All categories

2 years ago

SOMEONE PLEASE HELP ME URGENTLY WILL MARK AS BRAINLIEST

Physics

1 answer:

max2010maxim [7]2 years ago

4 0

Explanation:

I am not sure about this

You might be interested in

A beam of light traveling through a liquid (of index of refraction n1 = 1.47) is incident on a surface at an angle of θ1 = 59° w

frosja888 [35]

Answer:

(a) $n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}$

(b) $n_{2} = 1.349$

(d) $v_{2} = 2.22\times 10^{8}\ m/s$

Solution:

As per the question:

Refractive index of medium 1, $n_{1} = 1.47$

Angle of refraction for medium 1, $\theta_{1} = 59^{\circ}$

Angle of refraction for medium 2, $\theta_{1} = 69^{\circ}$

Now,

(a) The expression for the refractive index of medium 2 is given by using Snell's law:

$n_{1}sin\theta_{1} = n_{2}sin\theta_{2}$

where

$n_{2}$ = Refractive Index of medium 2

Now,

$n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}$

(b) The refractive index of medium 2 can be calculated by using the expression in part (a) as:

$n_{2} = \frac{1.47\times sin59^{\circ}}{sin69^{\circ}}$

$n_{2} = 1.349$

We know that:

$Refractive\ index,\ n = \frac{Speed\ of\ light\ in vacuum,\ c}{Speed\ of\ light\ in\ medium,\ v}$

Thus for medium 1

$n_{1} = \frac{c}{v_{1}$

$v_{1} = \frac{c}{n_{1} = \frac{3\times 10^{8}}{1.47} = 2.04\times 10^{8}\ m/s$

(d) To calculate the velocity of light in medium 2:

For medium 2:

$n_{2} = \frac{c}{v_{2}$

$v_{2} = \frac{c}{n_{1} = \frac{3\times 10^{8}}{1.349} = 2.22\times 10^{8}\ m/s$

5 0

2 years ago

Read 2 more answers

A candy-filled piñata is hung from a tree for Elia's birthday. During an unsuccessful attempt to break the 4.4-kg piñata, Tonja

Klio2033 [76]

Answer: v = 0.6 m/s

Explanation: Momentum Conservation Principle states that for a collision between two objects in an isolated system, the total momentum of the objects before the collision is equal to the total momentum of the objects after the collision.

Momentum is calculated as Q = m.v

For the piñata problem:

$Q_{i}=Q_{f}$

$m_{p}v_{p}_{i}+m_{s}v_{s}_{i}=m_{p}v_{p}_{f}+m_{s}v_{s}_{f}$

Before the collision, the piñata is not moving, so $v_{p}_{i}=0$ .

After the collision, the stick stops, so $v_{s}_{f}=0$ .

Rearraging, we have:

$m_{s}v_{s}_{i}=m_{p}v_{p}_{f}$

$v_{p}_{f}=\frac{m_{s}v_{s}_{i}}{m_{p}}$

Substituting:

$v_{p}_{f}=\frac{(0.54)(4.8)}{(4.4)}$

$v_{p}_{f}=$ 0.6

Immediately after being cracked by the stick, the piñata has a swing speed of 0.6 m/s.

3 0

3 years ago

At takeoff, a commercial jet has a speed of 72 m/s. Its tires have a diameter of 0.89 m. Part (a) At how many rev/min are the ti

34kurt

Answer:

a) Revolutions per minute = 2.33

b) Centripetal acceleration = 11649.44 m/s²

Explanation:

a) Angular velocity is the ratio of linear velocity and radius.

Here linear velocity = 72 m/s

Radius, r = 0.89 x 0. 5 = 0.445 m

Angular velocity

$\omega =\frac{72}{0.445}=161.8rad/s$

Frequency

$f=\frac{2\pi}{\omega}=\frac{2\times \pi}{161.8}=0.0388rev/s=2.33rev/min$

Revolutions per minute = 2.33

b) Centripetal acceleration

$a=\frac{v^2}{r}$

Here linear velocity = 72 m/s

Radius, r = 0.445 m

Substituting

$a=\frac{72^2}{0.445}=11649.44m/s^2$

Centripetal acceleration = 11649.44m/s²

6 0

2 years ago

Casey is playing with her yo-yo. She performs an "around the world" trick, in which the yo-yo travels up and around in a circle

USPshnik [31]

Answer:

It has the least potential energy at the bottom of its circular path.

Explanation:

It has the least potential energy at the bottom of its circular path.

Remember the equation

U = m*g*h

where U is the potential energy

m is the mass of the yo-yo

h is the height

4 0

2 years ago

You have a two-wheel trailer that you pull behind your ATV. Two children with a combined mass of 76.2 kg hop on board for a ride

marin [14]

a) The spring constant is 12,103 N/m

b) The mass of the trailer 2,678 kg

c) The frequency of oscillation is 0.478 Hz

d) The time taken for 10 oscillations is 20.9 s

Explanation:

When the two children jumps on board of the trailer, the two springs compresses by a certain amount

$\Delta x = 6.17 cm = 0.0617 m$

Since the system is then in equilibrium, the restoring force of the two-spring system must be equal to the weight of the children, so we can write:

$2mg = k'\Delta x$ (1)

where

m = 76.2 kg is the mass of each children

$g=9.8 m/s^2$ is the acceleration of gravity

$k'$ is the equivalent spring constant of the 2-spring system

For two springs in parallel each with constant k,

$k'=k+k=2k$

Substituting into (1) and solving for k, we find:

$2mg=2k\Delta x\\k=\frac{mg}{\Delta x}=\frac{(76.2)(9.8)}{0.0617}=12,103 N/m$

The period of the oscillating system is given by

$T=2\pi \sqrt{\frac{m}{k'}}$

where

And for the system in the problem, we know that

T = 2.09 s is the period of oscillation

m is the mass of the trailer

$k'=2k=2(12,103)=24,206 N/m$ is the equivalent spring constant of the system

Solving the equation for m, we find the mass of the trailer:

$m=(\frac{T}{2\pi})^2 k'=(\frac{2.09}{2\pi})^2 (24,206)=2,678 kg$

The frequency of oscillation of a spring-mass system is equal to the reciprocal of the period, therefore:

$f=\frac{1}{T}$

where

f is the frequency

T is the period

In this problem, we have

T = 2.09 s is the period

Therefore, the frequency of oscillation is

$f=\frac{1}{2.09}=0.478 Hz$

The period of the system is

T = 2.09 s

And this time is the time it takes for the trailer to complete one oscillation.

In this case, we want to find the time it takes for the trailer to complete 10 oscillations (bouncing up and down 10 times). Therefore, the time taken will be the period of oscillation multiplied by 10.

Therefore, the time needed for 10 oscillations is:

$t=10T=10(2.09)=20.9 s$

#LearnwithBrainly

7 0

3 years ago