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3 years ago

How does the uneaven heating of earths surface affects earths weather patterns

Physics

1 answer:

Cloud [144]3 years ago

8 0

Answer: it causes some parts of the earth to get more radiation than others.

Explanation: earth rotates around the sun on a tilted axis so the Rays of the sun cause earth to have more radiation than it needs.

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) Water falls from a height of 60m at the rate of 15kg/s to operate a turbine. The losses due to frictional force are 10% of ene

Angelina_Jolie [31]

Answer:

8100W

Explanation:

Let g = 10m/s2

As water is falling from 60m high, its potential energy from 60m high would convert to power. So the rate of change in potential energy is

$P = \dot{E} = \dot{m}gh = 15*10*60 = 9000 J/s$ or 9000W

Since 10% of this is lost to friction, we take the remaining 90 %

P = 9000*90% = 8100 W

3 0

3 years ago

In a physics lab experiment, a spring clamped to the table shoots a 22g ball horizontally. When the spring is compressed 21cm ,

worty [1.4K]

a square and a triangle I think

6 0

4 years ago

A car is traveling up a hill that is inclined at an angle θ above the horizontal. Determine the ratio of the magnitude of the no

padilas [110]

Explanation:

The weight of the car is equal to, $W_c=m\times g$ ...........(1)

Where

m is the mass of car

g is the acceleration due to gravity

The normal or vertical component of the force is, $F_N=mg\ cos\theta$

$F_N=mg\ cos(13)$ .............(2)

The horizontal component of the force is, $F_H=mg\ sin\theta$

Taking ratio of equation (1) and (2) as :

$\dfrac{F_N}{W_c}=\dfrac{mg\ cos\theta}{mg}$

$\dfrac{F_N}{W_c}=cos(13)$

$\dfrac{F_N}{W_c}=0.97$

$\dfrac{F_N}{W_c}=\dfrac{97}{100}$

Hence, this is the required solution.

5 0

3 years ago

Can somebody help me on finding the constants if this experiment?

ki77a [65]

The constants could be the day because the experiment is all on the same day or the person throwing the pumpkin because it will always be the same perosn

8 0

3 years ago

A student uses a spring loaded launcher to launch a marble vertically in the air. The mass of the marble is

GarryVolchara [31]

Answer:

Part a)

When spring compressed by 2 cm

H = 1.47 m

Part b)

When spring is compressed by 4 cm

H = 5.94 m

Explanation:

Part a)

As we know that the spring is compressed and released

so here spring potential energy is converted into gravitational potential energy at its maximum height

So we will have

$\frac{1}{2}kx^2 = mg(H + x)$

$0.5(220)(0.02)^2 = 0.003(9.81)(H + 0.02)$

so we have

$H = 1.47 m$

Part b)

Similarly when spring is compressed by 4 cm

then we have

$\frac{1}{2}kx^2 = mg(H + x)$

$0.5(220)(0.04)^2 = 0.003(9.81)(H + 0.04)$

so we have

$H = 5.94 m$

8 0

4 years ago