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3 years ago

What are the tow major topics studied by a physicist?

Physics

2 answers:

Valentin [98]3 years ago

6 0

Matter and energy are the two major topics studied by a physicist.

Answer: Option D

<u>Explanation:</u>

The universe is composed of different types of matter because of which evolution occurs. Due to interaction between matter, energy is released or absorbed as a result. Physics deals with the study of matter and energy and so it is the main area of concentration for the physicists.

Biology and ecology are related to biologists and environmentalists. Evolution is related to the paleontologists where matter forms the basis for the evolution of human kind.

Nutka1998 [239]3 years ago

5 0

D- Matter and Energy

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A purse at radius 2.30 m and a wallet at radius 3.45 m travel in uniform circular motion on the floor of a merry-go-round as the

ivolga24 [154]

Answer:

The acceleration of the wallet is $3\hat{i}+6\hat{j}$

Explanation:

Given that,

Radius of purse r= 2.30 m

Radius of wallet r'= 3.45 m

Acceleration of the purse $a=2\hat{i}+4.00\hat{j}$

We need to calculate the acceleration of the wallet

Using formula of acceleration

$a=r\omega^2$

Both the purse and wallet have same angular velocity

$\omega=\omega'$

$\sqrt{\dfrac{a}{r}}=\sqrt{\dfrac{a'}{r'}}$

$\dfrac{a}{r}=\dfrac{a'}{r'}$

$\dfrac{a'}{a}=\dfrac{r'}{r}$

$\dfrac{a'}{a}=\dfrac{3.45}{2.30}$

$\dfrac{a'}{a}=\dfrac{3}{2}$

$a'=\dfrac{3}{2}\times(2\hat{i}+4.00\hat{j})$

$a'=3\hat{i}+6\hat{j}$

Hence, The acceleration of the wallet is $3\hat{i}+6\hat{j}$

4 0

3 years ago

Solar panels convert light energy from sunlight into electrical energy. What material is most likely used in solar panels, and w

klemol [59]

Answer:

A metalloid is used because it is a semiconductor and can become more conductive when more light shines on it

Explanation:

The material used in a solar panel is a metalloid. It can often become conductive when more light shines on it.

Metalloids have properties that straddles between those of metals and non-metals.

In essence, they can be conductive or not under certain conditions.

The most important property they exhibit is that they can become more conductive when more light shines on them. This way more electrons are produced.

3 0

3 years ago

Read 2 more answers

At resonance, what is impedance of a series RLC circuit? less than R It depends on many other considerations, such as the values

denis23 [38]

Answer:

at resonance impedence is equal to resistance and quality factor is dependent on R L AND C all

Explanation:

we know that for series RLC circuit impedance is given by

$Z=\sqrt{R^2+\left ( X_L-X_C \}right )^2$

but we know that at resonance $X_L=X_C$

putting $X_L=X_C$ in impedance formula , impedance will become

Z=R so at resonance impedance of series RLC is equal to resistance only

now quality factor of series resonance is given by

$Q=\frac{\omega L}{R}=\frac{1}{\omega CR}=\frac{1}{R}\sqrt{\frac{L}{C}}$ so from given expression it is clear that quality factor depends on R L and C

3 0

3 years ago

Finding the spring constant as shown, spring 3, which has an unknown spring constant k3, replaces spring 2. the mass of the weig

nevsk [136]

Replaces spring 2. the mass of the weight and pulley are unchanged: m=5.8 kg and mp=1.7 kg

6 0

4 years ago

An electron and a 0.033 0-kg bullet each have a velocity of magnitude 495 m/s, accurate to within 0.010 0%. Within what lower li

lara31 [8.8K]

Answer:

1.170*10^-3 m

3.23*10^-32 m

Explanation:

To solve this, we apply Heisenberg's uncertainty principle.

the principle states that, "if we know everything about where a particle is located, then we know nothing about its momentum, and vice versa." it also can be interpreted as "if the uncertainty of the position is small, then the uncertainty of the momentum is large, and vice versa"

Δp * Δx = h/4π

m(e).Δv * Δx = h/4π

If we make Δx the subject of formula, by rearranging, we have

Δx = h / 4π * m(e).Δv

on substituting the values, we have

for the electron

Δx = (6.63*10^-34) / 4 * 3.142 * 9.11*10^-31 * 4.95*10^-2

Δx = 6.63*10^-34 / 5.67*10^-31

Δx = 1.170*10^-3 m

for the bullet

Δx = (6.63*10^-34) / 4 * 3.142 * 0.033*10^-31 * 4.95*10^-2

Δx = 6.63*10^-34 / 0.021

Δx = 3.23*10^-32 m

therefore, we can say that the lower limits are 1.170*10^-3 m for the electron and 3.23*10^-32 for the bullet

7 0

4 years ago