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REY [17]
4 years ago
15

A scalloped hammerhead shark swims at a steady speed of 1.5 m/s with its 88 cm -cm-wide head perpendicular to the earth's 56 μT

magnetic field. What is the magnitude of the emf induced between the two sides of the shark's head? Express your answer using two significant figures.
Physics
1 answer:
Nadya [2.5K]4 years ago
8 0

Answer:

7.4 x 10⁻⁵ T.

Explanation:

Emf induced = BLv

Where B is magnetic field , L is length of rod or any other object moving in magnetic field and v is velocity of moving rod.

Here B = 56 x 10⁻⁶ T

L = 88 X 10⁻² m

v = 1.5 m/s

emf induced = 56 x 10⁻⁶ x 88 x 10⁻² x 1.5

= 7.4 x 10⁻⁵ T.

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4 A magnet can exert a force of attraction or a force of repulsion on another magnet.
GenaCL600 [577]

Answer:

Push -repulsion

Pull - attraction

Explanation:

When two magnets are brought together, a push happens when a force of repulsion is experienced where the magnets move away from each other. This means their polarity is the same and this will cause the magnet to push away from each other.

When two magnets are brought together , a pull happens when a force of attraction is experienced where the magnets move close to each other. This means their polarity is different and thus causes the magnets to pull closer to each other.

7 0
3 years ago
Bài 4: Một vật dao động điều hòa với phương trình x = 4cos(4πt - π/3) cm.
ra1l [238]

Answer:

g

Explanation:

6 0
3 years ago
If a 2500 kg car is traveling with an acceleration of 20.2 m/s/s, what is the force acting on it?
a_sh-v [17]

Answer:

50500

Explanation:

F = m x a

F = 2500kg x 20.2 m/s/S

F = 50500

4 0
3 years ago
A 6.72 particle moves through a region of space where an electric field of magnitude 1270 N/C points in the positive direction,
Romashka [77]

Answer:

                    v_{y}  = -104 m/s

Explanation:

Using:

Force = electric field * charge

F=e*q

Force = magnitude of charge * velocity * magnetic field * sin tither

F_{x2}= |q|*v*B*sin \alpha

Force on particle due to electric field:

     F_{x1}= E*q = (1270N/C)*(-6.72*10^{-6} ) = -8.53*10^{-3}

Force on particle due to magnetic field:

F_{x2}= |q|*v*B*sin \alpha  = (6.72*10^{-6} )*(1.15)*(sin90)*v = (7.728*10^{-6})*(v)

F_{x2} is in the positive x direction as F_{x1} is in the negative x direction while net force is in the positive x direction.

Magnetic field is in the positive Z direction, net force is in the positive x direction.

According to right hand rule, Force acting on particle is perpendicular to the direction of magnetic field and velocity of particle. This would mean the force is along the y-axis. As this is a negatively charged particle, the direction of the velocity of the particle is reversed. Therefore velocity of particle, v, has to be in the negative y direction.

Now,

                    F_{xnet}- F_{x1 } = F_{x2 }

                    (6.13*10^{-3}) - (8.53*10^{-3} ) = (7.728*10^{-6})*(v)

                    v = (F_{xnet}  - F_{x1}) / (F_{x2} )

                        =((6.13*10^{-3} ) - (8.53*10^{-3})) / (7.728*10^{-6})

                       = (- 104.25) m/s

                      v_{y}  = -104 m/s

8 0
3 years ago
please help me guys please please please please please please please please please please please please please​
seropon [69]

Answer:

1.F = 256 N

2.a = 8 m/s/s

By looking at the given information, you know force, and an acceleration. Therefore you have enough information to use the first formula.

F = ma

256 N = m * 8 m/s/s

m = 256 N/8 m/s/s

m = 32 Kg

6 0
3 years ago
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