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erastova [34]
4 years ago
10

A 0.200 m wire is moved parallel to a 0.500 T

Physics
1 answer:
goldenfox [79]4 years ago
8 0

Answer:

The required emf moved across the wire is zero

Explanation:

For a moving charge particle, the magnetic force can be determined by using the formula;

\varepsilon = Bvlsin \theta

since the wire moves in parallel, the angle \theta between magnetic field and velocity = 0°

B = 0.500 T

v = 1.50 m/s

l = 0.200 m

∴

\varepsilon = (0.500  \ T )(1.50 \ m/s) \times (0.200 \ m)\times sin (0)

\varepsilon = 0.15\times sin (0)

\varepsilon = 0

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A forklift lifts 5 boxes from the ground to a height of 2 meters (m). The boxes push down with a force of 1000 newtons (N). How
Nata [24]
Hello!

Answer:

2000 J

Explanation

Work equation is expressed as:

W=F.d.Cos \alpha

Where:

F: Applied force
d: traveled distance
α: Angle between the direction of the force and the direction of the movement. (in this case, both of the direction are the same, so the angle is 0°)

By substituting:

F=1000N.2m.Cos(0)=2000N.m=2000 J

Have a nice day!
8 0
4 years ago
Suppose a 65.5 kg gymnast climbs a rope. What is the tension in the rope if she climbs at a constant speed
natali 33 [55]

The tension in the rope when the gymnast climbs it at constant speed is 641.9 N.

Given:

Mass of gymnast, m = 65.5 kg

The speed 'v' of gymnast is constant

Solution:

Consider the free-body diagram of the system as shown below.

Balancing forces along the vertical axis we get:

ΣFy = 0

Thus, we get:

F = ma     - (1)

where, m is mass of gymnast

            a is acceleration of gymnast (a = 0m/s², as the speed is constant)              

Also,

F = T - mg          -(2)

where, T is tension in the rope

           g is acceleration due to gravity

Equating (1) & (2), we get:

ma = T - mg

Re-arranging the equation, we get:

T = m(a+g)

Applying values in above equation we get:

T = (65.5 kg)(0 m/s²+9.8 m/s²)

T = 641.9 N

Therefore, the tension in the rope when the gymnast climbs it at constant speed is 641.9 N.

Learn more about tension here:

<u>brainly.com/question/14294135</u>

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7 0
2 years ago
An electric dipole consisting of charges of magnitude 2.00 nC separated by 8.40 μm is in an electric field of strength 1390 N/C.
amid [387]

Answer:

(a) The magnitude of the electric dipole moment is 1.68 x 10⁻¹⁴ C.m

(b) The difference between the potential energies ΔU, is 4.6704 x 10⁻¹¹ J

Explanation:

Given;

magnitude of charge, q = 2 nC = 2 x 10⁻⁹ C

distance of separation, d = 8.4 μm = 8.4 x 10⁻⁶ m

strength of electric field, E = 1390 N/C

(a) the magnitude of the electric dipole moment

p = qd

p = (2 x 10⁻⁹ C)(8.4 x 10⁻⁶ m)

p = 1.68 x 10⁻¹⁴ C.m

(b) the difference between the potential energies for dipole orientations parallel and anti-parallel to E

ΔU = U(180) - U(0)

ΔU = 2pE

ΔU = 2(1.68 x 10⁻¹⁴ )(1390)

ΔU = 4.6704 x 10⁻¹¹ J

6 0
3 years ago
What to do with a brand new mobile phone battery
Rashid [163]
Place it in the back of your phone where the old battery was

7 0
4 years ago
According to kinetic molecular theory, which statement best describes collisions between gas particles?
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A they are elastic.
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