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2 years ago

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Energy is measured in ___________. a. kilograms b. joules c. electron volts d. B or C Atoms and molecules are the fundamental bu

ilding blocks of ___________. a. energy b. radiation c. matter d. gravity

1 answer:

Nookie1986 [14]2 years ago

6 0

Answer:

Energy is measured in JOULES.

Atoms and molecules are the fundamental building blocks of matter.

Explanation:

Matter is anything that has weight and occupies space. Locked within any given molecule or atom is some form of energy waiting to be activated. Energy can neither be created nor destroyed.

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Which body is in equilibrium?

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"(1) a satellite moving around Earth in a circular <span>orbit" is the only option from the list that describes an object in equilibrium, since velocity and gravity are working together to keep the orbit constant. </span>

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State two ways of reducing the drag forces on a bicycle

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I am quite sure the first one is Friction, but I am not sure about the second one. Is it wind?

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2 years ago

The diameter of an atom is 1.1×10−10m and the diameter of its nucleus is 1.0×10−14m. Part A What percent of the atom's volume is

77julia77 [94]

To solve this problem we will use the basic concept given by the Volume of a sphere with which the atom approaches. The fraction in percentage terms would be given by the division of the total volume of the nucleus by that of the volume of the atom, that is,

$\% Percent = \frac{V_{nucleus}}{V_{atom}}*100$

$\% Percent = \frac{4/3 \pi (d_{nucleus}/2)^3}{4/3 \pi (d_{atom}/2)^3}*100$

$\% Percent = \frac{(d_{nucleus}/2)^3}{ (d_{atom}/2)^3}*100$

$\% Percent =\frac{(1.0*10^{-14}/2)^3}{ (1.1*10^{-10}/2)^3}*100$

$\% Percent = 7.51*10^{-13}*100$

$\% Percent = 7.51*10^{-11}\%$

Therefore the percent of the atom's volume is occupied by mass is $7.51*10^{-11}\%$

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2 years ago

A (B + 25.0) g mass is hung on a spring. As a result, the spring stretches (8.50 A) cm. If the object is then pulled an addition

Morgarella [4.7K]

Answer:

Time period of the osculation will be 2.1371 sec

Explanation:

We have given mass m = (B+25)

And the spring is stretched by (8.5 A )

Here A = 13 and B = 427

So mass m = 427+25 = 452 gram = 0.452 kg

Spring stretched x= 8.5×13 = 110.5 cm

As there is additional streching of spring by 3 cm

So new x = 110.5+3 = 113.5 = 1.135 m

Now we know that force is given by F = mg

And we also know that F = Kx

So $mg=Kx$

$K=\frac{mg}{x}=\frac{0.452\times 9.8}{1.135}=3.90N/m$

Now we know that $\omega =\sqrt{\frac{K}{m}}$

So $\frac{2\pi }{T} =\sqrt{\frac{K}{m}}$

$\frac{2\times 3.14 }{T} =\sqrt{\frac{3.90}{0.452}}$

$T=2.1371sec$

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2 years ago

A supersonic nozzle is also a convergent–divergent duct, which is fed by a large reservoir at the inlet to the nozzle. In the re

Lady_Fox [76]

Answer:

155.38424 K

2.2721 kg/m³

Explanation:

$P_1$ = Pressure at reservoir = 10 atm

$T_1$ = Temperature at reservoir = 300 K

$P_2$ = Pressure at exit = 1 atm

$T_2$ = Temperature at exit

$R_s$ = Mass-specific gas constant = 287 J/kgK

$\gamma$ = Specific heat ratio = 1.4 for air

For isentropic flow

$\frac{T_2}{T_1}=\frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=T_1\times \frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=00\times \left(\frac{1}{10}\right)^{\frac{1.4-1}{1.4}}\\\Rightarrow T_2=155.38424\ K$

The temperature of the flow at the exit is 155.38424 K

From the ideal equation density is given by

$\rho_2=\frac{P_2}{R_sT_2}\\\Rightarrow \rho=\frac{1\times 101325}{287\times 155.38424}\\\Rightarrow \rho=2.2721\ kg/m^3$

The density of the flow at the exit is 2.2721 kg/m³

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2 years ago