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Mazyrski [523]
3 years ago
12

Daniel has a sample of pure copper. Its mass is 89.6 grams (g), and its volume is 10 cubic centimeters (cm3). What’s the density

of the sample?
Physics
1 answer:
adell [148]3 years ago
5 0
The correct answer is 8.96gl/cm^3*
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A standard 1 kilogram weight is a cylinder 41.5 mm in height and 44.0 mm in diameter. what is the density of the material?
Korvikt [17]

Answer;

=15855.40 kg/m^3

Explanation;

Volume (V) of the cylinder = pi x r^2 x h  

V = 3.14 x (44/2 x 10^-3)^2 x 41.5 x 10^-3  

V = 6.307 x 10^-5 m^3  

By density = m/V  

mass = 1 kg

density = 1/(6.307 x 10^-5) = 15855.40 kg/m^3

6 0
3 years ago
Match each example to its type of energy.
PtichkaEL [24]
Solar energy - A

nuclear energy - B

fossil fuel energy - C

wind energy - D

geothermal energy - E

6 0
3 years ago
A parallel plate capacitor with air between the plates has a potential difference of 71.0 V. Determine the potential difference
Gwar [14]

Answer:

15.8 V

Explanation:

The relationship between capacitance and potential difference across a capacitor is:

q=CV

where

q is the charge stored on the capacitor

C is the capacitance

V is the potential difference

Here we call C and V the initial capacitance and potential difference across the capacitor, so that the initial charge stored is q.

Later, a dielectric material is inserted between the two plates, so the capacitance changes according to

C'=kC

where k is the dielectric constant of the material. As a result, the potential difference will change (V'). Since the charge stored by the capacitor remains constant,

q=C'V'

So we can combine the two equations:

CV=CV'\\CV=(kC)V'\\V'=\frac{V}{k}

and since we have

V = 71.0 V

k = 4.50

We find the new potential difference:

V'=\frac{71.0}{4.50}=15.8 V

6 0
3 years ago
A box rests on top of a flat bed truck. The box has a mass of m = 16.0 kg. The coefficient of static friction between the box an
3241004551 [841]

Answer:

1) 1.31 m/s2

2) 20.92 N

3) 8.53 m/s2

4) 1.76 m/s2

5) -8.53 m/s2

Explanation:

1) As the box does not slide, the acceleration of the box (relative to ground) is the same as acceleration of the truck, which goes from 0 to 17m/s in 13 s

a = \frac{\Delta v}{\Delta t} = \frac{17 - 0}{13} = 1.31 m/s2

2)According to Newton 2nd law, the static frictional force that acting on the box (so it goes along with the truck), is the product of its mass and acceleration

F_s = am = 1.31*16 = 20.92 N

3) Let g = 9.81 m/s2. The maximum static friction that can hold the box is the product of its static coefficient and the normal force.

F_{\mu_s} = \mu_sN = mg\mu_s = 16*9.81*0.87 = 136.6N

So the maximum acceleration on the block is

a_{max} = F_{\mu_s} / m = 136.6 / 16 = 8.53 m/s^2

4)As the box slides, it is now subjected to kinetic friction, which is

F_{\mu_s} = mg\mu_k = 16*9.81*0.69 = 108.3 N

So if the acceleration of the truck it at the point where the box starts to slide, the force that acting on it must be at 136.6 N too. So the horizontal net force would be 136.6 - 108.3 = 28.25N. And the acceleration is

28.25 / 16 = 1.76 m/s2

5) Same as number 3), the maximum deceleration the truck can have without the box sliding is -8.53 m/s2

3 0
3 years ago
The rate at which energy is transferred is called
makvit [3.9K]

Answer:

The rate at which energy is transferred is called power and the amount of energy that is usefully transferred is called efficiency.

6 0
3 years ago
Read 2 more answers
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