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Sedaia [141]
3 years ago
10

Convert 9.83 m/s with a radius of 0.85 m to rotations per minute (rpm)

Physics
1 answer:
marissa [1.9K]3 years ago
3 0

Answer: Conversion of 9.83 m/s with a radius of 0.85 m into rotations per minute (rpm) is 99.25 m^{-1}.

Explanation:

Given: Speed = 9.83 m/s

Convert m/s into m/minute as follows.

1 min = 60 sec\\9.83 m/s = 9.83 m/s \times \frac{60 s}{1 min}\\= 529.8 m/min

Radius = 0.85 m

Now, rotations take place in circular motion. So, the circumference of a circle is 2 \pi r.

Therefore, circumference is calculated as follows.

Circumference = 2 \times pi \times r\\= 2 \times 3.14 \times 0.85 m\\= 5.338 m

Formula used to calculate rotations per minute (rpm) are as follows.

rpm = \frac{speed (in meter per minute)}{circumference}\\= \frac{529.8 m/min}{5.338 m}\\= 99.25 m^{-1}

Thus, we can conclude that conversion of 9.83 m/s with a radius of 0.85 m into rotations per minute (rpm) is 99.25 m^{-1}.

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The Balmer series in hydrogen atom produces only infrared emission. O True O False
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Answer:

False

Explanation:

As we know that, the Balmer series gives the n values as,

n_{i}=2.

[tex]n_{f}=3,4,5,.....\infty.

Now the value of wavelength can be calculated as,

\frac{1}{\lambda}=R(\frac{1}{n_{i} }-\frac{1}{n_{f} } )z^{2}.

Here, R=109677 cm^{-1}.

And n_{f}=3.

Now,

\frac{1}{\lambda}=109677 cm^{-1}(\frac{1}{2}-\frac{1}{3}).

Therefore,

\lambda=\frac{6}{109677} cm\\\lambda=547\times 10^{-9} m\\ \lambda=547 nm

Therefore, the wavelength of Balmer series lies in visible region which is 547 nm.

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3 years ago
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A particle leaves the origin with an initial velocity v → = (3.00iˆ) m/s and a constant acceleration a → = (−1.00iˆ − 0.500jˆ) m
tatiyna

Answer:

the position vector (x,y) will be (1.5 m,-2.25 m) and the velocity vector (vx,vy) will be ( 0 m/s , -1.5 m/s) when x reaches its maximum x coordinate

Explanation:

Since the velocity is related with the acceleration and coordinates through

vx²=v₀x²+2*ax*x

where

vx = velocity in the x direction

v₀x = initial velocity in the x direction = 3 m/s

ax = acceleration in the x direction = −1.00 m/s²

x= coordinates in the x-axis

when x reaches its maximum coordinate , then vx=0

thus

vx²=v₀x²+2*ax*x

0 = (3 m/s)² + 2* (−1.00 m/s²)*x

x= 1.5 m

also for the time t

vx = v₀x + ax*t → t= (vx-v₀x)/ax = (0- 3 m/s)/  (−1.00 m/s²) = 3 seconds

for the y coordinates

y = y₀+v₀y*t + 1/2 ay*t²

where

v₀y = initial velocity in the y direction = 0 m/s

ay = acceleration in the x direction = −0.5 m/s²

y= coordinates in the y-axis

y₀= initial coordinate in the y-axis =0

then since y₀=0 and v₀y=0

y = 1/2*ay*t²

y = 1/2*ay*t² = 1/2*(−0.5 m/s²)*(3 s)² = -2.25 m

and

vy=v₀y+ ay*t= 0+(−0.5 m/s²)*(3 s)= (-1.5 m/s)

therefore the position vector (x,y) will be (1.5 m,-2.25 m)

and the velocity vector (vx,vy) will be ( 0 m/s , -1.5 m/s)

7 0
2 years ago
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