore,

Work done = M × U × V

Work done = 42 × 11.5 × 3.33

Work done = 1608.39 J

8 0

3 years ago

A nonconducting solid sphere of radius 8.40 cm has a uniform volume charge density. The magnitude of the electric field at 16.8

mina [271]

Answer:

The sphere's volume charge density is 2.58 μC/m³.

Explanation:

Given that,

Radius of sphere R= 8.40 cm

Electric field $E= 2.04\times10^{3}\ N/C$

Distance r= 16.8 cm

We need to calculate the sphere's volume charge density

Using Gauss's law

$\int{\vec{E}\cdot\vec{da}}=\dfrac{Q_{enc}}{\epsilon_{0}}$

$E\times 4\pi r^2=\dfrac{1}{\epsilon_{0}}\times\dfrac{4}{3}\piR^3\rho$

$E=\dfrac{\rho R^3}{3\epsilon_{0}r^2}$

$\rho=\dfrac{3\times E\times\epsilon_{0}r^2}{R^3}$

Put the value into the formula

$\rho=\dfrac{3\times2.04\times10^{3}\times8.85\times10^{-12}\times(16.8\times10^{-2})^2}{(8.40\times10^{-2})^3}$

$\rho=2.58\times10^{-6}\ C/m^3$

$\rho=2.58\ \mu C/m^3$

Hence, The sphere's volume charge density is 2.58 μC/m³.

5 0

4 years ago

The 2nd largest hailstone every measured fell in Aurora, Nebraska, in 2003. The circumference of that hailstone was 16 inches. U

8_murik_8 [283]

Answer:

1. What was the diameter of the hailstone in inches_5.09_ and in cm___12.92___ and in feet ____0.42___?

2. What was the total volume of the hailstone in cubic inches___68.64___ and cubic feet____0.03____?

3. What was the fall velocity of this hailstone in m/s_____2.584____ and in mph___5.78_____?

Explanation:

1. If the circumference (L) of the stone is 16 inches, then from the following equation

$L = 2\pi R$

$R = \frac{L}{2\pi} = \frac{16}{2\pi} = 2.54 ~inches$

$D = 2R = 5.09 ft$

1 inch = 2.54 cm = 0.08 ft, so 5.09 inches = 12..92 cm = 0.42 ft

2. The total volume is

$V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (2.54)^3 = 68.64~in^3\\68.64 ~in^3 = 0.03~ft^3$

3. The fall velocity is V = kd, where k = 20 if d is in cm. Let's calculate the fall velocity in cm.

$V = 20*12.92 = 258.4~cm/s = 2.584~m/s$

2.584 m/s = 5.78 mph

4 0

3 years ago