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ch4aika [34]
2 years ago
12

Waves Please need help fast

Physics
1 answer:
mixas84 [53]2 years ago
6 0
I believe its the third answer
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Explicar los avances que dieron los estudioas de robert hooke y antonie van leeuwnhook
Dafna11 [192]

Answer:

El microscopio y el descubrimiento de microorganismos. Antonie van Leeuwenhoek (1632-1723) fue una de las primeras personas en observar microorganismos, utilizando un microscopio de su propio diseño, e hizo una de las contribuciones más importantes a la biología. Robert Hooke fue el primero en usar un microscopio para observar seres vivos.

5 0
3 years ago
Which statement best explains the movement of electric current from the clouds to the ground during a lightning storm? .
olga2289 [7]
"The <span>ground is positively charged and the clouds are negatively charged " is the statement among the statements given in the question that </span><span>best explains the movement of electric current from the clouds to the ground during a lightning storm. The correct option among all the options that are given in the question is the third option or option "C". </span>
3 0
3 years ago
Read 2 more answers
A parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 67 m/s*2. If he reaches the gr
KIM [24]

Answer:

3.49 seconds

3.75 seconds

-43200 ft/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

s=ut+\frac{1}{2}at^2\\\Rightarrow 50=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{50\times 2}{9.81}}\\\Rightarrow t=3.19\ s

Time the parachutist falls without friction is 3.19 seconds

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 50+0^2}\\\Rightarrow v=31.32\ m/s

Speed of the parachutist when he opens the parachute 31.32 m/s. Now, this will be considered as the initial velocity

v=u+at\\\Rightarrow 11=31.32+9.81t\\\Rightarrow t=\frac{11-31.32}{-67}=0.3\ s

So, time the parachutist stayed in the air was 3.19+0.3 = 3.49 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=0t+\frac{1}{2}\times a\times t^2\\\Rightarrow \frac{s}{2}=\frac{1}{2}at^2

s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=u1.1+\frac{1}{2}\times a\times 1.1^2

Now the initial velocity of the last half height will be the final velocity of the first half height.

v=u+at\\\Rightarrow v=at

Since the height are equal

\frac{1}{2}at^2=u1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow \frac{1}{2}at^2=at1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow 0.5t^2-1.1t-0.605=0\\\Rightarrow 500t^2-1100t-605=0

t=\frac{11\left(1+\sqrt{2}\right)}{10},\:t=\frac{11\left(1-\sqrt{2}\right)}{10}\\\Rightarrow t=2.65, -0.45

Time taken to fall the first half is 2.65 seconds

Total time taken to fall is 2.65+1.1 = 3.75 seconds.

When an object is thrown with a velocity upwards then the velocity of the object at the point to where it was thrown becomes equal to the initial velocity.

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-240^2}{2\times \frac{8}{12}}\\\Rightarrow a=-43200\ ft/s^2

Magnitude of acceleration is -43200 ft/s²

5 0
3 years ago
Write the ratio of the following?<br><br>CaCO3<br>C2H6<br>Fe(NO3)3​
Triss [41]

Answer:

C2H6 up the road to be with its own in

4 0
3 years ago
According to the graph, what is the motorcycle’s Average velocity ?
borishaifa [10]
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Therefore, we have
Avg velocity = change in direction/change in time
Avg velocity = (150km - 30km)/(5h - 0h)
Avg velocity = 24km/hr south. 

3 0
3 years ago
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