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Licemer1 [7]
2 years ago
6

¿Con qué velocidad viaja una onda formada en una cuerda de 10 m de longitud y 1 kg de masa, si se le sostiene con una tensión de

40N?

Physics
1 answer:
slavikrds [6]2 years ago
7 0
This is a picture I took

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Why are most of the world’s deserts located between latitudes 10°n to 30°n and 10°s to 30°s.
RoseWind [281]

The bulk of the world's deserts are located at 30 degrees north latitude and 30 degrees south latitude, when the warm equatorial air begins to descend. The heavy, warm, descending air vaporises large amounts of water from the ground's surface. As a result, the environment is rather dry.

<h3>Why are the majority of the desert regions on Earth located between 20 and 30 degrees latitude?</h3>

The zones of falling air are those between 20 and 30 latitudes on the western borders of continents (high pressure and dry weather). As a result, the moisture continues to decrease as the air is compressed and warmed as it falls.

Where the scorching equatorial air starts to descend, the majority of the world's deserts are found between 30 degrees north and 30 degrees south latitude. Large volumes of water are vaporised off the surface of the ground by the thick, warming, falling air. As a result, the climate is extremely dry.

Learn more about latitude refer

brainly.com/question/1939015

#SPJ4

3 0
1 year ago
I need help plz and thank you this is do
love history [14]

Answer:

Oh I am sorry this is my first time on brainly i dont how to exit and sorry but dont know the answer

Explanation:

4 0
2 years ago
A car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.76 m/s2 for t1 = 20 s. At that point the
alex41 [277]

Answer:

(a)v_1 = a_1t_1 = 1.76 t_1

(b) It won't hit

(c) 110 m

Explanation:

(a) the car velocity is the initial velocity (at rest so 0) plus product of acceleration and time t1

v_1 = v_0 + a_1t_1 = 0 +1.76t_1 = 1.76t_1

(b) The velocity of the car before the driver begins braking is

v_1 = 1.76*20 = 35.2m/s

The driver brakes hard and come to rest for t2 = 5s. This means the deceleration of the driver during braking process is

a_2 = \frac{\Delta v_2}{\Delta t_2} = \frac{v_2 - v_1}{t_2} = \frac{0 - 35.2}{5} = -7.04 m/s^2

We can use the following equation of motion to calculate how far the car has travel since braking to stop

s_2 = v_1t_2 + a_2t_2^2/2

s_2 = 35.2*5 - 7.04*5^2/2 = 88 m

Also the distance from start to where the driver starts braking is

s_1 = a_1t_1^2/2 = 1.76*20^2/2 = 352

So the total distance from rest to stop is 352 + 88 = 440 m < 550 m so the car won't hit the limb

(c) The distance from the limb to where the car stops is 550 - 440 = 110 m

8 0
3 years ago
An object is moving to the left
dybincka [34]
The object is moving with constant speed means there is no change in speed Hence the acceleration is 0

Where, u is initial speed,

v is final speed

t is time taken and

a is acceleration

a = ( v-u) /t

a =(0)/t because u=v

a=0

Therefore, Acceleration is 0
4 0
2 years ago
PLZZZZZZZZZZZZZZZZZ I REALLY NEED THE ANSWER PLZZZZZZZZZZZZZZZ HELP ME ASAP!!!
Rashid [163]
I think it's RF technology
7 0
3 years ago
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