6. What velocity vector will move you 200 miles east in 4 hours travelingat a constant speed?

most waves approach the shore at an angle. however, they bend to be nearly parallel to the shore as they approach it because bla

Alex

Most waves approach the shore at an angle. However, they bend to be nearly parallel to the shore as they approach it because when a wave reaches a beach or coastline, it releases a burst of energy that generates a current, which runs parallel to the shoreline.

Most waves approach shore at an angle. As each one arrives, it pushes water along the shore, creating what is known as a longshore current within the surf zone.
Waves approach the coast at an angle because of the direction of prevailing wind.
The part of the wave in shallow water slows down, while the part of the wave in deeper water moves at the same speed.
Thus when wave reaches a beach or coastline, it releases a burst of energy that generates a current, which runs parallel to the shoreline.

To know more about waves visit:

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6 0

1 year ago

What do you think is the difference between being imaginative in doing science and Doing pseudoscience?

jenyasd209 [6]

I would say that being imaginative in science means finding new ways to find or analyse the data, but which do not make the data false or which do not manipulate the data. Being imaginative in science should not cross the boundary of making the science not reliable

Being imaginative in pseudoscience could mean manipulating the data or even making experiments in such a way as to only obtaining specific data or throwing away certain data.

7 0

3 years ago

Students in an introductory physics lab are performing an experiment with a parallel-plate capacitor made of two circular alumin

gogolik [260]

Answer:

$7.54\cdot 10^{-7} C$

Explanation:

The capacitance of a parallel-plate capacitor is given by

$C=\frac{\epsilon_0 A}{d}$

where $\epsilon_0$ is the vacuum permittivity, A is the surface area of the plates, d their separation.

We also know the following relationship

$C=\frac{Q}{V}$

where Q is the charge stored on the capacitor and V the potential difference between the plates.

Combining the two equations,

$\frac{\epsilon_0 A}{d}=\frac{Q}{V}$

We also know that for a uniform electric field (such as the one between the plates of a parallel-plate capacitor), we have

$V= Ed$

where E is the magnitude of the electric field. Substituting into the previous equation and re-arranging it,

$\frac{\epsilon_0 A}{d}=\frac{Q}{Ed}\\Q=\frac{\epsilon_0 A E d}{d}=\epsilon_0 A E$

For the capacitor in the problem:

$A=\pi r^2 = \pi (\frac{d}{2})^2 = \pi (\frac{0.19 m}{2})^2=0.0284 m^2$ is the area of the plates

$E=3\cdot 10^6 N/C$ is the maximum electric field before a spark is produced

Solving for Q, we find the maximum charge that can be added before that occurs:

$Q=(8.85\cdot 10^{-12})(0.0284)(3\cdot 10^6)=7.54\cdot 10^{-7} C$

3 0

3 years ago

Consider a wet banked roadway, where there is a coefficient of static friction of 0.30 and a coefficient of kinetic friction of

gulaghasi [49]

Answer:

a) v = 20.9 m/s

b) v = 8.46 m/s

Explanation:

Given:-

- The coefficient of static friction is us = 0.30

- The coefficient of static friction is uk= 0.25

- The radius of the curve R = 50m

- The bank Angle β = 25

Find:-

a) If the banking angle is β=25∘, what is the maximum speed the automobile can have before sliding up the banking?

b) What is the minimum speed the automobile can have before sliding down the banking?

Solution:-

- We will investigate the sliding-up case first. Develop a FBD as given in (attachment).

- Use Newton's second law of motion vertical to slope of bank where the car is in equilibrium:

Sum ( F_n ) = 0

N*cos(β) - m*g - Ff*sin(β) = 0

Where, Frictional Force Ff = us*N

N (cos(β) - us*sin(β)) = mg ... Eq 1

- Use Newton's second law of motion horizontal to slope of bank where the car is accelerating:

Sum ( F_h ) = m*a

Ff*cos(β) + Nsin(β) = m*v^2 / R

N (us*cos (β) + sin (β) ) = m*v^2 / R .... Eq 2

- Divide the two equations:

v^2 / gR = [ us*cos (β) + sin (β) ] / [ cos (β) - us*sin (β) ]

v^2 = [ 0.25*cos (25) + sin (25) ]*9.81*50 / [ cos (25) - 0.25*sin (25) ]

v = 20.9 m/s

- For the slip down case. We have, friction force Ff reversed hence us = -us. Then the v can be given as:

v^2 / gR = [ -us*cos (β) + sin (β) ] / [ cos (β) + us*sin (β) ]

v^2 = [ -0.25*cos (25) + sin (25) ]*9.81*50 / [ cos (25) + 0.25*sin (25) ]

v = 8.46 m/s

6 0

3 years ago

A beam of light traveling in air strikes a glass slab at an angle of incidence less than 90°. After entering the glass slab, wha

ad-work [718]

Answer:

True A and B

Explanation:

Let's propose the solution of the exercise before seeing the affirmations.

We use the law of refraction

n₁ sin θ₁ = n₂. Sin θ₂

Where n₁ and n₂ are the refractive indices of the two means, θ₁ and θ₂ are the angles of incidence and refraction, respectively

sin θ₂ = (n1 / n2) sin θ₁

Let's apply this equation to the case presented. The index of refraction and airs is 1 (n1 = 1)

Sin θ₂ = (1 / n2) sin θ₁

the angle θ₂ which is the refracted angle is less than the incident angle

Let's analyze the statements time

A. False. We saw that it deviates

B. True Approaches normal (vertical axis)

C. False It deviates, but it is not parallel to normal

D. False It deviates, but approaching the normal not moving away

E. True. Because its refractive index is higher than air,

8 0

3 years ago