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REY [17]
2 years ago
14

5. bending the electric wire a.change in size b.change in shape c.change in texture​

Physics
1 answer:
il63 [147K]2 years ago
8 0

Answer:

change in shape

hope this will help you

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In the same configuration of the previous problem 3, four long straight wires are perpendicular to the page, and their cross sec
faust18 [17]

Complete Question:

In the same configuration of the previous problem 3, four long straight wires are perpendicular to the page, and their cross sections form a square of edge length a = 13.5 cm. Each wire carries 7.50 A, and the currents are out of the page in wires 1 and 4 and into the page in wires 2 and 3.

a) Draw a diagram in a (x,y) plane of the four wires with wire 4 perpendicular to the origin. Indicate the current's directions.

b) Draw a diagram of all magnetic fields produced at the position of wire 3 by the other three currents.

c) Draw a diagram of all magnetic forces produced at the position of wire 3 by the other three currents.

d) What are magnitude and direction of the net magnetic force per meter of wire length on wire 3?

Answer:

force, 1.318 ₓ 10⁻⁴

direction, 18.435°

Explanation:

The attached file gives a breakdown step by step solution to the questions

7 0
3 years ago
A block of mass
Gennadij [26K]

(a) The work done by the applied force is 26.65 J.

(b) The work done by the normal force exerted by the table is 0.

(c) The work done by the force of gravity is 0.

(d) The work done by the net force on the block is 26.65 J.

<h3>Work done by the applied force</h3>

W = Fdcosθ

W = 14 x 2.1 x cos25

W = 26.65 J

<h3>Work done by the normal force</h3>

W = Fₙd

W = mg cosθ x d

W = (2.5 x 9.8) x cos(90) x 2.1

W = 0 J

<h3>Work done force of gravity</h3>

The work done by force of gravity is also zero, since the weight is at 90⁰ to the displacement.

<h3> Work done by the net force on the block</h3>

∑W = 0 + 26.65 J = 26.65 J

Thus, the work done by the applied force is 26.65 J.

The work done by the normal force exerted by the table is 0.

The work done by the force of gravity is 0.

The work done by the net force on the block is 26.65 J.

Learn more about work done here: brainly.com/question/8119756

#SPJ1

6 0
2 years ago
What is an example of velocity?What is an example of velocity?
givi [52]
Speed with direction
5 0
2 years ago
Read 2 more answers
A constant force of 12N is applied for 3.0s to a body initially at rest. The final velocity of the body is 6.0ms–1. What is the
sp2606 [1]
From the question,
u = 0m {s}^{ - 1}
v = 6m {s}^{ - 1}

t = 3s
F=12N



Using Impulse, the product of the constant force, F and time t equals the product of the mass of the body and change in velocity.

Ft =m(v-u)


12(3.0)=m(6.0- \: 0)
This implies that

36.0 = 6m
m =  \frac{36.0}{6.0}
\therefore \: m = 6.0kg


You can also use the equation of linear motion,
v = u + at
6 = 0 + a(3)
6 = 3a
a =  \frac{6}{3}

a = 2 {ms}^{ - 2}
But
F=ma
12 = m(2)
12 = 2m
\frac{12}{2}  = m
\therefore \: m = 6kg
4 0
3 years ago
WILL GIVE BRAINLIEST TO CORRECT ANSWER PLEASE HELP ME
koban [17]

Answer:

The total distance is 381.5 [m]

Explanation:

In order to solve this problem we must use the expressions of kinematics. The clue to solve this problem is that the motorcyclist starts from rest, i.e. its initial speed is zero.

v_{f} =v_{o} +(a*t)

where:

Vf = final velocity [m/s]

Vo = initial velocity = 0

a = acceleration = 2 [m/s²]

t = time = 7 [s]

Vf = 0 + (2*7)

Vf = 14 [m/s]

With this velocity, we can calculate the displacement using the following expression.

v_{f} ^{2} =v_{o} ^{2} +2*a*x

where

x = distance traveled [m]

14² = 0 + (2*7*x)

x = 196/(14)

x = 14 [m]

Note: The positive sign in the equations is because the car is accelerating, it means its velocity is increasing.

The other important clue to solve this problem in the second part is that the final velocity is now the initial velocity.

We must calculate the final velocity.

v_{f}= v_{i} +(a*t)

Vf = final velocity [m/s]

Vi = initial velocity = 14 [m/s]

a = desacceleration = 4 [m/s²]

t = time = 8 [s]

Vf = 24 + (4*8)

Vf = 56 [m/s]

With this velocity, we can calculate the displacement using the following expression.

v_{f} ^{2} =v_{o} ^{2} +2*a*x

where

x = distance traveled [m]

56² = 14² + (2*4*x)

x = 2940/(8)

x = 367.5 [m]

Note: The positive sign in the equations is because the car is accelerating, it means its velocity is increasing.

Therefore the total distance is Xt = 14 + 367.5 = 381.5 [m].

4 0
3 years ago
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