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adelina 88 [10]
2 years ago
9

Suppose the circumference of a bicycle wheel is 2 meters. If it rotates at 1 revolution per second when you are riding the bicyc

le, then your speed will be: A) 3.14 m/s. B) 1 m/s. C) 2 m/s. D) 3 m/s. E) 6.28 m/s.
Physics
1 answer:
Debora [2.8K]2 years ago
5 0

Answer:

(c) speed will be 2 m/sec

Explanation:

We have given circumference C = 2 meters

Angular velocity = 1 rev/sec =1×2π rad/sec

We know that circumference C=2\pi r , where r is the radius of circumference

So 2=2\pi \times r

r = 0.318 meter

Linear velocity v=r\omega =0.318\times 2\times \pi =2m/sec

So option C will be the correct option

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A rectangular loop of wire of width 10 cm and length 20 cm has a current of 2.5 A flowing through it. Two sides of the loop are
Dahasolnce [82]

Answer:

(a) 0.05 Am^2

(b) 1.85 x 10^-3 Nm

Explanation:

width, w = 10 cm = 0.1 m

length, l = 20 cm = 0.2 m

Current, i = 2.5 A

Magnetic field, B = 0.037 T

(A) Magnetic moment, M = i x A

Where, A be the area of loop

M = 2.5 x 0.1 x 0.2 = 0.05 Am^2

(B) Torque, τ = M x B x Sin 90

τ = 0.05 x 0.037 x 1

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2 years ago
A block of mass 3m is placed on a frictionless horizontal surface, and a second block of mass m is placed on top of the first bl
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By Newton's second law, assuming <em>F</em> is horizontal,

• the net <u>horizontal</u> force on the <u>larger</u> block is

<em>F</em> - <em>µmg</em> = 3<em>mA</em>

where <em>µmg</em> is the magnitude of friction felt by the larger block due to rubbing with the smaller one, <em>µ</em> is the coefficient of static friction between the two blocks, and <em>A</em> is the block's acceleration;

• the net <u>vertical</u> force on the <u>larger</u> block is

4<em>mg</em> - 3<em>mg</em> - <em>mg</em> = 0

where 4<em>mg</em> is the mag. of the normal force of the surface pushing up on the combined mass of the two blocks, 3<em>mg</em> is the weight of the larger block, and <em>mg</em> is the weight of the smaller block;

• the net <u>horizontal</u> force on the <u>smaller</u> block is

<em>µmg</em> = <em>ma</em>

where <em>µmg</em> is again the friction between the two blocks, but notice that this points in the same direction as <em>F</em>. It is the only force acting on the smaller block in the horizontal direction, so (b) static friction is causing the smaller block to accelerate;

• the net <u>vertical</u> force on the <u>smaller</u> block is

<em>mg</em> - <em>mg</em> = 0

where <em>mg</em> is the magnitude of both the normal force of the larger block pushing up on the smaller one, and the weight of the smaller block.

(You should be able to draw your own FBD's based on the forces mentioned above.)

(c) Solve the equations above for <em>A</em> and <em>a</em> :

<em>A</em> = (<em>F</em> - <em>µmg</em>) / (3<em>m</em>)

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5 0
1 year ago
A FBD of a rocket launching into space should include:
Vladimir [108]

Answer:

Explanation:

the force of the rocket engine pushing it up,  the force of gravity pulling it down,    maybe some force of air resistance as the rocket goes fast,   hmmm    Free Body Diagrams  (FBD)  should have any and all forces on the model,  unless they are negligible . or so slight they really make little difference in the total  outcome.  

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1 year ago
Two large non-conducting plates of surface area A = 0.25 m 2 carry equal but opposite charges What is the energy density of the
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Answer:

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Using E = q/A*eo

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Energy density = 1/2*eo*(E^2) = 1/2*eo*(q/A*eo)^2 = [q^2] / [2*(A^2)*eo]

= [(75*10^-6)^2] / [2*(0.25)^2*8.85*10^-12]

= 5.1*10^3 J/m^3

8 0
1 year ago
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tangare [24]

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6 0
1 year ago
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