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netineya [11]
3 years ago
6

A vector must always have both size and ..

Physics
1 answer:
DochEvi [55]3 years ago
5 0
Both magnitude and DIRECTION
For example,
• 12m East
• -2 miles
•9 meter north
• 8 miles up
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yarga [219]
Q1: At the highest point
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3 years ago
Two parallel conducting plates are separated by 9.2 cm, and one of them is taken to be at a potential of zero volts.What is the
True [87]

Answer:

E=54V/cm

\Delta V=496.8V between the plates.

Explanation:

The equation for change of voltage between two points separated a distance d inside parallel conducting plates (<em>which have between them constant electric field</em>) is:

\Delta V=Ed

So to calculate our electric field strength we use the fact that the potential 8.8 cm from the zero volt plate is 475 V:

E=\frac{\Delta V}{d}=\frac{475 V}{8.8cm}=54V/cm

And we use the fact that the plates are 9.2cm apart to calculate the voltage between them:

\Delta V=Ed=(54V/cm)(9.2cm)=496.8V

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3 years ago
which organ system maintains posture and produces heat? a. muscluar b. nervous c. skeletal d. integumentary
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Answer is a shoe this helps
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2 years ago
A motorist enters a freeway at 45 km/h and accelerates uniformly to 99 km/h. From the odometer in the car, the motorist knows th
Marat540 [252]

Answer:

a)  19440 km/h²

b) 10 sec

Explanation:

v₀ = initial velocity of the car = 45 km/h

v = final velocity achieved by the car = 99 km/h

d = distance traveled by the car while accelerating = 0.2 km

a = acceleration of the car

Using the kinematics equation

v² = v₀² + 2 a d

99² = 45² + 2 a (0.2)

a = 19440 km/h²

b)

t = time required to reach the final velocity

Using the kinematics equation

v = v₀ + a t

99 = 45 + (19440) t

t = 0.00278 h

t = 0.00278 x 3600 sec

t = 10 sec

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2 years ago
Consider horizontal parallel plates with a fixed potential difference. The upper plate has a voltage difference of 30 V with the
BlackZzzverrR [31]

relation between potential difference and electric field is given as

E . d = \Delta V

so here we know that

d = 3 cm

\Delta V = 30 V

E \times 0.03 = 30

E = 1000 N/C

So now when plates are separated to 4 cm distance carefully

the potential difference between them will change but the electric field between them will remain constant

So at distance of 4 cm also the electric field will be E = 1000 N/C

5 0
3 years ago
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