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3 years ago

How do mass and force affect the acceleration of an object?

2 answers:

6 0

The acceleration of an object is in the direction of the net force. ... net force doubles, then so does the size of the acceleration

joja [24]3 years ago

3 0

Gravity affects weight, it does not affect mass. Masses always remain the same. Newton's Second Law of Motion: Force = mass x acceleration The acceleration of an object is: a) directly proportional to the net force acting on the object. ... c) inversely proportional to the mass of the object.

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What is convectional current?.... .

fiasKO [112]

Answer:

Convection is the transfer of heat from one place to another along with the motion of the object's particles. Water is a poor conductor. However, when the bottom water is heated, it turns out that the top water is also hot. This means that there is another way of transferring heat to the water, namely convection.

8 0

2 years ago

A baseball player throws a baseball. The ball traveled 35 yards before it landed on the ground. The ball stayed in the air for 3

Free_Kalibri [48]

Average speed = (distance traveled) / (time to cover the distance)

= (35 yards) / (3.8 seconds)

= 9.21 yards per second

4 0

3 years ago

Read 2 more answers

A coil is wrapped with 300 turns of wire on the perimeter of a circular frame (radius = 8.0 cm). Each turn has the same area, eq

MAVERICK [17]

Answer:

Approximately 18 volts when the magnetic field strength increases from $\rm 20\; mT$ to $\rm 80\;mT$ at a constant rate.

Explanation:

By the Faraday's Law of Induction, the EMF $\epsilon$ that a changing magnetic flux induces in a coil is:

$\displaystyle \epsilon = N \cdot \frac{d\phi}{dt}$ ,

where

$N$ is the number of turns in the coil, and
$\displaystyle \frac{d\phi}{dt}$ is the rate of change in magnetic flux through this coil.

However, for a coil the magnetic flux $\phi$ is equal to

$\phi = B \cdot A\cdot \cos{\theta}$ ,

where

$B$ is the magnetic field strength at the coil, and
$A\cdot \cos{\theta}$ is the area of the coil perpendicular to the magnetic field.

For this coil, the magnetic field is perpendicular to coil, so $\theta = 0$ and $A\cdot \cos{\theta} = A$ . The area of this circular coil is equal to $\pi\cdot r^{2} = \pi\times 8.0\times 10^{-2}\approx \rm 0.0201062\; m^{2}$ .

$A\cdot \cos{\theta} = A$ doesn't change, so the rate of change in the magnetic flux $\phi$ through the coil depends only on the rate of change in the magnetic field strength $B$ . The size of the magnetic field at the instant that $B = \rm 50\; mT$ will not matter as long as the rate of change in $B$ is constant.

$\displaystyle \begin{aligned} \frac{d\phi}{dt} &= \frac{\Delta B}{\Delta t}\times A \\&= \rm \frac{80\times 10^{-3}\; T- 20\times 10^{-3}\; T}{20\times 10^{-3}\; s}\times 0.0201062\;m^{2}\\&= \rm 0.0603186\; T\cdot m^{2}\cdot s^{-1}\end{aligned}$ .

As a result,

$\displaystyle \epsilon = N \cdot \frac{d\phi}{dt} = \rm 300 \times 0.0603186\; T\cdot m^{2}\cdot s^{-1} \approx 18\; V$ .

9 0

3 years ago

If it requires 4.5 J of work to stretch a particular spring by 2.3 cm from its equilibrium length, how much more work will be re

saveliy_v [14]

Answer:

$\Delta W=24.1162\ J$

Explanation:

Given:

work done to stretch the spring, $W=4.5\ J$

length through which the spring is stretched beyond equilibrium, $\Delta x=2.3\ cm=0.023\ m$

additional stretch in the spring length, $\delta x=3.5\ cm=0.035\ m$

<u>We know the work done in stretching the spring is given as:</u>

$W=\frac{1}{2} \times k.\Delta x^2$

where:

k = stiffness constant

$4.5=0.5\times k\times 0.023^2$

$k=17013.2325\ N.m^{-1}$

Now the work done in stretching the spring from equilibrium to ( $\Delta x+\delta x$ ):

$W'=0.5\times k.(\Delta x+\delta x)^2$

$W'=0.5\times 17013.2325\times 0.058^2$

$W'=28.6162\ J$

So, the amount of extra work done:

$\Delta W=W'-W$

$\Delta W=28.6162-4.5$

$\Delta W=24.1162\ J$

4 0

3 years ago

Three importance of SI system

diamong [38]

Answer:

Firstly they are, by design, easy to use in most scientific and engineering calculations; you only ever have to consider multiples of 10. If I’m given a measurement of 3.4 kilometres, I can instantly see that it’s 3′400 metres, or 0.0034 Megametres, or 3′400′000 millimetres. It’s not even necessary to use arithmetic, I just have to remember the definitions of the prefixes (“kilo” is a thousand, “megametre” is a million, “milli” is a thousandth) and shift the decimal point across to the left or the right. This is especially useful when we’re considering areas, speeds, energies, or other things that have multiple units; for instance,

1 metre^2 = (1000millimetre)^2 = 1000000 mm^2.

If we were to do an equivalent conversion in Imperial, we would have

1 mile^2 = (1760 yards)^2

and we immediately have to figure out what the square of 1760 is! However, the fact that SI is based on multiples of 10 has the downside that we can’t consider division by 3, 4, 8, or 12 very easily.

Secondly they are (mostly) defined in terms of things that are (or, that we believe to be) fundamental constants. The second is defined by a certain kind of radiation that comes from a caesium atom. The metre is defined in terms of the second and the speed of light. The kelvin is defined in terms of the triple point of water. The mole is the number of atoms in 12 grams of carbon-12. The candela is defined in terms of the light intensity you get from a very specific light source. The ampere is defined using the Lorentz force between two wires. The only exception is the kilogram, which is still defined by the mass of a very specific lump of metal in a vault in France (we’re still working on a good definition for that one).

Thirdly, most of the Imperial and US customary units are defined in terms of SI. Even if you’re not personally using SI, you are probably using equipment that was designed using SI.

8 0

3 years ago