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3 years ago

Which winds are affected by specific landforms on earth's surface?

1 answer:

4 0

Hello!

The winds affected by specific landforms on earth's surface are: Local winds.

I hope my answer helped you out! :)

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A man is running on the straight road with the uniform velocity of 3 metre per second calculate the acceleration for produced hi

sp2606 [1]

Theoritically

the body moving with uniform velocity has acceleration zero.

Mathmatically,

u=3m/s

v=3m/s (since body is moving with uniform velocity)

a= v-u/t

3-3/t

0/t

0m/s.s

6 0

2 years ago

Describe when the chemical reaction occurs in a dry-cell battery

Step2247 [10]

Answer:

For a dry-cell battery to operate, oxidation will occur from the zinc anode and reduction will take place in the cathode. The most common type of cathode is a carbon graphite. Once reactants have been turned into products, the dry-cell battery will work to produce electricity.

Explanation:

The battery operates through electrochemical reactions called oxidation and reduction. These reactions involve the exchange of electrons between chemical species. If a chemical species loses one or more electrons, this is called oxidation. The opposite process, the gain of electrons, is called reduction.

8 0

2 years ago

A racecar accelerates from rest at 6.5 m/s2 for 4.1 s. How fast will it be going at the end of that time?

Pie

Answer:

The final velocity of the car is 26.65 m/s.

Explanation:

Given;

acceleration of the racecar, a = 6.5 m/s²

initial velocity of the car, u = 0

time of motion, t = 4.1 s

The final velocity of the car is given by;

v = u + at

where;

v is the final velocity of the car

suvstitute the givens

v = 0 + (6.5)(4.1)

v = 26.65 m/s.

Therefore, the final velocity of the car is 26.65 m/s.

6 0

2 years ago

A ball rolls down the hill which has a vertical height of 15 m. Ignoring friction what would be the gravitational potential ener

trasher [3.6K]

a) Potential energy: 147 m [J]

The gravitational potential energy of an object is given by

$U=mgh$

where

m is its mass

$g=9.8 m/s^2$ is the acceleration of gravity

h is the height of the object above the ground

In this problem,

h = 15 m

We call 'm' the mass of the ball, since we don't know it

So, the potential energy of the ball at the top of the hill is

$U=(m)(9.8)(15)=147 m$ (J)

b) Velocity of the ball at the bottom of the hill: 17.1 m/s

According to the law of conservation of energy, in absence of friction all the potential energy of the ball is converted into kinetic energy as the ball reaches the bottom of the hill. Therefore we can write:

$U=K=\frac{1}{2}mv^2$

where

v is the final velocity of the ball

We know from part a) that

U = 147 m

Substituting into the equation above,

$147 m = \frac{1}{2}mv^2$

And re-arranging for v, we find the velocity:

$v=\sqrt{2\cdot 147}=17.1 m/s$

5 0

3 years ago

Pressure in a liquid increases with its ..........

Novay_Z [31]

Answer: Pressure increases as the depth increases

Explanation: The pressure in a liquid is due to the weight of the column of water above. Since the particles in a liquid are tightly packed, this pressure acts in all directions.

6 0

3 years ago

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