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ASHA 777 [7]
3 years ago
8

Why do sea breezes occur during the day?

Physics
2 answers:
Sergio039 [100]3 years ago
7 0

Sea breezes occur during the day because solar radiation warms the land more than the water. This causes the warmer air over the land to rise. The resulting convection current causes wind to blow in from the sea.

Hatshy [7]3 years ago
6 0
The reason why is because the solar radiation warms land more than wind and because the unequal heating rates of land and water. Hope this helps :)
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1. How much force would you have to apply to a 15kg object in order to accelerate it!<br> a 2 m/s?
mihalych1998 [28]

30N

Explanation:

Given parameters:

Mass of object = 15kg

Acceleration = 2m/s

Unknown:

Force = ?

Solution:

Force is given as the product of mass and acceleration:

         F = m x a

   m is the mass

    a is the acceleration

Inputting the parameters:

     F = 15 x 2 = 30N

The unit of force is newtons, N .

Learn more:

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3 0
3 years ago
A ball is tossed up in the air. at its peak, it stops before beginning to fall. the ball at its peak has
Arte-miy333 [17]
The projectile (ball) reaches an instantaneous vertical speed (Vy) of zero at maximum height.

so, V(max height) = ¬Г(Vx)^2+(Vy)^2
in this case V(max height) = Vx, where Vy=0

The maximum height, Yf, can be solved using Vfy^2=Viy^2 + 2gy. At maximum height Vfy=0.
3 0
4 years ago
Read 2 more answers
In Ancient Greece, athletes competing in the long jump used handheld weights called halteres to lengthen their jumps. You are a
katovenus [111]

The halter add the distance to the jump in meters is 0.55 m.

<h3>What is projectile?</h3>

When an object is thrown at an angle from the horizontal direction, the object is said to be in projectile motion. The object which follows the projectile motion is called the projectile.

The magnitude of velocity u =10.3 m/s, angle of jumping θ = 22.8 degrees.

Components of velocity in x and y direction are

Vx = 10.3 cos 22.8 = 9.5 m/s

Vy = 10.3 sin 22.8 = 4 m/s

Maximum Range of athlete achieved using halter is given by

R = u²sin2θ /g

where, u = initial velocity, θ is the angle of projection and g is the gravitational acceleration.

Substituting the values, we get

R = (10.3)² sin(2 x 22.8 °) / 2 x 9.81

R = 7.75m

At the peak of jump you throw two 5.5 kg masses horizontally behind you such that their velocity is zero in the ground's reference frame.

The momentum is conserved in this situation,

(M+2m)Vxo =MVx'

Vx' = (M+2m)/M x Vxo'

Change in x component of velocity ΔVx = Vx' -Vxo

Vxo = 2m/M x Vx

Vxo = 2 x 5.5 /78 x 9.5

Vxo = 1.34 s

Maximum height gained when final velocity is zero

Vy = 0 = Vyo -gt

time t = Vyo/g = 4/9.8 = 0.41s'

Increase in range by using of halters is

ΔR = ΔVx' x t

ΔR = 1.34 x 0.41

ΔR =0.55m

Thus, the halter add the distance to the jump in meters is 0.55 m.

Learn more about projectile.

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