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iogann1982 [59]
3 years ago
7

Suppose a candle is burning, and wind is blowing on one side of the flame. Which principle explains why the flame bends toward t

he wind?
Physics
2 answers:
rjkz [21]3 years ago
8 0

Answer: Bernoulli’s

Explanation:

SOVA2 [1]3 years ago
4 0

Can u tell me the answer choices? Than I will help.

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Indica 3 aplicações dos ultrassons. *
Ainat [17]

Answer:

El ultrasonido se utiliza en muchos campos diferentes. Los dispositivos ultrasónicos se utilizan para detectar objetos y medir distancias. La ecografía o la ecografía se utilizan a menudo en medicina. En las pruebas no destructivas de productos y estructuras, el ultrasonido se utiliza para detectar defectos invisibles.

Espero que esto ayude, tenga un maravilloso día / noche, manténgase a salvo, felices fiestas y feliz navidad.

3 0
3 years ago
Science, who ever gets this will get a brainlest
anyanavicka [17]
The answer is A because there’s so much heat in the core and pressure because of all of the layers of the earth
6 0
3 years ago
Read 2 more answers
Runner 1 has a velocity of 10 m/s west. Runner 2 has a velocity of 7 m/s east. From the frame of reference of runner 2, what is
Vsevolod [243]

Answer:

<em>17 m/s west</em>

Explanation:

Runner 1 has velocity = 10 m/s west

runner 2 has velocity = 7 m/s east

From the frame of reference of runner 2, we can imagine runner 2 as standing still, and runner 1 moving away from him, towards the west with their combined velocity of

velocity = 10 m/s + 7 m/s = <em>17 m/s west</em>

5 0
3 years ago
A teacup on a spinning teacup ride takes 1.5 s to complete a revolution around the center of the platform. What is the centripet
arlik [135]
The correct answer is 53 meters per second squared 
5 0
3 years ago
Read 2 more answers
determine the greates possile acceleration of the 975 kg race car so that its front wheels do not leace the gorund
wolverine [178]

<u>Answer</u>:

The greatest possible acceleration of the car is a_G= 6.78 m/s^2

<u>Explanation</u>:

N_A+N_B-Mg = 0

-N_Aa +N_B(b-a)- \mu_s N_Bh - \mu_s N_Ah = 0

0.8N_B +0.8N_A = 975a_G

N_A+N_B = 9564.75 -------------(1)

-N_A(1.82) + N_B(2.20 -1.82) -0.9N_B(0.55)-0.8N_A(0.55)=0

-N_A(1.82) +0.38 N_B -0.44N_B -0.44N_A=0

-2.26N_A -0.06N_B= 0 ----------------(2)

Solving the equation (1) and(2)

N_A + N_B = 9564.75

-2.26N_A-0.06N_B=0

N_A = -260.85N

N_B = 9825.60N

\mu_s N_B + \mu_s N_A = 975a_G

0.8(9825.60)+0.8(-260.85) = 975a_Ga_G=\frac{7651.8}{975}a_G_1=7.4848m/s^2

Next lets assume that the front wheels contact with the ground N_A = 0

F_B = Ma_G

N_B = M_g

N_B - M_g = 0

N_B(b-a) –F_Bh = 0

F_B = 975a_G

N_B-975(9.8) = 0

N_B=9564.75N

9564.75(2.20 -1.82) -F_B(0.55)=0

\frac{3634.605}{0.55}=F_B

F_B = 6608.3

F_B = Ma_G

6608.3 = 975a_G

a_G = 6.7778 m/s^2

a_G_2 = 6.78m/s^2

Choosing the critical case

a_G = min(a_G_1 ,a_G_2)

a_G = min(7.848, 6.78)

a_G= 6.78 m/s^2

3 0
4 years ago
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