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Angelina_Jolie [31]
3 years ago
9

Are y'all into spirituality ?

Physics
2 answers:
Afina-wow [57]3 years ago
8 0
Yeah
why are you ???
Murrr4er [49]3 years ago
8 0

Answer:

I'm christian

Explanation:

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Sounds travel faster in Question 1 options: warmer air. cooler air. Temperature does not influence the speed of sound. a vacuum.
olya-2409 [2.1K]

Answer:

warmer air

Explanation:

the particles are more excited which increases the probability that the particles will bump into each other

8 0
2 years ago
A hailstone traveling with a velocity of 43 meters/second comes to a virtual stop 0.28 seconds after hitting water. What is the
lisabon 2012 [21]

Magnitude of acceleration = (change in speed) / (time for the change) .

Change in speed = (ending speed) - (starting speed)

                            =       zero            - (43 m/s)

                            =          -43 m/s .

Magnitude of acceleration = (-43 m/sec) / (0.28 sec)

                                          =  (-43 / 0.28)  (m/sec) / sec

                                          =        153.57...  m/s²

                                          =        1.5...  x 10²  m/s²  .

5 0
3 years ago
Read 2 more answers
What is the correct path of electrons?
MrRa [10]

Electrons are transferred sequentially between the two photosystems, with photosystem I acting to generate NADPH and photosystem II acting to generate ATP. The pathway of electron flow starts at photosystem II, which is homologous to the photosynthetic reaction center of R. viridis already described.

3 0
2 years ago
A cannonball is fired perfectly horizontally from the top of a 210 m tall cliff. It is fired with an initial velocity of 50 m/s.
pochemuha

Answer:

the horizontal distance covered by the cannonball before it hits the ground is 327.5 m

Explanation:

Given;

height of the cliff, h = 210 m

initial horizontal velocity of the cannonball, Ux = 50 m/s

initial vertical velocity of the cannonball, Uy = 0

The time for the cannonball to reach the ground is calculated as;

h = u_yt - \frac{1}{2} gt^2\\\\h = 0 - \frac{1}{2} gt^2\\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 210}{9.8} }\\\\t  = 6.55 \ s

The horizontal distance covered by the cannonball before it hits the ground is calculated as;

X = U_x \times \ t\\\\X = 50 \times \ 6.55\\\\X = 327.5 \ m

Therefore, the horizontal distance covered by the cannonball before it hits the ground is 327.5 m

8 0
3 years ago
What is the acceleration experienced by a car that takes 10s to reach 27m/s from rest
Lerok [7]

|acceleration|  =  (change in speed) / (time for the change)

Change in the car's speed = (27 - 0) = 27 m/s
Time for the change  =  10 sec

|acceleration| = (27 m/s) / (10 s)  =  2.7 m/s² .

That's the magnitude of the car's acceleration.
We don't know anything about its direction.
5 0
3 years ago
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