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spin [16.1K]
3 years ago
11

A 1000-n weight is hanging from a 2.0 m long aluminum rod. A 500-n weight is hanging from a 1.0 m long aluminum rod. The two alu

minum rods have different cross sectional areas. Which rod is under the greater tensile stress?
Physics
1 answer:
Hunter-Best [27]3 years ago
5 0

Answer:

Both rod have the same tensile stress

Explanation:

Given information,

The weight first rod, W_{1} = 1000 N

The length of first rod, l_{1} = 2.0 m

The weight second rod, W_{2} = 500 N

The length of second rod, l_{2 = 1.0 m

The equation of tensile stress, σ = \frac{F}{A}

where

σ = tensile stress (N/m^{2} or Pa)

F = Force (N)

A = Area (N/m^{2} or Pa)

so

σ1 =  \frac{W_{1} }{A_{1} }, A = 2πl

    = \frac{1000}{2\pi(2) }

    = \frac{250}{\pi } N/m^{2}

now calculate σ2

σ2 =  \frac{W_{2} }{A_{2} }

     = \frac{500}{2\pi(1) }

     = \frac{250}{\pi } N/m^{2}

σ1/σ2 = \frac{250}{\pi } / \frac{250}{\pi }

σ1/σ2 = 1

σ1 = σ2

Hence, the tensile stress of first and second rod are the same.

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Suppose that two objects attract each other with a gravitational force of 16 units. If the mass of both objects was doubled, and
Lady_Fox [76]

Answer:

  • Distance, such that displacement over time is taken into account doesn't change regardless of the length in x meters or mass in kg
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6 0
3 years ago
When a current flows in an aluminum wire of diameter 2.91 mm 2.91 mm , the drift speed of the conduction electrons is 0.000191 m
charle [14.2K]

Answer:

Number of electrons are flowing per second is 2.42 x 10¹⁹

Explanation:

The electric current flows through a wire is given by the relation :

I=envA   ....(1)

Here I is current, e is electronic charge, v is drift velocity of electrons and A is the Area of the wire.

But electric current is also define as rate of electrons passing through junction times their charge, i.e. ,

I=Ne      ....(2)

Here N is the rate of electrons passing through junction.

From equation (1) and (2).

eN = envA

N=nvA

But area of wire, A=\pi \frac{d^{2} }{4}

Here d is diameter of wire.

So, N = nv\pi \frac{d^{2} }{4}

Substitute 2.91 x 10⁻³ m for d, 0.000191 m/s for v and 6 x 10²⁸ m⁻³ for n in the above equation.

N = 6\times10^{28}\times 0.000191\times\pi \frac{(2.91\times10^{-3} )^{2} }{4}

N = 2.42 x 10¹⁹ s⁻¹  

8 0
3 years ago
Which of the following examples illustrates static friction?
vivado [14]

Answer:

A box sits stationary  on a ramp

Explanation:

Static friction is a force which keeps an object at rest as it is in the case of the box. It has to be overcome for the object to be set into motion.

Static force of friction is calculated as follows:

F= μη

F is static force of friction.

μ is the coefficient of static friction.

η is the normal force.

6 0
3 years ago
PLEASE HELP ME I JUST CAN'T FIGURE THIS ONE OUT
kupik [55]
Indeed the answer is c!!
4 0
3 years ago
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An ultrasonic tape measure uses frequencies above 20 MHz todetermine dimensions of structures such as buildings. It does so byem
ohaa [14]

Answer:  

a) 1m

b) 2μs

c) 3mm

Explanation:

3 0
3 years ago
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