2. Calculate the dipole moment of 5 & 3 columb charges separated by

Alborosie

Temperature is measured with a thermometer

8 0

4 years ago

A force in the +x-direction with magnitude F(x)=18.0N−(0.530N/m)x is applied to a 8.90 kg box that is sitting on the horizontal,

Lady_Fox [76]

Answer:

6.875 m/s

Explanation:

The force is variable which is given by

F(x) = 18 - 0.53 x

mass of the box, m = 8.9 kg

initially it is at rest at x = 0

Let the velocity is v after travelling a distance of 15 m.

According to the work energy theorem, the work done by all the forces is equal to the change in kinetic energy of the body.

Work done = change in kinetic energy

$\int \overrightarrow{F}.d\overrightarrow{x}=\Delta K.E$

$\int_{0}^{15} \left ( 18-0.53 x \right )dx=\frac{1}{2}\times m \left ( v^{2}-u^{2} \right )$

$\left ( 18x-0.265x^{2} \right )_{0}^{15}=\frac{1}{2}\times 8.9\times \left ( v^{2}-0^{2} \right )$

18 x 15 - 0.265 x 15 x 15 = 4.45 x v²

270 - 59.625 = 4.45 v²

v² = 47.275

v = 6.875 m/s

Thus, the final velocity of the box is 6.875 m/s.

4 0

3 years ago

Hi! I need help quick!

11Alexandr11 [23.1K]

Answer:

We usually travel in January or Feb for our tropical get-a-way. For those of you that have traveled in the fall and winter...

What is the real difference...Like in the floral and fauna, weather, and the capacity of the resorts between these 2 times of the year.

7 0

2 years ago

Which of the following is an example of a main sequence star

Westkost [7]

The star is the main sequence

3 0

3 years ago

Read 2 more answers

A stone is tied to a string and whirled at constant speed in a horizontal circle. The speed is then doubled without changing the

Tanya [424]

Answer:

Afterward the magnitude of the acceleration of the stone is four times as great

Explanation:

A stone tied to a string and whirled at a constant speed in a horizontal circle will have a centripetal acceleration given by

$a = \frac{v^{2} }{r}$

Where $a$ is the centripetal acceleration

$v$ is the speed

and $r$ is the radius of the circle

Here, the radius of the circle is the length of the string.

Now, from the question

The speed is then doubled without changing the length of the string,

Let the new speed be $v_{2}$ , that is

$v_{2} = 2v$

and without changing the length of the string means radius $r$ is constant.

To determine the magnitude of the acceleration of the stone afterwards,

Let the new acceleration be $a_{2}$ .

Then we can write that

$a_{2} = \frac{v_{2}^{2} }{r}$

From

$a = \frac{v^{2} }{r}$

$v = \sqrt{ar}$

Recall that

$v_{2} = 2v$

∴ $v_{2} = 2\sqrt{ar}$

Now, we will put the value of $v_{2}$ into

$a_{2} = \frac{v_{2}^{2} }{r}$

Then,

$a_{2} = \frac{(2\sqrt{ar}) ^{2} }{r}$

$a_{2} = \frac{4ar }{r}$

$a_{2} = 4a$

The new magnitude of the acceleration of the stone is four times the initial acceleration.

Hence,

Afterward the magnitude of the acceleration of the stone is four times as great

3 0

3 years ago