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3 years ago

When do sea breezes occur?

2 answers:

7 0

The answer is B.

Sea breezes only occurs during the night because the land is cooler than the sea.

This causes hot air to rise from the sea and move towards a cooler area, this is called convection currents.

During the night, the pressure above the land is greater than the pressure above the sea, therefore, wind is travelled from a cooler area to a hotter area.

The opposite occurs during the day, because the sea is cooler than the land during the day.

azamat3 years ago

3 0

The answer would be C, but cooling surface temperatures make it cooler at night, as the water tends to cool down as fast.

You might be interested in

a person walks 750m due north then 250 m due east of the entire walk takes 12 min find the person average velocity

Virty [35]

Answer:

v = 1.098 m/s

Explanation:

Given that,

A person walks 750m due north then 250 m due east of the entire walk takes 12 min.

We need to find the average velocity of the person.

Displacement,

$d=\sqrt{750^2+250^2} \\\\d=790.56\ m$

Average velocity = displacement/time

$v=\dfrac{790.56\ m}{12\times 60\ s}\\\\v=1.098\ m/s$

So, the average velocity of the person is 1.098 m/s.

6 0

3 years ago

It is winter in Puerto Rico. Compare the air temperatures beachgoers feel near the water

Leviafan [203]

Large amounts of water do have a big impact on the weather: indeed, it takes less energy to warm/cool land than water.
Therefore, places near large amounts of water tend to have smaller differences in temperature between summer and winter than places far from waters.

Hence, during winter in Puerto Rico, alongside the coast, the temperature will be higher than in the innermost parts of the island.

7 0

3 years ago

A bicycile traves with an average velocty 15km/h north for 20min what is his displasment

Mila [183]

Answer:

5000 m equivalent to 5 Km

Explanation:

Average velocity = $\frac{Displacement}{Time}$

so, Displacement =Average velocity × time

We should convert Km/h to m/s so Km/h ⇒ $\frac{1000}{60*60}$ m/s, also convert time to second so, 20min ⇒(20* 60)seconds

Displacement = (15 × $\frac{1000}{60*60}$ ) × (20×60) =5000m OR 5Km

6 0

3 years ago

The cabinet is mounted on coasters and has a mass of 45 kg. The casters are locked to prevent the tires from rotating. The coeff

stira [4]

Answer:

the force P required for impending motion is 132.3 N

the largest value of "h" allowed if the cabinet is not to tip over is 0.8 m

Explanation:

Given that:

mass of the cabinet m = 45 kg

coefficient of static friction μ = 0.30

A free flow body diagram illustrating what the question represents is attached in the file below;

The given condition from the question let us realize that ; the casters are locked to prevent the tires from rotating.

Thus; considering the forces along the vertical axis ; we have :

$\sum f_y =0$

The upward force and the downward force is :

$N_A+N_B = mg$

where;

$\mathbf { N_A \ and \ N_B}$ are the normal contact force at center point A and B respectively .

$N_A+N_B = 45*9.8$

$N_A+N_B = 441$ ------- equation (1)

Considering the forces on the horizontal axis:

$\sum f_x = 0$

$F_A +F_B = P$

where ;

$\mathbf{ F_A \ and \ F_B }$ are the static friction at center point A and B respectively.

which can be written also as:

$\mu_s N_A + \mu_s N_B = P$

$\mu_s( N_A + N_B) = P$

replacing our value from equation (1)

P = 0.30 ( 441)

P = 132.3 N

Thus; the force P required for impending motion is 132.3 N

b) Since the horizontal distance between the casters A and B is 480 mm; Then half the distance = 480 mm/2 = 240 mm = 0.24 cm

the largest value of "h" allowed for the cabinet is not to tip over is calculated by determining the limiting condition of the unbalanced torque whose effect is canceled by the normal reaction at $N_A$ and it is shifted to $N_B$ :