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3 years ago

What are the main factors that influence the climate of any location on earth?

Physics

1 answer:

icang [17]3 years ago

8 0

Answer:

Explanation:

The climate of any particular place is influenced by a host of interacting factors.

The factor that influence climate are

1. latitude

2. elevation,

3. nearby water

4. ocean currents

5. topography

6. vegetation

7. prevailing winds.

The primary cause of climate change is the burning of fossil fuels, such as oil and coal, which emits greenhouse gases into the atmosphere—primarily carbon dioxide. Other human activities, such as agriculture and deforestation, also contribute to the proliferation of greenhouse gases that cause climate change.

You might be interested in

How does the currently accepted model of the nucleus provide evidence of the existence of the strong nuclear force?

olganol [36]

Answer:

Protons and neutrons are all attracted to each other as a result - the strong nuclear force. This is an attractive force that only has an effect over a very short range in the nucleus.

4 0

3 years ago

During a tennis volley, a ball that arrives at a player at 40 m/s is struck by the racquet and returned at 40 m/s. The other pla

Butoxors [25]

Answer:Racquet force is twice of Player force

Explanation:

Given

ball arrives at a speed of $u=-40\ m/s$

ball returned with speed of $v=40\ m/s$

average Force imparted by racquet on the ball is given by

$F_{racquet}=\frac{m(v-u)}{\Delta t}$

where $m=mass\ of\ ball$

$\Delta t=$ time of contact of ball with racquet

$F_{racquet}=\frac{m(40-(-40))}{\Delta t}$

$F_{racquet}=\frac{80m}{\Delta t}-----1$

When it land on the player hand its final velocity becomes zero and time of contact is same as of racquet

$F_{player}=\frac{m(0-40)}{\Delta t}$

$F_{player}=\frac{-40m}{\Delta t}-----2$

From 1 and 2 we get

$F_{racquet}=-2F_{player}$

Hence the magnitude of Force by racquet is twice the Force by player

5 0

3 years ago

Two workers are sliding 290 kg crate across the floor. One worker pushes forward on the crate with a force of 430 N while the ot

m_a_m_a [10]

Answer:0.27

Explanation:

Given

One worker Pushes with force $F_1=430 N$

other Pulls it with a rope of rope $F_2=360 N$

mass of crate $m=290 kg$

both forces are horizontal and crate slides with a constant speed

Both forces are in the same direction so Friction will oppose the forces and will be equal in magnitude of sum of two forces because crate is moving with constant speed i.e. net force is zero on it

$f_r=F_1+F_2$

where $f_r$ is the friction force

$f_r=430+360$

$f_r=790 N$

$f_r=\mu N$

where $\mu$ is the coefficient of static friction

$N=mg$

$790=\mu 29\cdot 9.8$

$\mu =0.27$

6 0

3 years ago

Suppose that you have a 680 Ω, a 720 Ω and a 1.20 kΩ resistor. (a) What is the maximum resistance you can obtain by combining th

Delvig [45]

Explanation:

As the given data is as follows.

$R_{1} = 680 \ohm$ ohm $\ohm$ , $R_{2} = 720 \ohm$ ohm,

$R_{3} = 1.2 k\ohm$ = 1200 $\ohm$ (as 1 k ohm = 1000 m)

(a) We will calculate the maximum resistance by combining the given resistances as follows.

Max. Resistance = $R_{1} + R_{2} + R_{3}$

= $(680 + 720 + 1200) \ohm$ ohm

= 2600 ohm

or, = 2.6 $k\ohm$ ohm

Therefore, the maximum resistance you can obtain by combining these is 2.6 $k\ohm$ ohm.

(b) Now, the minimum resistance is calculated as follows.

Min. Resistance = $\frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}$

= $\frac{1}{680} + \frac{1}{720} + \frac{1}{1200}$

= $3.683 \times 10^{-3}$ ohm

Hence, we can conclude that minimum resistance you can obtain by combining these is $3.683 \times 10^{-3}$ ohm.

3 0

2 years ago

In the equation for the gravitational force between two objects, which quantity must be squared?

Alex787 [66]

Answer:

The distance between the two objects must be squared.

Explanation:

Gravitational force always act between two objects that have mass. The gravitational force is a weak force and attractive in nature.

The force of pull depends on the masses of the two objects and the distance between them.

The formula to calculate gravitational force between two objects having masses 'm' and 'M' and separated by a distance 'd' is given as:

$F_g=\frac{GmM}{d^2}$

Where, 'G' is called the universal gravitational constant and its value is equal to $6.674\times10^{-11}\ m^3 kg^{-1} s^{-2}$ .

Now, from the above formula, it is clear that, the force of gravitation is inversely proportional to the square of the distance between the two objects.

Thus, the quantity that must be squared in the equation of gravitational force between two objects is the distance 'd'.

7 0

3 years ago