All categories

3 years ago

Why do clouds tend to form over land with a sea breeze and over water with a land breeze?

Physics

1 answer:

Masja [62]3 years ago

6 0

Because the waves in the water with the fan like system.

You might be interested in

Please help only eight questions C:

Strike441 [17]

1 2 answer

2 3 answer

3 1 answer

4 3 answer

5 0

3 years ago

Read 2 more answers

Two teams are playing tug of war. Team A pulls to the right with a force of 450 N. Team B pulls to the left with a force of 415

melamori03 [73]

Answer:

Two teams are playing tug of war. Team A pulls to the right with a force of 450 N. Team B pulls to the left with a force of 415 N

4 0

3 years ago

The total magnification of a specimen being viewed with a 10X ocular lens and a 40X objective lens is _____.

raketka [301]

Answer:

total magnification = 400 X

Explanation:

given data

ocular lens = 10 X

objective lens = 40 X

to find out

total magnification

solution

we know that total magnification is express as

total magnification = Objective magnification × ocular magnification .................1

put here value we get

total magnification = 10 × 40

total magnification = 400 X

3 0

3 years ago

A circuit has a voltage drop of 54.0 V across a 30.0 o resistor that carries a current of 1.80 A. What is the power used by the

Blizzard [7]

Answer:

P = 97.2 W

Explanation:

Given that,

Voltage drop, V = 54 V

The resistance of the resistor, R = 20 Ohms

Current, I = 1.8 A

We need to find the power used by the resistor. The formula used to find the power is given by :

P = VI

Putting all the values,

P = 54 V × 1.8 A

P = 97.2 W

So, the power used by the resistor is 97.2 W.

5 0

3 years ago

Read 2 more answers

A ball is launched at an angle of 10 degrees from a 20 meter tall building with a speed of 4 m/s. How long is the ball in the ai

shutvik [7]

Answer:

d) None of the above

Explanation:

$v_{o}$ = inituial velocity of launch = 4 m/s

θ = angle of launch = 10 deg

Consider the motion along the vertical direction

$v_{oy}$ = initial velocity along vertical direction = 4 Sin10 = 0.695

m/s

$a_{y}$ = acceleration along the vertical direction = - 9.8 m/s²

y = vertical displacement = - 20 m

t = time of travel

using the equation

$y=v_{oy} t+(0.5)a_{y} t^{2}$

- 20 = (0.695) t + (0.5) (- 9.8) t²

t = 2.1 sec

consider the motion along the horizontal direction

x = horizontal displacement

$v_{ox}$ = initial velocity along horizontal direction = 4 Cos10 = 3.94 m/s

$a_{x}$ = acceleration along the horizontal direction = 0 m/s²

t = time of travel = 2.1 s

Using the kinematics equation

$x =v_{ox} t+(0.5)a_{x} t^{2}$

x = (3.94) (2.1) + (0.5) (0) (2.1)²

x = 8.3 m

Consider the motion along the vertical direction

$v_{oy}$ = initial velocity along vertical direction = 4 Sin10 = 0.695

m/s

$a_{y}$ = acceleration along the vertical direction = - 9.8 m/s²

$y_{o}$ =initial vertical position at the time of launch = 20 m

$y$ = vertical position at the maximum height = 20 m

$v_{fy}$ = final velocity along vertical direction at highest point = 0 m/s

using the equation

${v_{fy}}^{2}= {v_{oy}}^{2} + 2 a_{y}(y - y_{o})$

$0^{2}= 0.695^{2} + 2 (- 9.8)(y - 20)$

$y$ = 20.02 m

h = height above the starting height

h = $y$ - $y_{o}$

h = 20.02 - 20

h = 0.02 m

7 0

3 years ago