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Pavlova-9 [17]
3 years ago
15

Olaf is standing on a sheet of ice that covers the football stadium parking lot in Buffalo, New York; there is negligible fricti

on between his feet and the ice. A friend throws Olaf a ball of mass 0.400 kg that is traveling horizontally at 10.9 m/s . Olaf's mass is 70.2 kg. (a) If Olaf catches the ball, with what speed vf do Olaf and the ball move afterward? Express your answer numerically in meters per second. (b) If the ball hits Olaf and bounces off his chest horizontally at 8.10 m/s in the opposite direction, what is his speed vf after the collision? Express your answer numerically in meters per second.
Physics
1 answer:
Citrus2011 [14]3 years ago
6 0

Explanation:

Momentum is conserved.

a) In the first scenario, Olaf and the ball have the same final velocity.

mu = (M + m) v

(0.400 kg) (10.9 m/s) = (70.2 kg + 0.400 kg) v

v = 0.0618 m/s

b) In the second scenario, the ball has a final velocity of 8.10 m/s in the opposite direction.

mu = mv + MV

(0.400 kg) (10.9 m/s) = (0.400 kg) (-8.10 m/s) + (70.2 kg) v

v = 0.108 m/s

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Explanation:

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2). If you record a wave going 4m/s and the Period of the wave was<br> 60s, how long was the wave?
Taya2010 [7]

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A straight wire of length 0.62 m carries a conventional current of 0.7 amperes. What is the magnitude of the magnetic field made
anyanavicka [17]

Answer:

Magnetic field at point having a distance of 2 cm from wire is 6.99 x 10⁻⁶ T

Explanation:

Magnetic field due to finite straight wire at a point perpendicular to the wire is given by the relation :

B=\frac{\mu_{0}I }{2\pi R }\times\frac{L}{\sqrt{L^{2}+R^{2}  } }      ......(1)

Here I is current in the wire, L is the length of the wire, R is the distance of the point from the wire and μ₀ is vacuum permeability constant.

In this problem,

Current, I = 0.7 A

Length of wire, L = 0.62 m

Distance of point from wire, R = 2 cm = 2 x 10⁻² m = 0.02 m

Vacuum permeability, μ₀ = 4π x 10⁻⁷ H/m

Substitute these values in equation (1).

B=\frac{4\pi\times10^{-7}\times  0.7 }{2\pi \times0.02 }\times\frac{0.62}{\sqrt{(0.62)^{2}+(0.02) ^{2}  } }

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3 0
3 years ago
When light of wavelength 160 nm falls on a gold surface, electrons having a maximum kinetic energy of 2. 66 ev are emitted. Find
enyata [817]

When the light of wavelength is falling on gold surface, the electrons begin to exchange energies.

a)The work function in eV is Φ =5.097 eV.

b) The cut-off wavelength is λ₀ = 243.71 nm

c) The frequency is ν₀  =1.231 × 10¹⁵ Hz

<h3>What is work function?</h3>

The energy needed for a particle to escape and break through the surface.

The kinetic energy of the light emitted is 2.66 eV and wavelength of the light is 160 nm = 160 × 10⁻⁹ m.

a) The work function of the gold for given maximum kinetic energy is

Φ = hc / λ  - K.Emax

Substituting 6.626 × 10⁻³⁴ J.s for h, 3 × 10⁸ m/s for c and 2.66 eV for K.Emax, work function will be

Φ =8.16 × 10⁻¹⁹ J

1 eV = 1.6 × 10⁻¹⁹

The work function in eV is Φ =5.097 eV.

b) The cutoff wavelength is related to work function as

λ₀ = hc / Φ

Substitute the corresponding values into the equation, we get the cut off wavelength

λ₀ = 243.71 nm

c) The frequency corresponding to the cut-off wavelength is

ν₀ = c / λ₀

Substitute the corresponding values into the equation, we get the frequency,

ν₀  =1.231 × 10¹⁵ Hz

Therefore, the values for the following are

a)The work function in eV is Φ =5.097 eV.

b) The cut-off wavelength is λ₀ = 243.71 nm

c) The frequency is ν₀  =1.231 × 10¹⁵ Hz

Learn more about wave function.

brainly.com/question/17484291

#SPJ4

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