An ant takes 5 secs to complete one round

When an object radiates heat, the strength of this radiation far from the object decreases when distance from the source increas

Andrew [12]

Answer:

The source of cosmic background radiation filled the entire universe.

Explanation:

D:The source of cosmic background radiation filled the entire universe.

7 0

3 years ago

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Can anyone help with these questions please

Svet_ta [14]

Here... I've finished this course already

6 0

3 years ago

One recently discovered extrasolar planet, or exoplanet, orbits a star whose mass is 0.70 times the mass of our sun. This planet

Stels [109]

0.078 times the orbital radius r of the earth around our sun is the exoplanet's orbital radius around its sun.

Answer: Option B

<u>Explanation:</u>

Given that planet is revolving around the earth so from the statement of centrifugal force, we know that any

$\frac{G M m}{r^{2}}=m \omega^{2} r$

The orbit’s period is given by,

$T=\sqrt{\frac{2 \pi}{\omega r^{2}}}=\sqrt{\frac{r^{3}}{G M}}$

Where,

$T_{e}$ = Earth’s period

$T_{p}$ = planet’s period

$M_{s}$ = sun’s mass

$r_{e}$ = earth’s radius

Now,

$T_{e}=\sqrt{\frac{r_{e}^{3}}{G M_{s}}}$

As, planet mass is equal to 0.7 times the sun mass, so

$T_{p}=\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}$

Taking the ratios of both equation, we get,

$\frac{T_{e}}{T_{p}}=\frac{\sqrt{\frac{r_{e}^{3}}{G M_{s}}}}{\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}}$

$\frac{T_{e}}{T_{p}}=\sqrt{\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}}$

$\left(\frac{T_{e}}{T_{p}}\right)^{2}=\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}$

$\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}=\frac{r_{e}^{3}}{r_{p}^{3}}$

$\frac{r_{e}}{r_{p}}=\left(\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}$

Given $T_{p}=9.5 \text { days }$ and $T_{e}=365 \text { days }$

$\frac{r_{e}}{r_{p}}=\left(\left(\frac{365}{9.5}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=\left(\frac{133225}{90.25} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=(2108.82)^{\frac{1}{3}}$

$r_{p}=\left(\frac{1}{(2108.82)^{\frac{1}{3}}}\right) r_{e}=\left(\frac{1}{12.82}\right) r_{e}=0.078 r_{e}$

7 0

3 years ago

An alpha particle collides with an oxygen nucleus, initially at rest. The alpha particle is scattered at an angle of 25.0° above

Greeley [361]

Answer:

$v_{i}= 19\times 10^5\ m/s$

Explanation:

given,

scattering angle of alpha particle = 25.0° above its initial direction of motion

oxygen nucleus recoils at = 50.0° below this initial direction.

final speed of the oxygen = 2.08×10⁵ m/s

mass of alpha particle = 4.0 u

mass o oxygen nucleus = 16 u

momentum conservation along x- axis

$m_{a}v_{i} = m_a v_a cos\theta + m_o v_o cos\theta$

$4v_{i} = 4\times v_a cos25^0 + 16\times 2.08 \times 10^5 cos50^0$

$v_{i}= \dfrac{3.625\times v_a+ 21.39\times 10^5}{4}$ ....(1)

Along y-direction

$0 = m_av_a sin \theta - m_ov_o sin\theta$

$0 = 4\times v_a sin 25 - 16\times 2.08 \times 10^5 sin50^0$

$v_a = \dfrac{25.49 \times 10^5}{1.69}$

$v_a = 15.082\times 10^5\ m/s$

putting value in equation (1)

$v_{i}= \dfrac{3.625\times 15.082\times 10^5+ 21.39\times 10^5}{4}$

$v_{i}= 19\times 10^5\ m/s$

5 0

4 years ago

What causes the movement of liquid in a lava lamp

Andru [333]

Answer:

I think theres air that flows thru it

Explanation:

8 0

2 years ago

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