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pav-90 [236]
3 years ago
5

O How are electricity and magnetism related?

Physics
1 answer:
Nataly [62]3 years ago
6 0
Magnetism is the product of a moving charged particle. We can have electricity without magnetism but we can not have magnetism without electricity.An electro magnet is made so that we have a soft metal core and electricity around it. A bar magnet is a normal magnet in bar shape with permanent magnetism.
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Compute the dot product of the vectors u and v​, and find the angle between the vectors. Bold v equals 7 Bold i minus Bold j and
OLga [1]

Answer:

\theta = 106.3 degree

Explanation:

As we know that

\vec w = -\hat i + 7\hat j

\vec v = 7\hat i - \hat j

also we know that

\vec v. \vec w = -14

it is given as

\vec v. \vec w = (-\hat i + 7\hat j).(7\hat i - \hat j)

\vec v. \vec w = - 7 - 7 = -14

also we can find the magnitude of two vectors as

|v| = \sqrt{(-1)^2 + (7)^2}

|v| = \sqrt{50}

similarly we have

|w| = \sqrt{(7^2) + (-1)^2}

|w| = \sqrt{50}

now we know the formula of dot product as

\vec v. \vec w = |v||w| cos\theta

-14 = (\sqrt{50})^2cos\theta

\theta = cos^{-1}(\frac{-14}{50})

\theta = 106.3 degree

3 0
3 years ago
A mass m attached to a horizontal massless spring with spring constant k, is set into simple harmonic motion. its maximum displa
Lesechka [4]
At the point of maximum displacement (a), the elastic potential energy of the spring is maximum:
U_i= \frac{1}{2} ka^2
while the kinetic energy is zero, because at the maximum displacement the mass is stationary, so its velocity is zero:
K_i =0
And the total energy of the system is
E_i = U_i+K= \frac{1}{2}ka^2

Viceversa, when the mass reaches the equilibrium position, the elastic potential energy is zero because the displacement x is zero:
U_f = 0
while the mass is moving at speed v, and therefore the kinetic energy is
K_f =  \frac{1}{2} mv^2
And the total energy is
E_f = U_f + K_f =  \frac{1}{2} mv^2

For the law of conservation of energy, the total energy must be conserved, therefore E_i = E_f. So we  can write
\frac{1}{2} ka^2 =  \frac{1}{2}mv^2
that we can solve to find an expression for v:
v= \sqrt{ \frac{ka^2}{m} }
6 0
3 years ago
If no friction acts on a diver during a dive, then which of the following statements is true? A) The total mechanical energy of
EleoNora [17]
If no frictional work is considered, then the energy of the system (the driver at all positions is conserved.

Let
position 1 = initial height of the diver (h₁), together with the initial velocity (v₁).
position 2 = final height of the diver (h₂) and the final velocity (v₂).

The initial PE = mgh₁ and the initial KE  = (1/2)mv₁²
where g = acceleration due to gravity,
m = mass of the diver.
Similarly, the final PE and KE are respectively mgh₂ and (1/2)mv₂².
PE in position 1 is converted into KE due to the loss in height from position 1 to position 2.
 
Therefore
(KE + PE) ₁ = (KE + PE)₂

Evaluate the given answers.
A) The total mechanical energy of the system increases.
     FALSE

B) Potential energy can be converted into kinetic energy but not vice versa.
     TRUE

C) (KE + PE)beginning = (KE + PE) end.
     TRUE

D) All of the above.
     FALSE

4 0
3 years ago
Read 2 more answers
A very hard rubber ball (m = 0.5 kg) is falling vertically at 4 m/s just before it bounces on the floor. The ball rebounds back
saw5 [17]

Answer:

The force exerted by the floor is 80 N.

Explanation:

Given that,

Mass of ball = 0.5 kg

Velocity= 4 m/s

Time t = 0.05 s

When the ball rebounds then the kinetic energy is

K.E =\dfrac{1}{2}mv^2

Where, m = mass of ball

v = velocity of ball

Put the value into the formula

K.E=\dfrac{1}{2}\times0.5\times(4)^2

K.E = 4\ J

The average force exerted by the floor on the ball = change in kinetic energy over collision time

F = \dfrac{4}{0.05}

F=80\ N

Hence, The force exerted by the floor is 80 N.

4 0
3 years ago
A 1.0-m long wire is carrying a certain amount of current. The wire is placed perpendicular to a magnetic field of strength 0.20
Marysya12 [62]

For a current-carrying wire running perpendicular to a magnetic field, the magnetic force acting on the wire is given by:

F = ILB

F = magnetic force, I = current, L = wire length, B = magnetic field strength

Given values:

F = 0.60N, L = 1.0m, B = 0.20T

Plug in and solve for I:

0.60 = I(1.0)(0.20)

I = 3.0A

5 0
3 years ago
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