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1 year ago

A mass attached to a spring vibrates back and forth. At maximum displacement, the spring force and the

Physics

1 answer:

Svetradugi [14.3K]1 year ago

8 0

ANSWER

Velocity of the mass reaches zero

EXPLANATION

We want to identify what hapens to a mass attached toa a spring at maximum displacement.

When a mass attached to a spring is at its maximum position of displacement, the direction of the mass begins to change. This implies that the velocity of the mass will reach zero.

Hence, at maximum displacement, the velocity of the mass reaches zero.

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A wire is made from a material having a temperature coefficient of resistivity of 0.0003125 (°C^-1). In an experiment, we mainta

beks73 [17]

Answer:

The power decreases by 36%

Explanation:

Given:

At 20° C

Power, P₀ = 300 W

Potential difference, V = 150 volts

Now, power is given as

P = V²/R

where, R is the resistance

on substituting the values, we get

300 = 150²/R₀

R₀ = 75 Ω

Now, the variation of resistance with temperature is given as

R = R₀[1 + α(T - T₀)]

where, α is the temperature coefficient of resistivity = 0.0003125 (°C⁻¹)

now, at

T₀ = 20° C

R₀ = 75 Ω

for

T = 1820° C

we have

R = R₀[1 + α(T - T₀)]

substituting the values

we get

R = 75×[1 + 0.0003125 × (1820 - 20)]

R = 117.18 Ω

Now using the formula for power

We have,

P = V²/R

P = 150²/117.18 = 192 W

Therefore, the percentage change will be

= $\frac{P-P_0}{P_0}\times 100$

on substituting the values , we get

= $\frac{192-300}{300}\times 100$

= -36%

here, negative sign depicts the decrease in power

3 0

3 years ago

Read 2 more answers

explain y it is easier to loosen a tight but using a spanner with along handle than with a short handle

8090 [49]

Answer:

This is because using a long handled requires less force to the center of gravity and makes it easier to rotate than a short handled spanner

3 0

3 years ago

When a carpenter shuts off his circular saw, the 10.0 inch diameter blade slows from 4250 rpm to 0.00 in 4.00 s. (a) What is the

MaRussiya [10]

Answer:

(a) $\alpha=-111.26rad/s$

(b) $s=4450.6in$

Explanation:

First change the units of the velocity, using these equivalents $1rev=2\pi rad$ and $1 min =60s$

$4250rpm(\frac{2\pi rad}{1rev})(\frac{1 min}{60 s} )=445.06rad/s$

The angular acceleration $\alpha$ the time rate of change of the angular speed $\omega$ according to:

$\alpha=\frac{\Delta \omega}{\Delta t}$

$\Delta \omega=\omega_i-\omega_f$

Where $\omega_i$ is the original velocity, in the case the velocity before starting the deceleration, and $\omega_f$ is the final velocity, equal to zero because it has stopped.

$\alpha=\frac{\Delta \omega}{\Delta t} =\frac{\omega_i-\omega_f}{4}\frac{0-445.06}{4} =\frac{-445.06}{4} =-111.26rad/s$

b) To find the distance traveled in radians use the formula:

$\theta = \omega_i t + \frac{1}{2} \alpha t^2$

$\theta = 445.06 (4) + \frac{1}{2}(-111.26) (4)^2=1780.24-890.12=890.12rad$

To change this result to inches, solve the angular displacement $\theta$ for the distance traveled $s$ ( $r$ is the radius).

$\theta=\frac{s}{r} \\s=\theta r$

$s=890.12(5)=4450.6in$

c) The displacement is the difference between the original position and the final. But in every complete rotation of the rim, the point returns to its original position. so is needed to know how many rotations did the point in the 890.16 rad of distant traveled:

$\frac{890.12}{2\pi}=141.6667$

The real difference is in the 0.6667 (or 2/3) of the rotation. To find the distance between these positions imagine a triangle formed with the center of the blade (point C), the initial position (point A) and the final position (point B). The angle $\gamma=\frac{2\pi}{3}=\frac{360^o}{3}=120$ is between the two sides known. Using the theorem of the cosine we can find the missing side of the the triangle(which is also the net displacement):