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yarga [219]
3 years ago
6

Describe how a change in resistance would affect the current in a circuit

Physics
1 answer:
Elena-2011 [213]3 years ago
8 0
Resistance typically opposes flow of current.

When the Resistance is increased, the current would be reduced, but when the Resistance is reduced, the current would be increased in the circuit.
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Which equation describes the sum of the vectors plotted below?
worty [1.4K]

Answer:

It's the third one in the picture ...

r = 3x - 2y

Explanation:

5 0
3 years ago
In a choir practice room, two parallel walls are 5.70 m apart. The singers stand against the north wall. The organist faces the
ololo11 [35]

Answer:

4.98 m

Explanation:

Given that

Width of the mirror, d = 0.6 m

Organist distance to the mirror, s = 0.78 m

Distance between the singer and the organist, S = 5.7 + 0.78 = 6.48 m

Width of north wall, D?

Using the simple relationship

D/S = d/s, on rearranging

D = dS /s

D = (0.6 * 6.48) / 0.78

D = 3.888 / 0.78

D = 4.98 m

Therefore, we can conclude that the Width of north wall is 4.98 m

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3 years ago
Oil _____.
atroni [7]
Oil <span>must be mined from underground 

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4 0
3 years ago
Read 2 more answers
. At what velocity will the box in Problem 6 be traveling when it hits the ground? Use formula: Va = Vf - Vi / 2
VladimirAG [237]

Answer:

How much time does his victim on the ground below have to move out of harm's way? At what velocity will the safe hit the ground? sownt d-200m. n a = 100/2. Vi-o.

Explanation:

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3 years ago
A jet transport has a weight of 2.25 x 106 N and is at rest on the runway. The two rear wheels are 16.0 m behind the front wheel
Rudik [331]

Answer:

Explanation:

Given that,

Weight of jet

W = 2.25 × 10^6 N

It is at rest on the run way.

Two rear wheels are 16m behind the front wheel

Center of gravity of plane 10.6m behind the front wheel

A. Normal force entered on the ground by front wheel.

Taking moment about the the about the real wheel.

Check attachment for better understanding

So,

Clock wise moment = anti-clockwise moment

W × 5.4 = N × 16

2.25 × 10^6 × 5.4 = 16•N

N = 2.25 × 10^6 × 5.4 / 16

N = 7.594 × 10^5 N

B. Normal force on each of the rear two wheels.

Using the second principle of equilibrium body.

Let the rear wheel normal be Nr and note, the are two real wheels, then, there will be two normal forces

ΣFy = 0

Nr + Nr + N — W = 0

2•Nr = W—N

2•Nr = 2.25 × 10^6 — 7.594 × 10^5

2•Nr = 1.491 × 10^6

Nr = 1.491 × 10^6 / 2

Nr = 7.453 × 10^5 N

6 0
4 years ago
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