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zzz [600]
3 years ago
14

Need help with is if u can asap

Physics
1 answer:
jeka943 years ago
8 0
I am guessing by what I had it is a plz don't quote me on that I could be wrong
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A 1.5m wire carries a 7 A current when a potential difference of 87 V is applied. What is the resistance of the wire?
kramer

Answer:

Ohm's law states that I=V/R (Current=volts divided by resistance). Since we're looking for resistance, we'll rewrite it as R=V/I. Then just plug in the numbers; R=84/9, R= 9 1/3 or 28/3. The resistance of the wire is 9.33... or 9 1/3 ohm's, depending on how you wanna write it.

Hope it helped u if yes mark me BRAINLIEST!

Tysm!

I would appreciate it!

3 0
3 years ago
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A student is working in a lab to determine how time affects impulse. The student keeps the force the same in each trial but chan
pishuonlain [190]

yea some data is shown what is the question dude

4 0
3 years ago
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An unbanked (flat) curve of radius 150 m is rated for a maximum speed of 32.5 m/s. At what maximum speed, in m/s, should a flat
Rasek [7]

Answer:

The maximum speed is 21.39 m/s.

Explanation:

Given;

radius of the flat curve, r₁ = 150 m

maximum speed, v_{max} = 32.5 m/s

The maximum acceleration on the unbanked curve is calculated as;

a_c_{max} = \frac{V_{max}^2}{r} \\\\a_c_{max} = \frac{32.5^2}{150} \\\\a_c_{max} = 7.04 \ m/s^2

the radius of the second flat curve, r₂ = 65.0 m

the maximum speed this unbanked curve should be rated is calculated as;

a_c_{max} = \frac{V_{max}^2}{r_2} \\\\V_{max}^2 = a_c_{max}  \ \times \ r_2\\\\V_{max} = \sqrt{a_c_{max}  \ \times \ r_2} \\\\V_{max} =\sqrt{7.04 \ \times \ 65} \\\\V_{max} = 21.39 \ m/s

Therefore, the maximum speed is 21.39 m/s.

3 0
3 years ago
Has luster
ruslelena [56]

<span>The answer to this problem is magnesium. I hoped I helped someone with this</span>
3 0
4 years ago
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The man rolls the cupboard at a steady speed from the lorry to the house. The friction force in the wheels is 40 N. State the fo
tamaranim1 [39]

Answer:

40N

Explanation:

Using the newton's second law of motion

\sum Fx = max

Fm - Ff = max

Fm is the applied force

Ff is the frictional force

m is the mas of the cupboard

ax is the acceleration

Since the speed from the lorry is steady, ax = 0m/s^2

Also Ff = 40N

Substitute into  the formula;

Fm - 40 = m(0)

Fm - 40 = 0

Add 40 to both sides

Fm - 40 + 40 = 0 + 40

Fm = 40N

Hence the force with which the man applied to push is 40N

4 0
3 years ago
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