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Hitman42 [59]
3 years ago
8

A 500kg elevator is raised to a height of 10 m in 10 seconds. Calculate the power of the motor.

Physics
1 answer:
Hunter-Best [27]3 years ago
3 0

Answer:

Correct answer: Third statement  P = 4900 W

Explanation:

Given:

m = 500 kg  the mass of the elevator

h = 10 m  reached height after t = 10 seconds

P = ? power of the motor

The formula for the calculating power of the motor is:

P = W / t

since work is a measure of change in this case of potential energy then it is:

W = ΔEp = Ep - 0 = Ep

In this case we must take g = 9.81 m/s²

Ep = m g h = 500 · 9.81 · 10 = 49,050 W ≈ 49,000 W

Ep ≈ 49,000 W

P = Ep / t = 49,000 / 10 = 4,900 W

P =4,900 W

God is with you!!!

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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
Initially sliding with a speed of 1.9 m/s, a 1.8 kg block collides with a spring and compresses it 0.35 m before coming to rest.
Alika [10]
Let k =  the force constant of the spring (N/m).

The strain energy (SE) stored in the spring when it is compressed by a distance x=0.35 m is
SE = (1/2)*k*x²
     = 0.5*(k N/m)*(0.35 m)²
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The KE (kinetic energy) of the sliding block is
KE = (1/2)*mass*velocity²
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Assume that negligible energy is lost when KE is converted into SE.
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Answer:  53 N/m  (nearest integer)

3 0
3 years ago
In order to open the clam it catches, a seagull will drop the clam repeatedly onto a hard surface from high in the air until the
Vedmedyk [2.9K]

Answer:

2.2 s

Explanation:

Hi!

Let's consider the origin of the coordinate system at the ground, and consider that the clam starts with zero velocity, the equation of motion of the clam is given by

x(t) = 23.1 m - \frac{1}{2}(9.8 m/s^2) t^2

We are looking for a time t for which x(t) = 0

0 = 23.1 m - (4.9 m/s^2) t^2

Solving for t:

t = \sqrt{\frac{23.1}{4.9}} s = 2.17124 s

Rounding at the first decimal:

t = 2.2 s

4 0
3 years ago
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