Will an object with more mass roll faster down a hill?

Which of the following is not a way that astronomers can find how much dark matter there is in cluster of galaxies?

Tpy6a [65]

Answer:

b. Observe the radio waves coming from all dark matter; from the strength of the radio waves from each cluster, estimate the amount of dark matter needed to produce them.

Explanation:

The universe is thought to be made up of 85% dark matters. Dark matter is called dark because it does not appear to interact with the electromagnetic field, which means it doesn't absorb, reflect or emit electromagnetic radiation, and is therefore difficult to detect. This means that option b is wrong since radio wave is an electromagnetic wave. Dark matter is a form of matter that makes up about a quarter of the total mass–energy density of the universe. Dark matter was theorized due a variety of astrophysical observations and gravitational effects that cannot be explained by accepted theories of gravity unless there were more matter in the universe than can be seen.

6 0

3 years ago

What is the reflectivity of a glass surface (n =1.5) in air (n = 1) at an 45° for (a) S-polarized light and (b) P-polarized ligh

Goshia [24]

Answer:

a) $R_s = 0.092$

b) $R_p = 0.085$

Explanation:

given,

n =1.5 for glass surface

n = 1 for air

incidence angle = 45°

using Fresnel equation of reflectivity of S and P polarized light

$R_s=\left | \dfrac{n_1cos\theta_i-n_2cos\theta_t}{n_1cos\theta_i+n_2cos\theta_t} \right |^2\\R_p=\left | \dfrac{n_1cos\theta_t-n_2cos\theta_i}{n_1cos\theta_t+n_2cos\theta_i} \right |^2$

using snell's law to calculate θ t

$sin \theta_t = \dfrac{n_1sin\theta_i}{n_2}=\dfrac{sin45^0}{1.5}=\dfrac{\sqrt{2}}{3}$

$cos \theta_t =\sqrt{1-sin^2\theta_t} = \dfrac{sqrt{7}}{3}$

a) $R_s=\left | \dfrac{\dfrac{1}{\sqrt{2}}-\dfrac{1.5\sqrt{7}}{3}}{\dfrac{1}{\sqrt{2}}+\dfrac{1.5\sqrt{7}}{3}} \right |^2$

$R_s = 0.092$

b) $R_p=\left | \dfrac{\dfrac{\sqrt{7}}{3}-\dfrac{1.5}{\sqrt{2}}}{\dfrac{\sqrt{7}}{3}+\dfrac{1.5}{\sqrt{2}}} \right |^2$

$R_p = 0.085$

3 0

2 years ago

A 30.0-Î¼F capacitor is connected to a 49.0-Î© resistor and a generator whose rms output is 30.0 V at 60.0 Hz. (a) Find the rms

Natali5045456 [20]

Explanation:

Given that,

Capacitor = 30μC

Resistor = 49.0Ω

Voltage = 30.0 V

Frequency = 60.0 Hz

We need to calculate the impedance

Using formula of impedance

$Z=\sqrt{R^2+X_{c}^2}$ .....(I)

We need to calculate the value of $X_{c}$

Using formula of $X_{c}$

$X_{c}=\dfrac{1}{2\pi f c}$

$X_{c}=\dfrac{1}{2\times\pi\times60.0\times30\times10^{-6}}$

$X_{c}=88.42\ \Omega$

Put the value of $X_{c}$ into the formula of impedance

$Z=\sqrt{(49.0)^2+(88.42)^2}$

$Z=101.08\ \Omega$

(a). We need to calculate the rms current in the circuit

Using formula of rms current

$I_{rms}=\dfrac{V}{Z}$

$I_{rms}=\dfrac{30.0}{101.08}$

$I_{rms}=0.30\ A$

The rms current in the circuit is 0.30 A.

(b). We need to calculate the rms voltage drop across the resistor

Using formula of rms voltage

$V_{rms}=I_{rms}\times R$

Put the value into the formula

$V_{rms}=0.30\times49.0$

$V_{rms}=14.7\ V$

The rms voltage drop across the resistor is 14.7 V

(c). We need to calculate the rms voltage drop across the capacitor

Using formula of rms voltage

$V_{rms}=I_{rms}\times X_{C}$

$V_{rms}=0.30\times88.42$

$V_{rms}=26.53\ V$

The rms voltage drop across the capacitor is 26.53 V.

Hence, This is the required solution.

4 0

2 years ago

A spring is hanging from the ceiling. When a 250 gram of mass is attached to the free end, the spring elongates by 5 cm. The spr

lord [1]

Answer:

k = 49 N/m

Explanation:

Given that,

Mass, m = 250 g = 0.25 kg

When the mass is attached to the end of the spring, it elongates 5 cm or 0.05 m. We need to find the spring constant. Let it is k.

The force due to mass is balanced by its weight as follows :

mg=kx

$k=\dfrac{mg}{x}\\\\k=\dfrac{0.25\times 9.8}{0.05}\\\\k=49\ N/m$

So, the spring constant of the spring is 49 N/m.

8 0

3 years ago

First to answer gets brainliest

marissa [1.9K]

Letter na that would be the answer

4 0

3 years ago