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IrinaVladis [17]

3 years ago

14

BEST ANSWER GETS MARKED AS BRAINLIEST!!!! Can someone PLZ help me come up with a study method, no matter what I try, I just can'

t seem to think that studying helps at all. But on the back of my head i know it does help, but I don't know how to do so.

2 answers:

Leni [432]3 years ago

8 0

For physics, I would recommend to just keep doing practice problems and reviewing notes. Repetition of the same concepts will help drill it into your brain. Hope that helps!

____ [38]3 years ago

5 0

Have you tried Quizlet? It's a free flashcard/study website that can help. You can also try playing a really childish game, a memory game, where you add pairs of the term and definition and match them up. Hopefully this is helpful!

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Which of the following would be used in luminosity calculations?

Daniel [21]

Answer:

1) joule

2) $kgm^{2}/s^{2}$

3) $10\%$

Explanation:

1) Luminosity is the <u>amount of light emitted</u> (measured in Joule) by an object in a unit of<u> time</u> (measured in seconds). Hence in SI units luminosity is expressed as joules per second ( $\frac{J}{s}$ ), which is equal to Watts ( $W$ ).

This amount of light emitted is also called radiated electromagnetic power, and when this is measured in relation with time, the result is also called radiant power emitted by a light-emitting object.

Therefore, if we want to calculate luminosity the Joule as a unit will be used.

2) Work $W$ is expressed as force $F$ multiplied by the distane $d$ :

$W=F.d$

Where force has units of $kgm/s^{2}$ and distance units of $m$ .

If we input the units we will have:

$W=(kgm/s^{2})(m)$

$W=kgm^{2}/s^{2}$ This is 1Joule ( $1 J$ ) in the SI system, which is also equal to $1 Nm$

3) The formula to calculate the percent error is:

$\% error=\frac{|V_{exp}-V_{acc}|}{V_{acc}} 100\%$

Where:

$V_{exp}=7.34 (10)^{-11} Nm^{2}/kg^{2}$ is the experimental value

$V_{acc}=6.67 (10)^{-11} Nm^{2}/kg^{2}$ is the accepted value

$\% error=\frac{|7.34 (10)^{-11} Nm^{2}/kg^{2}-6.67 (10)^{-11} Nm^{2}/kg^{2}|}{6.67 (10)^{-11} Nm^{2}/kg^{2}} 100\%$

$\% error=10.04\% \approx 10\%$ This is the percent error

8 0

4 years ago

The main difference between a chest and a bounce pass is what?

snow_lady [41]

Answer: The main difference between the three is the mode of transmission. The chest pass is straight through the air towards your teammate. While the bounce pass is directed toward the ground and then at your teammate. Finally, the overhead pass is projected high in the air to avoid defenders.

Explanation:

4 0

3 years ago

Read 2 more answers

A monochromatic light beam is incident on a barium target that has a work function of 2.50 \mathrm{eV} . If a potential differen

leva [86]

The wavelength of the light beam required to turn back all the ejected electrons is 497 nm which is option (b).

Work function is a material property defined as the minimum amount of energy required to infinitely remove electrons from the surface of a particular solid.
The potential difference required to support all emitted electrons is called the stopping potential which is given by $v_0=\frac{K.E_m_a_x}{e}$ .....(1)
where $v_0$ is the stopping potential and e is the charge of the electron given by $1.6\times10^-^1^9$ .

It is given that work function (Ф) of monochromatic light is 2.50 eV.

Einstein photoelectric equation is given by:

$K.E_m_a_x=E-\phi$ ....(2)

where K.E(max) is the maximum kinetic energy.

Substituting (1) into (2) , we get

$ev_0=E-\phi\\1.6\times10^{-19} \times1=E-2.50\\E=1.6\times10^{-19}+2.50\\E=2.50eV$

As we know that $E=\frac{hc}{\lambda}$ ....(3)

where Speed of light, $c = 3\times10^8 m/s$ and Planck's constant , $h = 6.63\times 10^-^1^9Js = 4.14\times 10^-^1^5 eVs$

From equation (3) , we get

$\lambda=\frac{hc}{E} \\\\\lambda=\frac{ 4.14\times 10^-^1^5 \times 3 \times10^8}{2.50} \\\\\lambda=\frac{1240\times10^-^9}{2.50} \\\\\lambda=496.8\times10^-^9\\\\\lambda=497nm$

Learn about more einstein photoelectric equation here:

brainly.com/question/11683155

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8 0

1 year ago

The magnitude of the angular momentum of the two-satellite system is best represented by

zloy xaker [14]

The magnitude of the angular momentum of the two-satellite system is best represented as, L=m₁v₁r₁-m₂v₂r₂.

<h3>What is angular momentum.?</h3>

The rotational analog of linear momentum is angular momentum also known as moment of momentum or rotational momentum.

It is significant in physics because it is a conserved quantity. the total angular momentum of a closed system remains constant. Both the direction and magnitude of angular momentum are conserved.

The magnitude of the angular momentum of the two-satellite system is best represented as;

L=∑mvr

L=m₁v₁r₁-m₂v₂r₂

Hence, the magnitude of the angular momentum of the two-satellite system is best represented as, L=m₁v₁r₁-m₂v₂r₂.

To learn more about the angular momentum, refer to the link;

brainly.com/question/15104254

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3 0

2 years ago

Shrinking Loop. A circular loop of flexible iron wire has an initial circumference of 168 cm , but its circumference is decreasi

DedPeter [7]

Answer:

103.1 V

Explanation:

We are given that

Initial circumference=C=168 cm

$\frac{dC}{dt}=-15cm/s$

Magnetic field,B=0.9 T

We have to find the magnitude of the emf induced in the loop after exactly time 8 s has passed since the circumference of the loop started to decrease.

Magnetic flux= $\phi=BA=B(\pi r^2)$

Circumference,C= $2\pi r$

$r=\frac{C}{2\pi}$

$r=\frac{168}{2\pi}$ cm

$\frac{dr}{dt}=\frac{1}{2\pi}\frac{dC}{dt}=\frac{1}{2\pi}(-15)=-\frac{15}{2\pi} cm/s$

$\int dr=-\int \frac{15}{2\pi}dt$

$r=-\frac{15}{2\pi}t+C$

When t=0

$r=\frac{168}{2\pi}$

$\frac{168}{2\pi}=C$

$r=-\frac{15}{2\pi}t+\frac{168}{2\pi}$

E= $-\frac{d\phi}{dt}=-\frac{d(B\pi r^2)}{dt}=-2\pi rB\frac{dr}{dt}$

$E=-2\pi(-\frac{5}{2\pi}t+\frac{168}{2\pi})B\times -\frac{15}{2\pi}$

t=8 s

B=0.9

$E=2\pi\times \frac{15}{2\pi}\times 0.9(-\frac{15}{2\pi}(8)+\frac{168}{2\pi})$

$E=103.1 V$

6 0

3 years ago