In a summer storm, the wind is blowing with a velocity of 24 m/s north. Suddenly in 3 seconds the wind's velocity is 6 m/s. What is the wind's deceleration?

Answer:

Shift towards the normal

Explanation:

Refraction is defined as the change in direction of light rays when passing through from a medium to another.

The ray can either pass through a less dense medium to a denser medium or from a denser medium to a less dense medium. The light ray bends towards or away from the normal ray(ray perpendicular to the plane) depending whether the travels from less dense to denser or otherwise.

Note that if a ray travels from less dense medium which have a low refractive index like air to a more dense medium like water which have a higher refractive index than air, the refracted ray tends to bend towards the normal, otherwise they bend away from the normal.

Answer:

Shift towards the normal

Explanation:

Answer:

They would find [tex]3.14\times 10^{10}[/tex]galaxies.

Explanation:

Given that,

The number of density of galaxies in the universe is 3×10⁻⁶⁸ galaxies /m³.

Assuming that, the astronomers are observing at the center of sphere.

So, they can observe the sphere of space whose radius 10¹⁰ light years.

1 light year = 10¹⁶ meters

10¹⁰ light years =10¹⁰ .10¹⁶ meters

                         =10²⁶meters

The volume of the space is

=[tex]\frac43 \pi r^3[/tex]

[tex]=\frac43 \pi (10^{26})^3[/tex] m³

[tex]=\frac43 \pi 10^{78}[/tex] m³.

The number of galaxies

= Volume of the space × density

[tex]=(3\times 10^{-68}\ galaxies /m^3)\times(\frac43 \pi . 10^{78}\ m^3)[/tex]

[tex]=10^{10}\pi[/tex] galaxies

= [tex]3.14\times 10^{10}[/tex]galaxies

They would find [tex]3.14\times 10^{10}[/tex]galaxies.

If astronomers could observe all galaxies out to a distance of 10^10 light-years, they would find approximately 10^100 galaxies.

To find the number of galaxies that astronomers would find if they could observe all galaxies out to a distance of 1010 light-years, we can use the average number density of galaxies in the universe. The number density is given as 3 × 10-68 galaxies/m3. We can convert the distance to meters by multiplying by 1016 (since there are 1016 meters in 1 light-year).

Next, we can calculate the volume of space that astronomers would be observing. The volume can be found by multiplying the distance cubed (in meters) by 4/3π. The number of galaxies can then be calculated by multiplying the volume by the number density.

Substituting the given values into the equation, we have:

Volume = (4/3) × π × (1016)3 m3

Number of galaxies = Number density × Volume

After calculating the volume and multiplying it by the number density, we find that astronomers would find approximately 10100 galaxies if they could observe all galaxies out to a distance of 1010 light-years.

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Answer: B (Longitudinal waves travel through solids, liquids, and gases, Transverse waves only travel through  solids)

Explanation: A longitudinal wave alternately compresses the medium and stretches it out. Solids, liquids and gasses all push back when they are compressed— so they are all able to store energy this way and thus transmit the wave. But, Transverse waves can only go through solids because they have enough shear strength, but liquids and gases don't. ( Transverse waves are the transfer of energy in a motion that is perpendicular to the direction the wave is traveling. Only solids are able to switch it's motion to travel through the wave.)

Answer:

b

Explanation:

Answer:

- 58 cm

Explanation:

refractive index, n = 1.6

radius of curvature of left face, R1 = - 65 cm

Radius of curvature of the right face, R2 = 75 cm

Use the lens maker's formula

[tex]\frac{1}{f}=\left ( n-1 \right )\times \left ( \frac{1}{R_{1}}-\frac{1}{R_{2}} \right )[/tex]

[tex]\frac{1}{f}=\left ( 1.6-1 \right )\times \left ( -\frac{1}{65}-\frac{1}{75} \right )[/tex]

[tex]\frac{1}{f}=\left ( 0.6 \right )\times \left ( \frac{-75-65}{75\times 65}\right )[/tex]

f = - 58 cm

Thus, the focal length of the lens is - 58 cm.

