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Explain how land breeze and sea breezes occur?

marshall27 [118]

Explanation:

during the daytime the land gets hot and the hot air blows up to the sky and to feel the space the cool you are from the Sea blows to the land this phenomenon is called sea breeze similarly at night the hot air of sea goes up n cool air from land blows to sea which is called land breexe

please mark me the brainliest

5 0

3 years ago

Read 2 more answers

A block of mass 2 kg slides down a frictionless ramp of length 1.3 m tilted at an angle 25o to the horizontal. At the bottom of

marin [14]

Answer:

Diagrams in pictures

Explanation:

Using energy I can get

m g h = 1/2 m v^2

So the velocity at the end of the ramp is the squareroot of two times the initial height of the box times the gravity constant.

(H= 1,3m sin25)

V=2,32m/s

V= a t

And

X= v t +1/2 a t^2

Knowing v=2,32 m/s and x= 1,3 m

I can get

a= 6,21m/s2

F= m a

I can get the force of the box when it collides with the spring

F= 12, 425 N

The force the spring makes on the box then is

F = -12,425N = -k d

Then the spring's constant is k= 51,75N/m

To make the two diagrams I need the functions of time when the box slows down

I use the same two equations

V= a t

And

X= v t + 1/2 a t^2

Being now 2,32 my initial velocity and 0 my final velocity, and my distance 0,24 m.

I get there the time t=0,0689 seconds and the acceleration a= -33,67 m/s2 (negative because it's slowing down).

Then,

V(t)= - 33,67 m/s2 t for time between 0 and 0,689 sec

X(t)= 2,32 m/s t + 1/2 33,67 m/s2 t^2.

for time between 0 and 0,689 sec

Diagrams and equations are in the pictures

7 0

3 years ago

A block, M1=10kg, slides down a smooth, curved incline of height 5m. It collides elastically with another block, M2=5kg, which i

erma4kov [3.2K]

Answer:

2.86 m

Explanation:

Given:

M₁ = 10 kg

M₂ = 5 kg

$\mu_k$ = 0.5

height, h = 5 m

distance traveled, s = 2 m

spring constant, k = 250 N/m

now,

the initial velocity of the first block as it approaches the second block

u₁ = √(2 × g × h)

or

u₁ = √(2 × 9.8 × 5)

or

u₁ = 9.89 m/s

let the velocity of second ball be v₂

now from the conservation of momentum, we have

M₁ × u₁ = M₂ × v₂

on substituting the values, we get

10 × 9.89 = 5 × v₂

or

v₂ = 19.79 m/s

now,

let the velocity of mass 2 when it reaches the spring be v₃

from the work energy theorem, we have

Work done by the friction force = change in kinetic energy of the mass 2

or

$0.5\times5\times9.8\times2 = \frac{1}{2}\times5\times( v_3^2-19.79^2)$

or

v₃ = 20.27 m/s

now, let the spring is compressed by the distance 'x'

therefore, from the conservation of energy

we have

Energy of the spring = Kinetic energy of the mass 2

or

$\frac{1}{2}kx^2=\frac{1}{2}mv_3^2$

on substituting the values, we get

$\frac{1}{2}\times250\times x^2=\frac{1}{2}\times5\times20.27^2$

or

x = 2.86 m

8 0

3 years ago

Two simple pendulum of slightly different length , are set off oscillating in step is a time of 20s has elasped , during which t

adell [148]

Answer:

Length of longer pendulum = 99.3 cm

Length of shorter pendulum = 82.2 cm

Explanation:

Since the longer pendulum undergoes 10 oscillations in 20 s, its period T = 20 s/10 = 2 s.

From T = 2π√(l/g), the length of the pendulum. l = T²g/4π²

substituting T = 2s and g = 9.8 m/s² we have

l = T²g/4π²

= (2 s)² × 9.8 m/s² ÷ 4π²

= 39.2 m ÷ 4π²

= 0.993 m

= 99.3 cm

Now, for the shorter pendulum to be in step with the longer pendulum, it must have completed some oscillations more than the longer pendulum. Let x be the number of oscillations more in t = 20 s. Let n₁ = number of oscillations of longer pendulum and n₂ = number of oscillations of longer pendulum.

So, n₂ = n₁ + x. Also n₁ = t/T₁ and n₂ = t/T₂ where T₂ = period of shorter pendulum.

t/T₂ = t/T₁ + x

1/T₂ = 1/T₁ + x (1)

Also, the T₂ = t/n₂ = t/(n₁ + x) (2)

From (1) T₂ = T₁/(T₁ + x) (3)

equating (2) and (3) we have

t/(n₁ + x) = T₁/(T₁ + x)

substituting t = 20 s and n₁ = 10 and T₁ = 2s, we have

20 s/(10 + x) = 2/(2 + x)

10/(10 + x) = 1/(2 + x)

(10 + x)/10 = (2 + x)

(10 + x) = 10(2 + x)

10 + x = 20 + 10x

collecting like terms

10x - x = 20 - 10

9x = 10

x = 10/9

x = 1.11

x ≅ 1 oscillation

substituting x into (2)

T₂ = t/n₂ = t/(n₁ + x)

= 20/(10 + 1)

= 20/11

= 1.82 s

Since length l = T²g/4π²

substituting T = 1.82 s and g = 9.8 m/s² we have

l = T²g/4π²

= (1.82 s)² × 9.8 m/s² ÷ 4π²

= 32.46 m ÷ 4π²

= 0.822 m

= 82.2 cm

6 0

3 years ago

(a) How many fringes appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit

Ainat [17]

Answer:

a

The number of fringe is z = 3 fringes

b

The ratio is $I = 0.2545I_o$

Explanation:

a

From the question we are told that

The wavelength is $\lambda = 600 nm$

The distance between the slit is $d = 0.117mm = 0.117 *10^{-3} m$

The width of the slit is $a = 35.7 \mu m = 35.7 *10^{-6}m$

let z be the number of fringes that appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit pattern is and this mathematically represented as

$z = \frac{d}{a}$

Substituting values

$z = \frac{0.117*10^{-3}}{35.7 *10^{-6}}$

z = 3 fringes

b

From the question we are told that the order of the bright fringe is n = 3

Generally the intensity of a pattern is mathematically represented as

$I = I_o cos^2 [\frac{\pi d sin \theta}{\lambda} ][\frac{sin (\pi a sin \frac{\theta}{\lambda } )}{\pi a sin \frac{\theta}{\lambda} } ]$

Where $I_o$ is the intensity of the central fringe

And Generally $sin \theta = \frac{n \lambda }{d}$

$I = I_o co^2 [ \frac{\pi (\frac{n \lambda}{d} )}{\lambda} ] [\frac{\frac{sin (\pi a (\frac{n \lambda}{d} ))}{\lambda} }{\frac{\pi a (\frac{n \lambda}{d} )}{\lambda} } ]$

$I = I_o cos^2 (n \pi)[\frac{\frac{sin(\pi a (\frac{n \lambda}{d} ))}{\lambda} )}{ \frac{ \pi a (\frac{n \lambda }{d} )}{\lambda} } ]$

$I = I_o cos^2 (3 \pi) [\frac{sin (\frac{3 \pi }{6} )}{\frac{3 \pi}{6} } ]$

$I = I_o (1)(0.2545)$

$I = 0.2545I_o$

6 0

4 years ago