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riadik2000 [5.3K]
3 years ago
15

A marble is placed at the bottom of a semi-spherical bowl, as shown in the figure. The marble is then displaced from the bottom

of the bowl to a position about halfway from the top of the bowl. The marble is then released from rest such that the marble always remains in contact with the bowl. Students observe that the marble rolls back and forth as it oscillates about the bottom of the bowl. Which of the following statements best explains why the marble undergoes oscillatory motion?
a)The sides of the bowl become steeper at positions farther from the bottom of the bowl.b) The net force exerted on the marble always has a component directed toward the bottom of the bowl. c) The normal force exerted on the marble decreases with increasing distance from the bottom of the bowl. D ) The gravitational force exerted on the marble is constant in magnitude and direction.​

Physics
1 answer:
Andrew [12]3 years ago
8 0

Answer:

It is D

Explanation:

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245 ml

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3 years ago
A wire 6.90 m long with diameter of 2.15 mm has a resistance of 0.0320 Ω. Find the resistivity of the material of the wire. rho
spayn [35]
<h2>Answer:</h2>

1.68 x 10⁻⁸Ωm

<h2>Explanation:</h2>

The resistance (R) of a wire is related to its length(L), its material resistivity(ρ) and its crossectional area(A) as follows;

R = ρL/A               ------------------------(i)

Where;

A = πd² / 4              [where d = diameter of the wire]

From the question;

L = 6.90m

d = 2.15mm = 0.00215m

R = 0.0320Ω

First calculate the crossectional area (A) of the wire as follows;

A = πd² / 4      

[Take π = 3.142]

d = 0.00215m

∴ A = 3.142 x (0.00215)² / 4

∴ A = 0.000003631m²

Now, substitute the values of A, L, and R into equation (i) as follows;

R = ρL/A

0.0320 = ρ x 6.90 / 0.000003631

0.0320 = 1900302.95 x ρ

Solve for ρ;

=> ρ = 0.0320 / 1900302.95

=> ρ = 1.68 x 10⁻⁸Ωm

Therefore, the resistivity of the material of the wire is 1.68 x 10⁻⁸Ωm

4 0
3 years ago
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Answer:

B=9.1397*10^-4 Tesla

Explanation:

To find the velocity first we put kinetic energy og electron is equal to potential energy of electron

K.E=P.E

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where :

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v is the velocity

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v=\sqrt{\frac{2*e*V}{m} }    eq 1

Radius of electron moving in magnetic field is given by:

R=\frac{m*v}{q*B}       eq 2

where:

m is the mass of electron

v is the velocity

q=e=charge of electron

B is the magnitude of magnetic field

Put v from eq 1 into eq 2

R=\frac{m*\sqrt{\frac{2*e*V}{m} } }{e B}

B=\sqrt{\frac{2*m*V}{e*R^{2} } }

B=\sqrt{\frac{2*(9.31*10^{-31})*(2.12*10^{3})  }{(1.60*10^{-19})*(0.170)^{2}  } }

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3 0
3 years ago
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