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3 years ago

Can someone help me with this?

Physics

1 answer:

oksian1 [2.3K]3 years ago

5 0

Answer:

Yes

Explanation:

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Is the refraction different entering medium that has a higher index of refraction compared to entering amedium that has a lower

oee [108]

Answer:

Yes

Explanation:

The speed of light when it travels through glass, diamond, etc, the light travels at different speed from the speed of light. Speed of the light in material is related to the index of refraction.

The change in speed which occurs when the light passes from one medium to the another is responsible for bending of the light which is called as refraction.

When the light goes into a medium with the higher index of the refraction, light bends towards normal. Conversely, if the light traveling goes from higher refractive index to lower refractive index, it will bend away from the normal.

Hence, the refraction is different in both the scenario.

6 0

3 years ago

Suppose the electric field in some region is found to be E = kr3 ˆr, in spherical coordinates (k is some constant). (a) Find the

Assoli18 [71]

Answer:

Part a)

$\rho = 3\epsilon_0 k r^2$

Part b)

$Q = 4\pi \epsilon_0kR^5$

Explanation:

Part a)

As we know that electric field intensity due to some given charge distribution is given as

$E = kr^3 \hat r$

now electric flux through a spherical surface of radius r is given as

$\phi = E. A$

$\phi = kr^3(4\pi r^2)$

now by Guass law we know that

$E.A = \frac{q}{\epsilon_0}$

$q = 4\pi \epsilon_0kr^5$

now volume charge density is given as

$\rho = \frac{q}{\frac{4}{3}\pi r^3}$

$\rho = 3\epsilon_0 k r^2$

Part b)

Total charge inside the radius R is given as

$Q = 4\pi \epsilon_0kR^5$

7 0

3 years ago

The isotope of an atom containing 40 protons and 51 neutrons suddenly has 2 neutrons added to it. What isotope is created?

timama [110]

The isotope of an atom containing 40 protons and 51 neutrons suddenly has 2 neutrons added to it
That is X-93 so it will be
Zirconium-93
hope it helps

7 0

3 years ago

Two asteroids collide and stick together. The first asteroid has mass of 1.50 × 104 kg andis initially moving at 0.77 × 103 m/s.

KengaRu [80]

Answer:

Magnitude 900m/s, direction 12.8° respect to the velocity of the first asteroid.

Explanation:

This is a perfectly inelastic collision, because the two asteroids stick together at the end. That means that the kinetic energy doesn't conserves, but the linear momentum does. But, since the velocities of the asteroids have different directions, we have to break down them in components. For convenience, we will take the direction of the first asteroid as x-axis, and its perpendicular direction (in the plane of the two velocity vectors) as y-axis. So, we have that:

$p_{1ox}+p_{2ox}=p_{fx}\\\\p_{2oy}=p_{fy}$

And, since $p=mv$ , we get:

$m_1v_{1o}+m_2v_{2o}\cos\theta=(m_1+m_2)v_{fx}\\\\m_2v_{2o}\sin\theta=(m_1+m_2)v_{fy}$

Solving for v_fx and v_fy, and calculating their values, we get:

$v_{fx}=\frac{m_1v_{1o}+m_2v_{2o}\cos\theta}{m_1+m_2}\\\\\implies v_{fx}=\frac{(1.50*10^{4}kg)(0.77*10^{3}m/s)+(2.00*10^{4}kg)(1.02*10^{3}m/s)\cos20\°}{1.50*10^{4}kg+2.00*10^{4}kg}=878m/s\\\\\\v_{fy}=\frac{m_2v_{2o}\sin\theta}{m_1+m_2}\\\\\implies v_{fy}=\frac{(2.00*10^{4}kg)(1.02*10^{3}m/s)\sin20\°}{1.50*10^{4}kg+2.00*10^{4}kg}=199m/s$

Now, the final speed can be calculated using the Pythagorean Theorem:

$v_f=\sqrt{v_{fx}^{2}+v_{fy}^{2}} \\\\\implies v_f=\sqrt{(878m/s)^{2}+(199m/s)^{2}}=900m/s$

And the direction $\beta=\arctan \frac{v_{fy}}{v_{fx}}\\ \\\implies \beta=\arctan\frac{199m/s}{878m/s}=12.8\°$ can be obtained using trigonometry:

$\beta=\arctan \frac{v_{fy}}{v_{fx}}\\ \\\implies \beta=\arctan\frac{199m/s}{878m/s}=12.8\°$

That means that the final velocity of the two asteroids has a magnitude of 900m/s and a direction of 12.8° with respect to the velocity of the first asteroid.

7 0

3 years ago

Plutonium-239 is a radioactive isotope commonly used as fuel in nuclear reactors. The half-life of plutonium-239 is 24,100 year

r-ruslan [8.4K]

Answer:

72,300 years.

Explanation:

Initial mass of this sample: 504 grams;

Current mass of this sample: 63 grams.

What's the ratio between the current and the initial mass of this sample? In other words, what fraction of the initial sample hasn't yet decayed?

$\displaystyle \frac{\text{Current Mass}}{\text{Initial Mass}} = \rm \frac{63\; g}{504\; g} = \frac{1}{8}$ .

The value of this fraction starts at 1 decreases to 1/2 of its initial value after every half-life. How many times shall 1/2 be multiplied to 1 before reaching 1/8? $2^{3} = 8$ . It takes three half-lives or $3\times 24100 = 72300$ years to reach that value.

In certain questions the denominator of the fraction is large. It might not even be an integer power of 2. The base-x logarithm function on calculators could help. Evaluate

$\displaystyle \log_{\frac{1}{2}}{\frac{1}{8}} = 3$ to find the number of half-lives required. In case the base-x logarithm function isn't available, but the natural logarithm function $\ln()$ is, apply the following expression (derived from the base-changing formula) to get the same result:

$\displaystyle \frac{\displaystyle\ln{\left(\frac{1}{8}\right)}}{\displaystyle \ln{\left(\frac{1}{2}\right)}}$ .

3 0

3 years ago