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tia_tia [17]
3 years ago
11

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?

Physics
2 answers:
love history [14]3 years ago
6 0

Answer:

R\approx12.43 \,\, \Omega

Explanation:

We can use Ohm's Law to find the resistance R of a wire that carries a current I under a given potential difference:

V=I\,\,R\\R = \frac{V}{I} \\R=\frac{87}{7} \\R\approx12.43 \,\, \Omega

kramer3 years ago
3 0

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!

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Which statement about the atomic nucleus is correct? a. The nucleus is made of protons and neutrons and has a negative charge. b
Anon25 [30]
"b. The nucleus is made of protons and neutrons and has a positive charge" is the only option that is true about the atomic nucleus. The protons give it a positive charge.
5 0
3 years ago
A bike rider was moving to the right at a constant speed. Suddenly the wind then starts blowing against her with 3 N of force to
stiks02 [169]

Answer:

The correct option is D

Explanation:

This question can be better understood when discussed using the Newton's first law of motion which states that an object would continue to move with a uniform speed (in a straight line) unless acted upon by an external force. What happens here (in the question) is that the bike rider would have continued moving at a constant speed (to the right) if not for the opposing force of the wind that acted against her (to the left). <u>This wind/force would cause her speed to reduce or slow down (as posited by the law)</u>.

8 0
3 years ago
a load of 800 newton is lifted by an effort of 200 Newton. if the load is placed at a distance of 10 cm from the fulcrum. what w
nataly862011 [7]

Answer:

40 cm

Explanation:

We are given that

Load=800 N

Effort=200 N

Load  distance=10 cm

We have to find the effort distance.

We know that

load\times load\;distance=Effort\times effort\;distance

Using the formula

800\times 10=200\times effort\;distance

Effort distance=\frac{800\times 10}{200}

Effort distance=\frac{8000}{200}

Effort distance=40 cm

Hence,  the effort distance will be 40 cm.

7 0
3 years ago
A block of ice with mass 5.50 kg is initially at rest on a frictionless, horizontal surface. A worker then applies a horizontal
Yuki888 [10]

Answer:

A) 3.13 m/s

B) 5.34 N

C) W = 26.9 J

Explanation:

We are told that the position as a function of time is given by;

x(t) = αt² + βt³

Where;

α = 0.210 m/s² and β = 2.04×10^(−2) m/s³ = 0.0204 m/s³

Thus;

x(t) = 0.21t² + 0.0204t³

A) Velocity is gotten from the derivative of the displacement.

Thus;

v(t) = x'(t) = 2(0.21t) + 3(0.0204t²)

v(t) = 0.42t + 0.0612t²

v(4.5) = 0.42(4.5) + 0.0612(4.5)²

v(4.5) = 3.1293 m/s ≈ 3.13 m/s

B) acceleration is gotten from the derivative of the velocity

a(t) = v'(t) = 0.42 + 2(0.0612t)

a(4.5) = 0.42 + 2(0.0612 × 4.5)

a(4.5) = 0.9708 m/s²

Force = ma = 5.5 × 0.9708

F = 5.3394 N ≈ 5.34 N

C) Since no friction, work done is kinetic energy.

Thus;

W = ½mv²

W = ½ × 5.5 × 3.1293²

W = 26.9 J

6 0
3 years ago
particle of mass 59 g and charge 51 µC is released from rest when it is 32 cm from a second particle of charge −14 µC. Determine
AleksAgata [21]

Answer:

The initial acceleration of the 59g particle is 1062.7\frac{m}{s^{2}}

Explanation:

Newton's second laws relates acceleration (a), net force(F) and mass (m) in the next way:

F=ma (1)

We already know the mass of the particle so we should find the electric force on it to use on (1), the magnitude of the electric force between two charged objects by Columb's law is:

F=k\frac{\mid q_{1}q_{2}\mid}{r^{2}}

with q1 and q2 the charge of the particles, r the distance between them and k the constant k=9.0\times10^{9}\,\frac{Nm^{2}}{C^{2}}. So:

F=(9.0\times10^{9})\frac{\mid (51\times10^{-6})(-14\times10^{-6})\mid}{0.32^{2}}

F=62.7 N

Using that value on (1) and solving for a

a=\frac{F}{m}=\frac{62.7}{0.059}=1062.7\frac{m}{s^{2}}

4 0
3 years ago
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