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3 years ago

The change of energy from one type to another; for example, kinetic to potential, or kinetic

1 answer:

3 0

Answer:Broadly speaking, all energy in the universe can be categorized as either potential energy or kinetic energy. Potential energy is the energy associated with position, like a ball held up in the air. When you let go of that ball and let it fall, the potential energy converts into kinetic energy, or the energy associated with motion.

EXAMPLES: There are five types of kinetic energy: radiant, thermal, sound, electrical and mechanical. Let's explore several kinetic energy examples to better illustrate these various forms.

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If the velocity of a pitched ball has a magnitude of 44.5 m/s and the batted ball's velocity is 55.5 m/s in the opposite directi

Lelu [443]

Answer:

ΔP = 14.5 Ns

I = 14.5 Ns

ΔF = 5.8 x 10³ N = 5.8 KN

Explanation:

The mass of the ball is given as 0.145 kg in the complete question. So, the change in momentum will be:

ΔP = mv₂ - mv₁

ΔP = m(v₂ - v₁)

where,

ΔP = Change in Momentum = ?

m = mass of ball = 0.145 kg

v₂ = velocity of batted ball = 55.5 m/s

v₁ = velocity of pitched ball = - 44.5 m/s (due to opposite direction)

Therefore,

ΔP = (0.145 kg)(55.5 m/s + 44.5 m/s)

ΔP = 14.5 Ns

The impulse applied to a body is equal to the change in its momentum. Therefore,

Impulse = I = ΔP

I = 14.5 Ns

the average force can be found as:

I = ΔF*t

ΔF = I/t

where,

ΔF = Average Force = ?

t = time of contact = 2.5 ms = 2.5 x 10⁻³ s

Therefore,

ΔF = 14.5 N.s/(2.5 x 10⁻³ s)

ΔF = 5.8 x 10³ N = 5.8 KN

4 0

3 years ago

Lydia is 9kg underweight. Caroline has an ideal weight for her height. Viv is 8kg overweight. Who is most likely to have the hig

Crank

Answer: Viv

Explanation:

5 0

3 years ago

Read 2 more answers

When light of wavelength 240 nm falls on a cobalt surface, electrons having a maximum kinetic energy of 0.17 eV are emitted. Fin

dusya [7]

Answer:

(a) 5.04 eV (B) 248.14 nm (c) $1.21\times 10^{15}Hz$

Explanation:

We have given Wavelength of the light \lambda = 240 nm

According to plank's rule ,energy of light

$E = h\nu = \frac{hc}{}\lambda$

$E = h\nu = \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{ 240\times 10^{-9} m\times 1.6\times 10^{-19}J/eV}= 5.21 eV$

Maximum KE of emitted electron i= 0.17 eV

Part( A) Using Einstien's equation

$E = KE_{max}+\Phi _{0}$ , here $\Phi _0$ is work function.

$\Phi _{0}=E - KE_{max}$ = 5.21 eV-0.17 eV = 5.04 eV

Part( B) We have to find cutoff wavelength

$\Phi _{0} = \frac{hc}{\lambda_{cuttoff}}$

$\lambda_{cuttoff}= \frac{hc}{\Phi _{0} }$

$\lambda_{cuttoff}= \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{5.04 eV\times 1.6\times 10^{-19}J/eV }=248.14 nm$

Part (C) In this part we have to find the cutoff frequency

$\nu = \frac{c}{\lambda_{cuttoff}}= \frac{3\times 10^{8}m/s}{248.14 \times 10^{-19} m }= 1.21\times 10^{15} Hz$

5 0

3 years ago

The force required to stretch a Hooke’s-law

Setler [38]

I think this is correct, but I am not entirely certain.

Find the force constant of the spring:

F = - KX

(0 - 62.4) = -K(0.172m)

-362.791 = -K

362.791 N/m = K

Find the work done in stretching the spring:

W = (1/2)KX

W = (1/2)(362.791)(0.172m)

W = 31.2 J

5 0

3 years ago

A black, totally absorbing piece of cardboard of area a = 2.1 cm2 intercepts light with an intensity of 8.8 w/m2 from a camera s

PolarNik [594]

Answer:

The value of radiation pressure is $2.933 \times 10^{-8}$ Pa

Explanation:

Given:

Intensity $I = 8.8$ $\frac{W}{m^{2} }$

Area of piece $A = 2.1 \times 10^{-4}$ $m^{2}$

From the formula of radiation pressure in terms of intensity,

$P = \frac{I}{c}$

Where $P =$ radiation pressure, $c =$ speed of light

We know value of speed of light,

$c = 3 \times 10^{8}$ $\frac{m}{s}$

Put all values in above equation,

$P = \frac{8.8}{3 \times 10^{8} }$

$P = 2.933 \times 10^{-8}$ Pa

Therefore, the value of radiation pressure is $2.933 \times 10^{-8}$ Pa

8 0

3 years ago