All categories

2 years ago

How does proximity to oceans affect Temperatures and rainfall amounts?

Physics

1 answer:

vampirchik [111]2 years ago

4 0

Answer:

Because they absorb and emit heat more steadily than land, oceans help moderate temperatures in coastal areas, making winters warmer and summers cooler. Winds that blow in from across the oceans bring more rainfall and higher humidity.

Explanation:

You might be interested in

Water from a vertical pipe emerges as a 10-cm-diameter cylinder and falls straight down 7.5 m into a bucket. The water exits the

cupoosta [38]

The diameter of the column of the water as it hits the bucket is 4.04 cm

The equation of continuity occurs in the fluid system and it asserts that the inflow and the outflow of the volume rate at the inlet and at the outlet of the system are equal.

By using the kinematics equation to determine the speed of the water in the bucket and applying the equation of continuity to estimate the diameter of the column, we have the following;

Using the kinematics equation:

$\mathbf{v_f ^2 = v_i^2 + 2gh}$

$\mathbf{v_f ^2 =(2.0)^2 + 2\times 9.8 \times 7.5}$

$\mathbf{v_f ^2 =151 m/s}$

$\mathbf{v_f =\sqrt{151 m/s}}$

$\mathbf{v_f =12.29 \ m/s}$

From the equation of continuity:

$\mathbf{A_iV_i = A_fV_f}$

$\mathbf{\pi r^2_iV_i = \pi r^2_fV_f}$

$\mathbf{ r^2_iV_i = r^2_fV_f}$

$\mathbf{ (\dfrac{10}{2})^2\times 2.0 = r_f^2 \times 12.29}$

$\mathbf{ 50 = 12.29 \times r_f^2}$

$\mathbf{ r_f= \sqrt{\dfrac{50}{12.29} }}$

$\mathbf{ V_f= 2.02 \ cm }$

Since diameter = 2r;

∴

The diameter of the column of the water is:

= 2(2.02) cm

= 4.04 cm

Learn more about the equation of continuity here:

brainly.com/question/10822213

6 0

2 years ago

Until a train is a safe distance from the station, it must travel at 5 m/s. Once the train is on open track, it can speec

nadezda [96]

Answer:

The acceleration of the train is 5 m/s².

Explanation:

Given:

let the initial velocity of a train = 5 m/s and

final velocity of a train = 45 m/s

time taken = 8 s

To find:

acceleration: ?

Solution:

We define acceleration as change in velocity per unit time that is the difference between the final velocity and initial velocity divided by time.

$Acceleration = \frac{\textrm{final velocity} - \textrm{initial velocity}}{time} \\$

On substituting the above values we get the required acceleration

$Acceleration = \frac{45 - 5}{8}\\ =\frac{40}{8}\\ =5\ m/s^{2}$

Therefore,the acceleration of the train is 5 m/s².

4 0

3 years ago

Convert the following angles in degrees to radians: (a) 300° (b) 18° (c) 105°

almond37 [142]

Answer:

a) 5.23599

b) 0.314159

c) 1.8326

Explanation:

5 0

2 years ago

Read 2 more answers

A car is traveling 16 m/s East. If the car then speeds up at a constant acceleration, what is the direction of the car’s acceler

konstantin123 [22]

The direction is east since its constant

8 0

2 years ago

Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.

SIZIF [17.4K]

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

(b) m1 = 915kg;

m2 = 605kg;

m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

$F_{pull}=m_{T}.a$

$m_{T}=\frac{F_{pull}}{a}$

$m_{T}=\frac{3615}{1.516}$

$m_{T}=2384.5$

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

For $m_{1}$ :

The only force acting On the $m_{1}$ box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

$m_{1} = \frac{F_{12}}{a}$

$m_{1} = \frac{1387}{1.516}$

$m_{1}$ = 915kg

For $m_{2}$ :

There are two forces acting on $m_{2}$ : tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

$m_{2} = \frac{F_{23}-F_{12}}{a}$

$m_{2} = \frac{2304-1387}{1.516}$

$m_{2}$ = 605kg

For $m_{3}$ :

$m_{3} = m_{T} - (m_{1}+m_{2})$

$m_{3} = 2384.5-1520.0$

$m_{3}$ = 864.5kg

8 0

3 years ago