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aleksklad [387]

3 years ago

7

When a certain rubber band is stretched a distance x, it exerts a restoring force of magnitude f = ax, where a is a constant. th

e work done by a person in stretching this rubber band from x = 0 to x = l is:?

1 answer:

Veseljchak [2.6K]3 years ago

4 0

Given:
F = ax
where
x = distance by which the rubber band is stretched
a = constant

The work done in stretching the rubber band from x = 0 to x = L is
$W=\int_{0}^{L} Fdx = \int_{0}^{L}ax \, dx = \frac{a}{2} [x^{2} ]_{0}^{L} = \frac{aL^{2}}{2}$

Answer: $\frac{aL^{2}}{2}$

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Two 10-cm-diameter charged rings face each other, 25.0cm apart. Both rings are charged to +20.0nC. What is the electric field st

Black_prince [1.1K]

Answer:

Part A:

$E_{midpoint}=0$

Part B:

$E_{center}=2711.7558 N/C$

Explanation:

Part A:

Formula of Electric Field Strength:

$E=\frac{1}{4\pi\epsilon}\frac{xQ}{(x^2+R^2)^{3/2}}$

Where:

x is the distance from the ring

R is the radius of the ring

$\epsilon$ is constant permittivity of free space=8.854*10^-12 farads/meter

Q is the charge

For right Ring E at the midpoint can be calculated as:

x for right plate=25/2=12.5 cm=0.125 m

Radius=R=10/2=5 cm=0.05 m

$E_{right}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.125)*(20*10^{-19})}{((0.125)^2+(0.05)^2)^{3/2}}\\E_{right}=9208.1758 N/C$

For Left Ring E at the midpoint can be calculated as:

Since charge on both plates is +ve and same in magnitude, the electric field will be same for both plates.

$E_{left}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.125)*(20*10^{-19})}{((0.125)^2+(0.05)^2)^{3/2}}\\E_{left}=9208.1758 N/C$

Electric Field at midpoint:

Both rings have same magnitude but the direction of fields will be opposite as they have same charge on them.

$E_{midpoint}=E_{left}-E_{right}\\E_{midpoint}=9208.1758-9208.1758\\E_{midpoint}=0$

Part B:

At center of left ring:

Due to left ring Electric field at center is zero because x=0.

$E_{left}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0)*(20*10^{-19})}{((0)^2+(0.05)^2)^{3/2}}\\E_{left}=0 N/C$

Due to right ring Electric field at center of left ring:

Now: x=25 cm= o.25 m (To the center of left ring)

$E_{right}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.25)*(20*10^{-19})}{((0.25)^2+(0.05)^2)^{3/2}}\\E_{right}=2711.7558 N/C$

Electric Field Strength at center of left ring is same as that of right ring.

$E_{center}=2711.7558 N/C$

5 0

3 years ago

A block of size 20m x 10 mx 5 m exerts a force of 30N. Calculate the

Orlov [11]

Answer:

We know that force applied per unit area is called pressure.

Pressure = Force/ Area

When force is constant than pressure is inversely proportional to area.

1- Calculating the area of three face:

A1 = 20m x 10 m =200 Square meter

A2 = 10 mx 5 m = 50 Square meter

A3 = 20m x 5 m = 100 Square meter

Therefore A1 is maximum and A2 is minimum.

2- Calculate pressure:

P = F/ A1 = 30 / 200 = 0.15 Nm⁻² ( minimum pressure)

P = F / A2 = 30 / 50 = 0.6 Nm⁻² ( maximum pressure)

Hence greater the area less will be the pressure and vice versa.

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2 years ago

Two blocks of masses m and M are connected by a string and pass over a frictionless pulley. Mass m hangs vertically, and mass M

horsena [70]

Answer:

$sin\theta - \mu_k cos\theta = \frac{m}{M}$

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Explanation:

Force of friction on M mass so that it will move down the inclined plane is given as

$F_f = \mu Mgcos\theta$

now if it is moving down the inclined plane at constant speed

so we will have

$Mgsin\theta - T - \mu mgcos\theta = 0$

on other side the mass "m" will go up at constant speed

so we have

$T - mg = 0$

so we have

$Mgsin\theta = \mu Mgcos\theta + mg$

so we have

$sin\theta - \mu_k cos\theta = \frac{m}{M}$

for special case when m = M

then we have

$sin\theta - \mu_k cos\theta = 1$

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2 years ago

HELP ME PLEASE If an object is placed under a force of 20 N, it accelerates at a rate of 5 m/s^2. If the force is increased to 5

lara [203]

Answer:

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Explanation:

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2 years ago

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A Meteorite Strikes On October 9, 1992, a 27-pound meteorite struck a car in Peekskill, NY, leaving a dent 22 cm deep in the tru

vagabundo [1.1K]

Answer:

Acceleration of the meteorite, $a=-38409.09\ m/s^2$

Explanation:

It is given that,

A Meteorite after striking struck a car, v = 0

Initial speed of the Meteorite, u = 130 m/s

Distance covered by Meteorite, s = 22 cm = 0.22 m

We need to find the magnitude of its deceleration. It can be calculated using the third equation of motion as :

$a=\dfrac{v^2-u^2}{2s}$

$a=\dfrac{-(130)^2}{2\times 0.22}$

$a=-38409.09\ m/s^2$

So, the deceleration of the Meteorite is $-38409.09\ m/s^2$ . Hence, this is the required solution.

7 0

2 years ago