Ask question

Ask question

All categories

2 years ago

11

Which event typically causes upwelling? Wind blows warm water away from the shore. Warm water rises to the surface of the ocean.

Neap tides move cold water away from the shore. Cold water falls to the bottom of the ocean and pushes warm water up.

2 answers:

bixtya [17]2 years ago

7 0

Answer:

A. Wind blows warm water away from the shore.

Explanation:

just took the test

padilas [110]2 years ago

7 0

Answer:

AAAAAAAAAAAA sure about my answer

Explanation:

brainliest please

You might be interested in

When a vertical beam of light passes through a transparent medium, the rate at which its intensity I decreases is proportional t

antoniya [11.8K]

Answer:

Intensity of beam 18 feet below the surface is about 0.02%

Explanation:

Using Lambert's law

Let dI / dt = kI, where k is a proportionality constant, I is intensity of incident light and t is thickness of the medium

then dI / I = kdt

taking log,

ln(I) = kt + ln C

I = Ce^kt

t=0=>I=I(0)=>C=I(0)

I = I(0)e^kt

t=3 & I=0.25I(0)=>0.25=e^3k

k = ln(0.25)/3

k = -1.386/3

k = -0.4621

I = I(0)e^(-0.4621t)

I(18) = I(0)e^(-0.4621*18)

I(18) = 0.00024413I(0)

Intensity of beam 18 feet below the surface is about 0.2%

3 0

2 years ago

A projectile has an initial horizontal velocity of 15 meters per second and an initial vertical velocity of 25 meters per second

Artyom0805 [142]

Answer:

75 m

Explanation:

The horizontal motion of the projectile is a uniform motion with constant speed, since there are no forces acting along the horizontal direction (if we neglect air resistance), so the horizontal acceleration is zero.

The horizontal component of the velocity of the projectile is

$v_x = 15 m/s$

and it is constant during the motion;

the total time of flight is

t = 5 s

Therefore, we can apply the formula of the uniform motion to find the horizontal displacement of the projectile:

$d= v_x t =(15 m/s)(5 s)=75 m$

5 0

2 years ago

Given the indices of refraction n1 and n2 of material 1 and material 2, respectively, rank these scenarios on the basis of the p

lisov135 [29]

Answer:

$c>d>f=a>b>e$

Explanation:

When a pair of medial has greater difference between the their individual refractive indices with respect to vacuum then it has a greater deviation between the refracted ray and the incident ray.

According to the Snell's law:

$\rm refractive\ index\ (n)=\frac{speed\ of\ light\ in\ the\ incident\ medium}{speed\ of\ light\ in\ the\ refracted\ medium}$

a)

$n_1-n_2=1.33-1.00\\=0.33$

b)

$n_2-n_1=1.46-1.33$

$=0.23$

c)

$n_2-n_1=2.42-1.33\\=1.09$

d)

$n_2-n_1=1.46-1.00\\=0.46$

e)

$n_1-n_2=1.50-1.33\\=0.17$

f)

$n_2-n_1=1.33-1.00\\=0.33$

$c>d>f=a>b>e$

5 0

2 years ago

A pump lifts water from a lake to a large tank 20 m above the lake. How much work against gravity does the pump do as it transfe

Aleonysh [2.5K]

Answer:

980 kJ

Explanation:

Work = change in energy

W = mgh

W = (1000 kg/m³ × 5.0 m³) (9.8 m/s²) (20 m)

W = 980,000 J

W = 980 kJ

The pump does 980 kJ of work.

3 0

2 years ago

Read 2 more answers

Scientists studying an anomalous magnetic field find that it is inducing a circular electric field in a plane perpendicular to t

yarga [219]

Answer

The rate at which the magnetic field is changing is $[\frac{dB}{dt} ] = 0.000467 T/s$

Explanation

From the question we are told that

The electric field strength is $E = 3.5mV/m = 3.5 *10^{-3} \ V/m$

The radius is $r = 1.5 \ m$

The rate of change of the magnetic field is mathematically represented as

$\frac{d \phi }{dt} = \int\limits^{} {E \cdot dl}$

Where $dl$ is change of a unit length

$\frac{d \phi}{dt} = A * \frac{dB}{dt}$

Where A is the area which is mathematically represented as

$A = \pi r^2$

So

$E \int\limits^{} { dl} = ( \pi r^2) (\frac{dB}{dt} )$

$E L = ( \pi r^2) (\frac{dB}{dt} )$

where L is the circumference of the circle which is mathematically represented as

$L = 2 \pi r$

So

$E (2 \pi r ) = (\pi r^2 ) [\frac{dB}{dt} ]$

$E = \frac{r}{2} [\frac{dB}{dt} ]$

$[\frac{dB}{dt} ] = \frac{E}{ \frac{r}{2} }$

substituting values

$[\frac{dB}{dt} ] = \frac{3.5 *10^{-3}}{ \frac{15}{2} }$

$[\frac{dB}{dt} ] = 0.000467 T/s$

8 0

2 years ago