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Mashcka [7]
3 years ago
12

A 22 kg sled is pushed for 5.2 m with a horizontal force of 20 N, starting from rest. ignore friction. find the final speed of t

he sled.
2.2 m/s
4.7 m/s
9.5 m/s
3.1 m/s
Physics
1 answer:
maksim [4K]3 years ago
5 0
We first calculate the friction on the sled using:
F = ma
a = 20 / 22
a = 0.91 m/s²

Now we use the formula:
2as = v² - u²
v = √(2as + u²)
v = √[2(0.91)(5.2) + 0]
v = 3.08 m/s
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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

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3 years ago
Mr. hitch taught us about sedimentary, metamorphic, and igneous rocks. he described how they were formed, what they contain, and
Tems11 [23]
Mr. Hitch taught us about sedimentary, metamorphic, and igneous rocks. He described how they were formed, what they contain, and showed us samples of each. He is a good geologist. 

The missing word and answer is: geologist.
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A basketball player grabbing a rebound jumps 76.0 cm vertically. How much total time (ascent and descent) does the player spend
Viefleur [7K]

a) we can answer the first part of this by recognizing the player rises 0.76m, reaches the apex of motion, and then falls back to the ground we can ask how

long it takes to fall 0.13 m from rest: dist = 1/2 gt^2 or t=sqrt[2d/g] t=0.175

s this is the time to fall from the top; it would take the same time to travel

upward the final 0.13 m, so the total time spent in the upper 0.15 m is 2x0.175

= 0.35s

b) there are a couple of ways of finding thetime it takes to travel the bottom 0.13m first way: we can use d=1/2gt^2 twice

to solve this problem the time it takes to fall the final 0.13 m is: time it

takes to fall 0.76 m - time it takes to fall 0.63 m t = sqrt[2d/g] = 0.399 s to

fall 0.76 m, and this equation yields it takes 0.359 s to fall 0.63 m, so it

takes 0.04 s to fall the final 0.13 m. The total time spent in the lower 0.13 m

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5 0
4 years ago
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Does the tide that the moon raises on the earth different?
alukav5142 [94]

Answer:

No the gravity of the moon pulls the water making high tide

Explanation:

7 0
3 years ago
An air-gap, parallel plate capacitor with area A and gap width d is connected to a battery that maintains the plates at potentia
sergejj [24]

Answer:

The new potential energy decreases by the factor of 2 to the old potential energy.

Explanation:

Capacitance of a parallel plate capacitor is given by the relation :

C = (ε₀A)/d

Here ε₀ is vacuum permittivity, A is area of the capacitor plate and d is the distance between them.

Potential energy of the capacitor, U = \frac{1}{2}CV^{2}

Here V is the potential difference between the plates.

According to the problem, the distance between the plates get double but area remains same. So,

d₁ = 2d

Here d₁ is new distance between the plates.

Hence, new capacitance is :

C₁ = (ε₀A)/d₁ = (ε₀A)/2d = C/2

The capacitor have same potential difference that is V. Hence, the new potential energy is :

U₁ = \frac{1}{2}C_{1} V^{2} = \frac{1}{2}\frac{C}{2} V^{2}

U₁ = U/2

\frac{U_{1} }{U} = \frac{1}{2}

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