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adelina 88 [10]
1 year ago
14

What type of meter is connected in

Physics
1 answer:
madreJ [45]1 year ago
6 0

Answer / Explanation:

B

Potential difference is measured using a device called a voltmeter . Just like ammeters, some types have a pointer on a dial, but most have a digital display. However, unlike an ammeter, you must connect the voltmeter in parallel to measure the potential difference across a component in a circuit.

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When unpolarized light is incident on a sheet of polarizing material with a transmission axis oriented vertically, what percenta
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Answer:

If the light were incident upon two polarizers at right angles, no light would get thru - thus each polarizer must block 50% of the light.

One polarizer would allow 50% of the light to pass.

4 0
2 years ago
when a charge of 1 C has an electric PE of 1 J, it has an electric potential of 1 V. When a charge of 2 C has an electric PE of
matrenka [14]
1 volt. 
1/1 = 1
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3 years ago
A train travels 65 kilometers in 5 hours, and then 61 kilometers in 4 hours. What is its average speed? _______km/hr
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The trains average speed is 13 km/hr.
4 0
3 years ago
Read 2 more answers
A river flows due east at 1.70 m/s. A boat crosses the river from the south shore to the north shore by maintaining a constant v
ad-work [718]

Answer:

a)

v = 14.1028 m/s  

∅ = 83.0765° north of east

b)

the required distance is 40.98 m

Explanation:

Given that;

velocity of the river u = 1.70 m/s

velocity of boat v = 14.0 m/s

Now to get the velocity of the boat relative to shore;

( north of east), we say

a² + b² = c²

(1.70)² + (14.0)² = c²

2.89 + 196 = c²

198.89 = c²

c = √198.89

c = 14.1028 m/s  

tan∅ = v/u = 14 / 1.7 =  8.23529

∅ = tan⁻¹ ( 8.23529 ) = 83.0765° north of east

Therefore, the velocity of the boat relative to shore is;

v = 14.1028 m/s  

∅ = 83.0765° north of east

b)  

width of river = 340 m,

ow far downstream has the boat moved by the time it reaches the north shore in meters = ?

we say;

340sin( 90° - 83.0765°)

⇒ 340sin( 6.9235°)

= 40.98 m

Therefore, the required distance is 40.98 m

5 0
3 years ago
The tires of a car make 75 revolutions as the car reduces its speed uniformly from 95 km/hr to 55 km.hr. the tires have a diamet
sveta [45]
<span>95 km/h = 26.39 m/s (95000m/3600 secs) 55 km/h = 15.28 m/s (55000m/3600 secs) 75 revolutions = 75 x 2pi = 471.23 radians radius = 0.80/2 = 0.40m v/r = omega (rad/s) 26.39/0.40 = 65.97 rad/s 15.28/0.40 = 38.20 rad/s s/((vi + vf)/2) = t 471.23 /((65.97 + 38.20)/2) = 9.04 secs (vf - vi)/t = a (38.20 - 65.97)/9.04 = -3.0719 The angular acceleration of the tires = -3.0719 rad/s^2 Time is required for it to stop (0 - 38.20)/ -3.0719 = 12.43 secs How far does it go? 65.97 - 38.20 = 27.77 M</span>
7 0
3 years ago
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