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3 years ago

If a change in speed “Δv” occurs and the mass starts at rest, would the change in speed also equal the final speed of the mass?

Physics

1 answer:

stealth61 [152]3 years ago

3 0

Yes, if the mass starts at rest, <u>the change in speed will be equal the final speed</u>, because:

Δv = Vf - Vo

How Vo (Initial velocity) is equal zero, we simplificate:

Δv = Vf

Then, the change of the speed, if the mass starts at rest, will be equal to final velocity.

Greetings.

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A material through which a current does not move easily is a(n) _____.

barxatty [35]

<span>A material through which a current does not move easily is called
an insulator.

Technically, charges CAN move through an insulator, but they lose
a lot of energy doing it, so the current that flows through the insulator
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Another way to look at it: Insulators have high resistance.
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6 0

3 years ago

The acceleration of a particle traveling along a straight line isa=14s1/2m/s2, wheresis in meters. Ifv= 0,s= 1 m whent= 0, deter

Yuliya22 [10]

Answer:

0.78m/s

Explanation:

We are given that

Acceleration= $a=\frac{1}{4}s^{\frac{1}{2}}m/s^2$

v=0, s=1 when t=0

We have to find the particle's velocity at s=2m

We know that

$a=\frac{dv}{dt}=\frac{dv}{ds}\times \frac{ds}{dt}=\frac{dv}{ds}v$

$vdv=ads$

$\int_{0}^{v} vdv=\int_{1}^{s}0.25s^{\frac{1}{2}}ds$

$\frac{v^2}{2}=0.25\times \frac{2}{3}(s^{\frac{3}{2})^{s}_{1}$

By using formula: $\int x^ndx=\frac{x^{n+1}}{n+1}+C$

$\frac{v^2}{2}=0.25\times \frac{2}{3}(s^{\frac{3}{2}}-1$

Substitute s=2

$\frac{v^2}{2}=\frac{0.50}{3}((2^{1.5})-1)$

$\frac{v^2}{2}=\frac{0.50}{3}\times 1.83$

$v^2=2\times 0.305=0.61$

$v=\sqrt{0.61}=0.78m/s$

Hence, the velocity of particle at s=2m=0.78m/s

3 0

4 years ago

A 0.25 kg ball is suspended from a light 0.65 m string as shown. The string makes an angle of 31° with the vertical. Let U = 0 w

steposvetlana [31]

Explanation:

a) The height of the ball h with respect to the reference line is

$h = L - L\cos{31°} = L(1 - \cos{31°})$

so its initial gravitational potential energy $U_0$ is

$U = mgh = mgL(1 - \cos{31°})$

$\:\:\:\:\:=(0.25\:\text{kg})(9.8\:\text{m/s}^2)(0.65\:\text{m})(1 - \cos{31})$

$\:\:\:\:\:=0.23\:\text{J}$

b) To find the speed of the ball at the reference point, let's use the conservation law of energy:

$\Delta{K} + \Delta{U} = 0 \Rightarrow K_0 + U_0 = K + U$

We know that the initial kinetic energy $K_0,$ as well as its final gravitational potential energy $U$ are zero so we can write the conservation law as

$mgL(1 - \cos{31°}) = \frac{1}{2}mv^2$

Note that the mass gets cancelled out and then we solve for the velocity v as

$v = \sqrt{2gL(1 - \cos{31°})}$

$\:\:\:\:\:= \sqrt{2(9.8\:\text{m/s}^2)(0.65\:\text{m})(1 - \cos{31°})}$

$\:\:\:\:\:= 1.3\:\text{m/s}$

5 0

3 years ago

Read 2 more answers

A wave with a frequency of 60 Hz is traveling along a string whose linear mass density is 230 g/m and whose tension is 65 N. If

matrenka [14]

To develop this problem we will use the concepts related to Speed in a string that is governed by Tension (T) and linear density (µ)

$V = \sqrt{\frac{T}{\mu}}$

Our values are given as:

$f = 60Hz\\\mu = 230 g/m = 0.230kg/m\\T = 65N\\P = 75w$

Replacing we have that the velocity is

$V = \sqrt{\frac{T}{\mu}}$

$V = \sqrt{\frac{65}{0.230}}$

$V = 16.81m/s$

From the theory of wave propagation the average power wave is given as

$P =\frac{1}{2} \mu \omega^2 A^2 V$

Where,

A = Amplitude

$\omega = 2\pi f \rightarrow$ Angular velocity

$A^2 = \frac{2P}{\mu \omega^2 V}$

$A^2 = \frac{2P}{\mu (2\pi f)^2 V}$

Replacing,

$A^2 = \sqrt{\frac{2(75)}{(0.230)(2\pi 60)^2(16.81)}}$

$A = 0.0165m$

Therefore the amplitude of the wave should be 0.0165m

8 0

3 years ago

An object, with mass 32 kg and speed 26 m/s relative to an observer, explodes into two pieces, one 5 times as massive as the oth

Brut [27]

Answer:

$\Delta K = 2164.053\,J$

Explanation:

Let consider the observer as an inertial reference frame. The object is modelled after the Principle of Momentum Conservation:

$(32\,kg)\cdot (26\,\frac{m}{s} ) = (5.333\,kg)\cdot (0\,\frac{m}{s} )+(26.665\,kg )\cdot v$

The speed of the more massive piece is:

$v = 31.202\,\frac{m}{s}$

The kinetic energy added to the system is:

$\Delta K = \frac{1}{2}\cdot [(5.333\,kg)\cdot (0\,\frac{m}{s} )^{2}+(26.665\,kg )\cdot (31.202\,\frac{m}{s} )^{2}]-\frac{1}{2}\cdot (32\,kg)\cdot (26\,\frac{m}{s} )^{2}$

$\Delta K = 2164.053\,J$

3 0

3 years ago