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kvasek [131]
3 years ago
11

How can you use PE in music class and how can you use music in PE class?

Physics
1 answer:
atroni [7]3 years ago
7 0

Answer:

fa fe fi fo fu hhhhhjjjjjkhvg

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A resistor and a capacitor are connected in series to an ideal battery of constant terminal voltage. At the moment contact is ma
Vanyuwa [196]

Answer:

At the moment contact is made with the battery, the voltage across the resistor is equal to the batteries terminal voltage

Explanation;

Because at series connection the battery and resistor have equal voltage

8 0
3 years ago
Help me with questions 1-8. here's the link to the textbook
SVETLANKA909090 [29]
Ok so basically I’m here for the points cuz I gotta quiz do in 10 minutes sorry
5 0
3 years ago
Diagram the cross section of a graduated cylinder, illustrating how to read the meniscus.
Gre4nikov [31]
<span>When taking the reading you must see where the lower meniscus lies . The value at which the lower meniscus lies is your reading .</span>
3 0
3 years ago
Read 2 more answers
A car travels 60 miles due West first then turns back and travels 120 miles due East in 3 hours. What is...
ella [17]

Answer:

<h2>A. 180 miles</h2><h2>B. 60 miles</h2><h2 />

Explanation:

In this problem, we are required to solve for the total distance that the car travelled. and the displacement

A) the distance travelled by car

this can be gotten by summing all the distances the car has travelled.

i,e total distance= 60 miles+120 miles

total distance= 180 miles

B) the displacement of the car

the displacement can be gotten by  subtracting the final distance from the initial distance

final distance = 120 miles

initial distance= 60 miles

displacement= 120-60= 60 miles

7 0
3 years ago
In the model of the hydrogen atom due to Niels Bohr, the electron moves around the proton at a speed of 3.3 × 106 m/s in a circl
irga5000 [103]

Answer:

1.5048\times 10^{-23}\ Am^2

Explanation:

q = Charge of proton = 1.6\times 10^{-19}\ C

r = Radius of circle = 5.7\times 10^{-11}\ m

v = Velocity of proton = 3.3\times 10^6\ m/s

Magnetic moment is given by

M=\frac{1}{2}qrv\\\Rightarrow M=\frac{1}{2}1.6\times 10^{-19}\times 5.7\times 10^{-11}\times 3.3\times 10^6\\\Rightarrow M=1.5048\times 10^{-23}\ Am^2

The magnetic moment associated with this motion is 1.5048\times 10^{-23}\ Am^2

5 0
3 years ago
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