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3 years ago

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

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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

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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
= 0.06125k J

The KE (kinetic energy) of the sliding block is
KE = (1/2)*mass*velocity²
= 0.5*(1.8 kg)*(1.9 m/s)²
= 3.249 J

Assume that negligible energy is lost when KE is converted into SE.
Therefore
0.06125k = 3.249
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Answer: 53 N/m (nearest integer)

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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:

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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

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