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sattari [20]
3 years ago
10

What is the slope of the line if the rise of a line on a distance versus-time graph is 900 meters and the run is 3 minutes?

Physics
1 answer:
Alexandra [31]3 years ago
6 0
600/3 = 200
the slope is 200m/min
 
OR

600/ (3/60) =
600 x 60/3 =
600 x 20 = 12,000 meters per hour 

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Find the wavelength of the third line in the lyman series, and identify the type of em radiation.
Natalija [7]

The wavelength of the third line in the Lyman series, and identify the type of EM radiation

In this series, the spectral lines are obtained when an electron makes a transition from any high energy level (n=2,3,4,5... ). The wavelength of light emitted in this series lies in the ultraviolet region of the electromagnetic spectrum.

1 / lambda = R(h)* ( \frac{1}{(n1)^{2} } -   \frac{1}{(n2)^{2} })

                 = 109678 ( \frac{1}{1^{2} } -  \frac{1}{3^{2} } )

                 = 109678 (8/9)

   Lambda = 9 / (109678 * 8 )

                  = 102.6 * 10^{-9} m = 102.6 nm

To learn more about Lyman series here

brainly.com/question/5762197

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8 0
2 years ago
how to A golf ball of mass 0.045 kg is hit off the tee at a speed of 45 m/s. The golf club was in contact with the ball for 3.5
Alexandra [31]

Answer: a) 12857.1 m/s/s b) 578.6 N

Explanation:

Impulse = change in momentum

Ft = mV2 - mV1

V = AT, 45 / .0035 = 12857.1 m/s/s

(b) .045 x 12857.1 = 578.6 N

4 0
3 years ago
Read 2 more answers
Pls help and thank u need asap!
aleksandr82 [10.1K]

Answer:

ill help if u help me?its 4am

5 0
3 years ago
The gravitational acceleration is 9.81 m/s2 here on Earth at sea level. What is the gravitational acceleration at a height of 35
azamat

To solve this problem it is necessary to apply the definition of severity of Newtonian laws in which it is specified that gravity is defined by

g= \frac{GM}{R^2}

Where

G= Gravitational Constant

M = Mass of Earth

R= Radius from center of the planet

According to the information we need to find the gravity 350km more than the radius of Earth, then

g_{ss} = \frac{GM}{R+h^2}

g_{ss} = \frac{6.67*10^{-11}*5.972*10^{24}}{(6371*10^3+350*10^3)^2}

g_{ss} = 8.82m/s^2

Therefore the gravitational acceleration at 350km is 8.82m/s^2

5 0
3 years ago
If the plotted points on a speed-time graph do not form a straight line, what do you know about the object's acceleration?
slega [8]
Remember, that while sped is constant, acceleration is not. Acceleration is when velicity changes. So the graph which shows the slop <span>of a velocity vs time describes acceleration.
</span>If we have the straight line on the graph it means that the slope is always the same whereas the <span>non-linear graphs has a variable slope that changes depending on your point in the graph.
</span>To conclude - if your graph is not a straight line it has variable acc at many points.<span>

</span>
6 0
3 years ago
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