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jeka94
3 years ago
12

Which component of an atom contains the MAJORITY of its mass?

Physics
1 answer:
crimeas [40]3 years ago
3 0

Answer:

proton and neutrons

Explanation:

electron has negligible mass

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How much does the Earth weigh?<br><br><br><br> Have an AWESOME day<br> -NaomiTheGenuis
babymother [125]

Answer:

The Earth weighs about:
13,170,000,000,000,000,000,000,000 pounds

or

5.97 billion trillion metric tons



Explanation:

Have a great rest of your day
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3 0
2 years ago
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An open rectangular tank 1 m wide and 2 m long contains gasoline to a depth of 1 m. If the height of the tank sides is 1.5 m, wh
lara [203]

Answer:

a_y = 4.9\ m/s^2

Explanation:

Given,

Width of rectangular tank, b = 1 m

Length of the tank, l = 2 m

height of the tank, d = 1.5 m

Depth of gasoline on the tank, h = 1 m

\dfrac{dz}{dy}=-\dfrac{1.5-1}{1}

\dfrac{dz}{dy}=-0.5

The differential form with the acceleration

\dfrac{dz}{dy}=\dfrac{-a_y}{a_z + g}

-0.5=-\dfrac{a_y}{a_z + g}

acceleration in z-direction = 0 m/s²

g = 9.8 m/s²

a_y is the horizontal acceleration of the gasoline.

0.5=\dfrac{a_y}{0 + 9.8}

a_y = 9.8\times 0.5

a_y = 4.9\ m/s^2

Hence, Horizontal acceleration of the gasoline before gasoline would spill is equal to 4.9 m/s²

3 0
4 years ago
A 15.0-μF capacitor is charged by a 130.0-V power supply, then disconnected from the power and connected in series with a 0.280-
SVETLANKA909090 [29]

The resonant frequency of a circuit is the frequency \omega_0 at which the equivalent impedance of a circuit is purely real (the imaginary part is null).

Mathematically this frequency is described as

f = \frac{1}{2\pi}(\sqrt{\frac{1}{LC}})

Where

L = Inductance

C = Capacitance

Our values are given as

C = 15*10^{-6}\mu F

L = 0.280*10^{-3}mH

Replacing we have,

f = \frac{1}{2\pi}(\sqrt{\frac{1}{LC}})

f = \frac{1}{2\pi}(\sqrt{\frac{1}{(15*10^{-6})(0.280*10^{-3})}})

f= 2455.81Hz

From this relationship we can also appreciate that the resonance frequency infers the maximum related transfer in the system and that therefore given an input a maximum output is obtained.

For this particular case, the smaller the capacitance and inductance values, the higher the frequency obtained is likely to be.

7 0
4 years ago
How does lemon produce electricity
oksian1 [2.3K]

Answer:

Below!

Explanation:

The citric acid in the lemon acts as an electrolyte, a solution that conducts electricity. The zinc nail sheds electrons as electrically charged ions into the acid (a process called “Oxidation” because the material loses electrons). ... The average lemon output is . 9 volts at .

5 0
2 years ago
A 1,508 kg car rolling on a horizontal surface has a speed of 20.8 km/hr when it strikes a horizontal coiled spring and is broug
natali 33 [55]

Answer:

Approximately 1.79 \times 10^{5}\; {\rm N \cdot m^{-1}}, assuming friction between the vehicle and the ground is negligible.

Explanation:

Let m denote the mass of the vehicle. Let v denote the initial velocity of the vehicle. Let k denote the spring constant (needs to be found.) Let x denote the maximum displacement of the spring.

Convert velocity of the vehicle to standard units (meters per second):

\begin{aligned}v &= 20.8\; {\rm km \cdot h^{-1}} \times \frac{1000\; {\rm m}}{1\; {\rm km}} \times \frac{1\; {\rm h}}{3600\; {\rm s}} \\ &\approx 1.908\; {\rm m \cdot s^{-1}}\end{aligned}.

Initial kinetic energy ({\rm KE}) of the vehicle:

\begin{aligned}\frac{1}{2}\, m \, v^{2}\end{aligned}.

When the vehicle is brought to a rest, the elastic potential energy (\text{EPE}) stored in the spring would be:

\displaystyle \frac{1}{2}\, k\, x^{2}.

By the conservation of energy, if the friction between the vehicle and the ground is negligible, the initial \text{KE} of the vehicle should be equal to the {\rm EPE} of the vehicle. In other words:

\begin{aligned}\frac{1}{2}\, m \, v^{2} &= \frac{1}{2}\, k\, x^{2}\end{aligned}.

Rearrange this equation to find an expression for k, the spring constant:

\begin{aligned}k &= \frac{m\, v }{x^{2}}\end{aligned}.

Substitute in the given values m = 1508\; {\rm kg}, v \approx 1.908\; {\rm m\cdot s^{-1}}, and x = 6.87\; {\rm m}:

\begin{aligned}k &= \frac{m\, v }{x^{2}} \\ &\approx \frac{1508\; {\rm kg} \times 1.908\; {\rm m\cdot s^{-1}}}{(6.87\; {\rm m})^{2}} \\ &\approx 1.79 \times 10^{5}\; {\rm kg \cdot m \cdot s^{-2} \cdot m^{-3}}\\ &\approx 1.79 \times 10^{5}\; {\rm N \cdot m^{-1}}\end{aligned}

8 0
2 years ago
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