Answer:

a) v=-6m/s

b) negative direction

c) 6m/s

d) decreasing

e) for t=2s

f) Yes

Explanation:

The particle position is given by:

[tex]x=4-12t+3t^2[/tex]

a) the velocity of the particle is given by the derivative of x in time:

[tex]v=\frac{dx}{dt}=-12+6t[/tex]

and for t=1s you have:

[tex]v=\frac{dx}{dt}=-12+6(1)^2=-6\frac{m}{s}[/tex]

b) for t=1s you can notice that the particle is moving in the negative x direction.

c) The speed can be computed by using the formula:

[tex]|v|=\sqrt{(-12+6t)^2}=\sqrt{(-12+6)^2}=6\frac{m}{s}[/tex]

d) Due to the negative value of the velocity in a) you can conclude that the speed is decreasing.

e) There is a time in which the velocity is zero. You can conclude that because if t=2 in the formula for v in a), v=0

[tex]v=0=-12+6(t)\\\\t=\frac{12}{6}=2[/tex]

f) after t=3s the particle will move in the negative direction, this because it is clear that 4+3t^2 does not exceed -12t.

Answer:

a)  v (1)  = -6 m/s

b) negative x-direction

c) s ( 1 ) = 6 m/s

d) The speed decreases at t increases from 0 to 2 seconds.

e) At t = 2 s, the velocity is 0

f) No

Explanation:

Given:-

- The position function of the particle:

                     x (t) = 4 - 12t + 3t^2

Find:-

what is its velocity at t = 1 s? (b) Is it moving in the positive or negative direction of x just then? (c) What is its speed just then? (d) Is the speed increasing or decreasing just then? (Try answering the next two questions without further calculation.) (e) Is there ever an instant when the velocity is zero? If so, give the time t; if not, answer no. (f) Is there a time after t = 3 s when the particle is moving in the negative direction of x? If so, give the time t; if not, answer no.

Solution:-

- The velocity function of the particle v(t) can be determined from the following definition:

                              v (t) = d x(t) / dt

                              v (t) = -12 + 6t

- Evaluate the velocity at time t = 1 s:

                              v (1) = -12 + 6(1)

                              v (1)  = -6 m/s

- The negative sign of the velocity at time t = 1s shows that the particle is moving in the negative x-direction.

- The speed ( s ( t )is the absolute value of velocity at time t = 1s:

                            s ( t ) = abs ( v ( t ) )

                            s ( 1 ) = abs ( v ( 1 ) )  

                            s ( 1 ) = abs ( -6 )

                            s ( 1 ) = 6 m/s

- The speed of the particle at time t = 0,

                            s ( t ) = abs ( -12 + 6t )

                            s ( 0 ) = abs (-12 + 6 (0) )  

                            s ( 0 ) = abs ( -12 )

                            s ( 0 ) = 12 m/s

- The speed of the particle at time t = 2,

                            s ( t ) = abs ( -12 + 6t )

                            s ( 2 ) = abs (-12 + 6 (2) )  

                            s ( 2 ) = abs (  0 )

                            s ( 2 ) = 0 m/s

- Hence, the speed of the particle decreases from s ( 0 ) = 12 m/s to s ( 2 ) = 0 m/s in the time interval t = 0 to t = 2 s.

- As the speed decreases as time increases over the interval t = 0 , t = 2 s the velocity v(t) also approaches 0, at time t = 2 s. s ( 2 ) = 0 m/s.

- We will develop an inequality when v (t) is positive:

                            v (t) = -12 + 6t > 0

                            6t > 12

                            t > 2

- So for all values of t > 2 the velocity of the particle is always positive.

Final answer:

The rocket's top speed is 46 m/s.

Explanation:

The rocket's top speed can be found using the equation v = vo + at, where v is the final velocity, vo is the initial velocity, a is the acceleration, and t is the time. In this case, the initial velocity (vo) is 0 m/s since the rocket starts at rest, the acceleration (a) is 10 m/s², and the time (t) is 4.6 s.

Plugging these values into the equation, we get:

v = 0 m/s + (10 m/s²)(4.6 s) = 46 m/s.

Therefore, the rocket's top speed is 46 m/s.