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3 years ago

How can a change in technology affect scientific knowledge?

1 answer:

7 0

When we advance our technology we increase the things we can learn and scientists can increase the amount of scientific knowledge we have.

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How do the gravitational potential energy and the kinetic energy of the ball change as the ball rolls down the ramp?

belka [17]

Explanation:

The gravitational potential energy is given by :

P = mgh

The kinetic energy of an object is given by :

$K=\dfrac{1}{2}mv^2$

As the ball reaches the bottom of the ramp, its potential energy decreases and kinetic energy increases.

It imply that, when the ball at the top most height, its gravitational potential energy is maximum and zero kinetic energy and when ball reaches the bottom of the ramp, it will have maximum kinetic energy and zero potential energy.

3 0

3 years ago

Describe one way you can increase the strength of an electromagnet.

Vsevolod [243]

Answer:

(1) By increasing the number of loops of wire around the iron core

(2) increasing the current or voltage.

5 0

3 years ago

A non uniform rod has mass

Doss [256]

Answer:

$r_{cm} = L/3$

Explanation:

Mass: M, Length: L.

$\sigma (x) = b(L-x)$

The formula that gives center of mass is

$\vec{r}_{cm} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2 + ...}{m_1 + m_2 + ...} = \frac{\Sigma m_i \vec{r}_i}{\Sigma m_i}$

In the case of a non-uniform mass density, this formula converts to

$\vec{r}_{cm} = \frac{\int\limits^L_0 {x\sigma(x)} \, dx }{\int\limits^L_0 {\sigma(x)} \, dx }$

where the denominator is the total mass and the nominator is the mass times position of each point on the rod.

We have to integrate the mass density over the total rod in order to find the total mass. Likewise, we have to integrate the center of mass of each point (xσ(x)) over the total rod. And if we divide the integrated center of mass to the total mass, we find the center of mass of the rod:

$\vec{r}_{cm} = \frac{\int\limits^L_0 {x\sigma(x)} \, dx }{\int\limits^L_0 {\sigma(x)} \, dx } = \frac{\int\limits^L_0 {xb(L-x)} \, dx }{\int\limits^L_0 {b(L-x)} \, dx } = \frac{b\int\limits^L_0{(xL - x^2)} \, dx }{b\int\limits^L_0 {(L-x)} \, dx } = \frac{\frac{x^2L}{2} - \frac{x^3}{3}}{Lx - \frac{x^2}{2}}\left \{ {{x=L} \atop {x=0}} \right.$

Here x's are cancelled. Otherwise, the denominator would be zero.

$r_{cm} = \frac{\frac{xL}{2}-\frac{x^2}{3}}{L-\frac{x}{2}}\left \{ {{x=L} \atop {x=0}} \right. = \frac{\frac{L^2}{2}-\frac{L^2}{3}}{L-\frac{L}{2}} = \frac{\frac{L^2}{6}}{\frac{L}{2}} = \frac{L}{3}$

8 0

3 years ago

lbvjy [14]

Answer: 3.0

Explanation:

6 0

3 years ago

what is a possible unit for the product VI, where V is the potential difference across a resistor and I is the current through t

liq [111]

Recall this equation for a device in a direct current circuit:
P = IV
P is the power dissipated by the device, I is the current through the device, and V is the voltage drop of the device.

If we choose to use the ampere as the unit of current and the volt as the unit of voltage, then the product of the current and the voltage will give the power with watts as the unit.

5 0

4 years ago