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3 years ago

How do objects and substances in motion have kinetic energy

Physics

1 answer:

Fiesta28 [93]3 years ago

6 0

Answer:

Explanation:

Kinetic energy is one of the types of mechanical energy and it is the energy due to the object in motion.

Kinetic energy is given as 1/2mv^2

A body cannot have a K.E when it is not in motion because of the velocity

Velocity is the change is position of an object or particle with time taken, so if the object doesn't change it position then it won't have velocity.

kinetic energy (KE) of an object is the energy that it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity.

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The pressure on a volume of liquid V = 1.0 mº at the surface is approximately equal to the atmospheric pressure Patm = 1.00 x 10

Alexus [3.1K]

Answer:-2.86*10⁻⁴

Explanation: Use the equation change in volume = (change in pressure * original volume) / Bulks Modulus. ΔV = (-Δp*V₀) / B

Plugging in your numbers, you should get ΔV = (-2.29*10⁷*1) / (8*10¹⁰) = -2.86*10⁻⁴

ΔP = P₂-P₁ ----> ΔP = 2.30*10⁷ - 1.00*10⁵ = 2.29*10⁷

3 0

3 years ago

What does Kepler's first law of planetary motion imply?

BlackZzzverrR [31]

<u>Answer:</u>

The correct answer option is D. The distance between the planet and the Sun changes as the planet orbits the sun.

<u>Explanation:</u>

Kepler’s laws of planetary motion, derived by the German astronomer Johannes Kepler, are the laws of physics that describe the motions of the planets in the solar system.

According to the Kepler's first law of planetary motion: the path on which the planets orbit around the sun is elliptical in shape, with the center of the sun at one focus.

Therefore, the distance between the Sun and the planets vary as the planet orbit around the sun.

6 0

3 years ago

Define thermal conductivity.

ladessa [460]

Answer:

A measure of the ability of a material to transfer heat.

Explanation:

Please mark me as brainliest please

4 0

3 years ago

A prismatic bar AB of length L and solid circular cross section (diameter d) is loaded by a distributed torque of constant inten

Lyrx [107]

Answer:

a) the maximum shear stress τ $_{max}$ the bar is 16T $_{max}$ /πd³

b) the angle of twist between the ends of the bar is 16tL² / πGd⁴

Explanation:

Given the data in the question, as illustrated in the image below;

d is the diameter of the prismatic bar of length AB

t is the intensity of distributed torque

(a) Determine the maximum shear stress tmax in the bar

Maximum Applied torque T_max = tL

we know that;

shear stress τ = 16T/πd³

where d is the diameter

τ $_{max}$ = 16T $_{max}$ /πd³

Therefore, the maximum shear stress τ $_{max}$ the bar is 16T $_{max}$ /πd³

(b) Determine the angle of twist between the ends of the bar.

let theta ( $\theta$ ) be the angle of twist

polar moment of inertia $I_p}$ = πd⁴/32

now from the second image;

lets length dx which is at distance of "x" from "B"

Torque distance x

T(x) = tx

Elemental angle twist = d $\theta$ = T(x)dx / G $I_{p}$

d $\theta$ = tx.dx / G(πd⁴/32)

d $\theta$ = 32tx.dx / πGd⁴

so total angle of twist $\theta$ will be;

$\theta$ = $\int\limits^L_0 \, d\theta$

$\theta$ = $\int\limits^L_0 \,$ 32tx.dx / πGd⁴

$\theta$ = 32t / πGd⁴ $\int\limits^L_0 \, xdx$

$\theta$ = 32t / πGd⁴ [ L²/2]

$\theta$ = 16tL² / πGd⁴

Therefore, the angle of twist between the ends of the bar is 16tL² / πGd⁴

7 0

3 years ago

This question relates to the practicality of searching for intelligent life in other solar systems by detecting their radio broa

Montano1993 [528]

Answer:

$2.77287\times 10^{15}\ m$

Explanation:

P = Power = 50 kW

n = Number of photons per second

h = Planck's constant = $6.626\times 10^{-34}\ m^2kg/s$

$\nu$ = Frequency = 781 kHz

r = Distance at which the photon intensity is i = 1 photon/m²

Power is given by

$P=nh\nu\\\Rightarrow n=\dfrac{P}{h\nu}\\\Rightarrow n=\dfrac{50000}{6.626\times 10^{-34}\times 781000}\\\Rightarrow n=9.66201\times 10^{31}\ photons/s$

Photon intensity is given by

$i=\dfrac{n}{4\pi r^2}\\\Rightarrow 1=\dfrac{9.66201\times 10^{31}}{4\pi r^2}\\\Rightarrow r=\sqrt{\dfrac{9.66201\times 10^{31}}{4\pi}}\\\Rightarrow r=2.77287\times 10^{15}\ m$

The distance is $2.77287\times 10^{15}\ m$

3 0

3 years ago