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2 years ago

A wire of length L and cross-sectional area A has resistance R.

Physics

1 answer:

Leona [35]2 years ago

6 0

Answer:

The resistance R stretched of the wire if it is stretched to twice its original length is [Δ(2L) / A ]* 2

Explanation:

Original Resistance R = ΔL / A

Given length is doubled

New length L' = 2L

New Volume is constant, resistivity and density of material is also assumed to be constant

AL = A'L'

AL = A'(2L)

A' = AL / 2L

A' = A / 2

∴ new Resistance R' = ΔL' / A'

= Δ(2L) / A/2

= [Δ(2L) / A ]* 2

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Two 51 g blocks are held 30 cm above a table. As shown in the figure, one of them is just touching a 30-long spring. The blocks

vivado [14]

The concept of this question can be well understood by listing out the parameters given.

The mass of the block = 51 g = 51 × 10⁻³ kg
The distance of the block from the table = 30 cm
Length of the spring = 30 cm

The purpose is to determine the spring constant.

Let us assume that the two blocks are Block A and Block B.

At point A on block A, the initial velocity on the block is zero

i.e. u = 0

We want to determine the time it requires for Block A to reach the table. The can be achieved by using the second equation of motion which can be expressed by using the formula.

$\mathsf{S = ut + \dfrac{1}{2}gt^2}$

From the above formula,

The distance (S) = 30 cm; we need to convert the unit to meter (m).

Since 1 cm = 0.01 m
Then, 30cm = 0.3 m

The acceleration (g) due to gravity = 9.8 m/s²

∴

inputting the values into the equation above, we have;

$\mathsf{0.3 = (0)t + \dfrac{1}{2}*(9.80)*(t^2)}$

$\mathsf{0.3 = \dfrac{1}{2}*(9.80)*(t^2)}$

$\mathsf{0.3 =4.9*(t^2)}$

By dividing both sides by 4.9, we have:

$\mathsf{t^2 = \dfrac{0.3}{4.9}}$

$\mathsf{t^2 = 0.0612}$

$\mathsf{t = \sqrt{0.0612}}$

$\mathsf{t =0.247 \ seconds}$

However, block B comes to an instantaneous rest on point C. This is achieved by the dropping of the block on the spring. During this process, the spring is compressed and it bounces back to oscillate in that manner. The required time needed to get to this point C is half the period, this will eventually lead to the bouncing back of the block with another half of the period, thereby completing a movement of one period.

By applying the equation of the time period of a simple harmonic motion.

$\mathsf{T = 2 \pi \sqrt{\dfrac{m}{k}}}$

where the relation between time (t) and period (T) is:

$\mathsf{t = \dfrac{T}{2}}$

T = 2t

T = 2(0.247)

T = 0.494 seconds

$\mathsf{T = 2 \pi \sqrt{\dfrac{m}{k}}}$

By making the spring constant k the subject of the formula:

$\mathsf{\dfrac{T}{2 \pi } = \sqrt{ \dfrac{m}{k}}}$

$\Big(\dfrac{T}{2 \pi }\Big)^2 = { \dfrac{m}{k}$

$\dfrac{T^2}{(2 \pi)^2 }= { \dfrac{m}{k}$

$\mathsf{ T^2 *k = 2 \pi^2*m} \\ \\ \mathsf{ k = \dfrac{2 \pi^2*m}{T^2}}$

$\mathsf{ k =\Big( \dfrac{(2 \pi)^2*(51 \times 10^{-3})}{(0.494)^2} \Big) N/m}$

$\mathbf{ k =8.25 \ N/m}$

Therefore, we conclude that the spring constant as a result of instantaneous rest caused by the compression of the spring is 8.25 N/m.

Learn more about simple harmonic motion here:

brainly.com/question/17315536?referrer=searchResults

6 0

2 years ago

A block of mass m = 1.5 kg is released from rest at a height of H = 0.81 m on a curved frictionless ramp. At the foot of the ram

kogti [31]

Answer:

0.31 m

Explanation:

m = mass of the block = 1.5 kg

H = height from which the block is released on ramp = 0.81 m

k = spring constant of the spring = 250 N/m

x = maximum compression of the spring

using conservation of energy

Spring potential energy gained by spring = Potential energy lost by block

(0.5) k x² = mgH

(0.5) (250) x² = (1.5) (9.8) (0.81)

x = 0.31 m

7 0

3 years ago

An elephant produces a 10Hz sound wave. Assuming the speed of sound in air is 344/s, determine the wavelength of this infrasonic

user100 [1]

Answer:

34.5 m

Explanation:

Given data : An elephant produces 10 Hz sound wave. Assuming the speed of sound in air is 345 m/s. To find : What is the wavelength of sound ? Answer : The wavelength of the sound is 34.5 m.

4 0

2 years ago

You toss a bowling ball straight up into the air with a speed of 2.1 m/s. How long does it take the bowling ball to reach its hi

lara [203]

Time taken by the bowling ball to reach its highest point= 0.214 s

initial velocity= Vi=2.1 m/s

Final velocity= Vf=0 as the velocity at the highest point is zero.

acceleration= g= -9.8 m/s²

using the kinematic equation Vf= Vi + at

0= 2.1 + (-9.8)t

t= -2.1/-9.8

t=0.214 s

Thus the time taken by the bowling ball to reach its highest point is 0.214 s

6 0

3 years ago

Hi i need help. thNKS

Ede4ka [16]

Answer:

Explanation:

Groups go down, periods go across :)

3 0

2 years ago