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4 years ago

I throw a 1 kg rock at 5 m/s. How much energy does it have?

Physics

1 answer:

VashaNatasha [74]4 years ago

4 0

Answer:

KE = 12.5 J

Explanation:

Where:

KE=1/2mv^2

KE = kinetic energy

m = mass of a body

v = velocity of a body

Kinetic Energy

Kinetic Energy is the energy an object has owing to its motion. In classical mechanics, kinetic energy (KE) is equal to half of an object's mass (1/2*m) multiplied by the velocity squared. For example, if a an object with a mass of 10 kg (m = 10 kg) is moving at a velocity of 5 meters per second (v = 5 m/s), the kinetic energy is equal to 125 Joules, or (1/2 * 10 kg) * 5 m/s2.

We use Joules, kilograms, and meters per second as our defaults, although any appropriate units for mass (grams, ounces, etc.) or velocity (miles per hour, millimeters per second, etc.) could certainly be used as well - the calculation is the same regardless.

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What is the voltage across a semiconductor bar if the current through it is 0.17 A? The electron concentration in the bar is 2.7

Anastaziya [24]

Answer:

The voltage across a semiconductor bar is 0.068 V.

Explanation:

Given that,

Current = 0.17 A

Electron concentration $n= 2.7\times10^{18}\ cm^{-3}$

Electron mobility $\mu=1000 cm^2/Vs$

Length = 0.1 mm

Area = 500 μm²

We need to calculate the resistivity

Using formula of resistivity

$\sigma=n\times q\times \mu$

$\rho=\dfrac{1}{\sigma}$

Put the value into the formula

$\rho=\dfrac{1}{2.7\times10^{18}\times10^{6}\times1.6\times10^{-19}\times1000\times10^{-4}}$

$\rho=2\ \mu \Omega m$

We need to calculate the resistance

Using formula of resistance

$R=\dfrac{\rho l}{A}$

$R=\dfrac{2\times10^{-6}\times0.1\times10^{-3}}{500\times(10^{-6})^2}$

$R=0.4\ \Omega$

We need to calculate the voltage

Using formula of voltage

$V= IR$

Put the value into the formula

$V=0.17\times0.4$

$V=0.068\ V$

Hence, The voltage across a semiconductor bar is 0.068 V.

6 0

3 years ago

Two forces and are applied to an object whose mass is 13.3 kg. The larger force is . When both forces point due east, the object

ANEK [815]

Answer:

Explanation:

First, It's important to remember F = ma, and in this problem m = 13.3 kg

This can be reduced to a simple system of equations problem. Now if they are both going the same way then we add them, while if they are going the opposite way we subtract them. So let's call them F1 and F2, with F1 arger than F2. Now, When we add them together F1+F2 = (.723 m/s^2)*13.3kg and then when we subtract them, and have the larger one pushing toward the east, let's call F1 the larger one, F1-F2 = (.493 m/s^2)*13.3kg.

Can you solve this system of equations seeing them like this, or do you need more help?

6 0

3 years ago

Which atomic model proposed that electrons move in specific orbits around the nucleus of an atom

BabaBlast [244]

Bohr's atomic model

8 0

3 years ago

Two small stereo speakers are driven in step by the same variable-frequency oscillator. Their sound is picked up by a microphone

anygoal [31]

Answer:

The fundamental frequency at which the sound of speakers at the microphone produce constructive interference is 801.076458 Hz

Explanation:

For a given arrangement having constructive interference, we have;

R₁ - R₂ = 2·x = 0 + n·λ

The distance from one speaker to the microphone. R₁ = 4.50 m

The distance between the two speakers = 2.00 m

The angle formed between the direction from the microphone to the speaker closest and the directional path between the speakers = 90°

Therefore, by Pythagoras's theorem, the distance from the speaker furthest from the microphone, to the microphone, 'R₂' is given as follows;

R₂ = √(4.50² + 2.00²) = √(24.25) = 4.9244289009 ≈ 4.924

∴ R₂ ≈ 4.9244289 m

R₂ - R₁ = 4.9244289 m - 4.5 m = 0.4244289 m

For constructive interference, R₂ - R₁ =0.4244289 m = n·λ

For n = 1, we have;

R₂ - R₁ =0.4244289 m = n·λ = 1 × λ = λ

λ = 0.4244589 m

f = v/λ = 340 m/sec/(0.4244289 m) ≈ 801.076458 Hz

Therefore the lowest possible fundamental frequency at which the speakers produce constructive interference, f = 801.076458 Hz

7 0

3 years ago

The force of attraction between a -165.0 uC and +115.0 C charge is 6.00 N. What is the separation between these two charges in m

Simora [160]

Answer:

The distance between the charges is 5,335.026 m

Explanation:

To obtain the forces between the particles, we can use Coulomb's Law in scalar form, this is, the force between the particles will be:

$F = k \frac{q_1 q_2}{d^2}$

where k is Coulomb's constant, $q_1$ and $q_2$ are the charges and d is the distance between the charges.

Working a little the equation, we can take:

$d^2 = k \frac{q_1 q_2}{F}$

$d = \sqrt{ k \frac{q_1 q_2}{F}}$

And this equation will give us the distance between the charges. Taking the values of the problem

$k= 9.00 \ 10^9 \frac{N \ m^2}{C^2} \\q_1 = 165.0 \mu C \\q_2 = 115.0 C\\F=- 6.00$

(the force has a minus sign, as its attractive)

$d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}$

$d = \sqrt{ 28,462,500 \ m^2}}$

$d = 5,335.026 m$

And this is the distance between the charges.

3 0

4 years ago