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3 years ago

The magnetic field 0.02 m from a wire is 0.1 t. what is the magnitude of the magnetic field 0.01 m from the same wire?

Physics

2 answers:

larisa86 [58]3 years ago

6 0

Answer:

The answer choice is D. 0.2T

Explanation:

juin [17]3 years ago

3 0

Answer:

0.2 T

Explanation:

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40N in either direction is the answer

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3 years ago

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PLEASE HURRY 20 PTS

Crank

Answer:

A. The electric field points to the left because the force on a negative charge is opposite to the direction of the field.

Explanation:

The electric force exerted on a charge by an electric field is given by:

where

F is the force

q is the charge

E is the electric field

We see that if the charge is negative, q contains a negative sign, so the force F and the electric field E will have opposite signs (which means they have opposite directions). This is due to the fact that the direction of the lines of an electric field shows the direction of the electric force experienced by a positive charge in that electric field: therefore, a negative charge will experience a force into opposite direction.

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3 years ago

Starting from zero, the electric current takes 2 seconds to reach half its maximum possible value in an RL circuit with a resist

Leno4ka [110]

Answer:

time=4s

Explanation:

we know that in a RL circuit with a resistance R, an inductance L and a battery of emf E, the current (i) will vary in following fashion

$i(t)=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})$ , where $i$ max= $\frac{E}{R}$

Given that, at i(2)= $\frac{imax}{2} =\frac{E}{2R}$

⇒ $\frac{E}{2R}=\frac{E}{R}(1-e^\frac{-2}{\frac{L}{R}})$

⇒ $\frac{1}{2}=1-e^\frac{-2}{\frac{L}{R}}$

⇒ $\frac{1}{2}=e^\frac{-2}{\frac{L}{R}}$

Applying logarithm on both sides,

⇒ $log(\frac{1}{2})=\frac{-2}{\frac{L}{R}}$

⇒ $log(2)=\frac{2}{\frac{L}{R}}$

⇒ $\frac{L}{R}=\frac{2}{log2}$

Now substitute $i(t)=\frac{3}{4}imax=\frac{3E}{4R}$

⇒ $\frac{3E}{4R}=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})$

⇒ $\frac{3}{4}=1-e^\frac{-t}{\frac{L}{R}}$

⇒ $\frac{1}{4}=e^\frac{-t}{\frac{L}{R}}$

Applying logarithm on both sides,

⇒ $log(\frac{1}{4})=\frac{-t}{\frac{L}{R}}$

⇒ $log(4)=\frac{t}{\frac{L}{R}}$

⇒ $t=log4\frac{L}{R}$

now subs. $\frac{L}{R}=\frac{2}{log2}$

⇒ $t=log4\frac{2}{log2}$

also $log4=log2^{2}=2log2$

⇒ $t=2log2\frac{2}{log2}$

⇒ $t=4$

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3 years ago

A 900 kg car speeds up from 30 M/S to 80 m/s how much momentum did the car gain?

AlexFokin [52]

The change in momentum of the car is 45,000 kg m/s

Explanation:

The change in momentum of an object is given by:

$\Delta p = m(v-u)$

where

m is the mass of the object

u is its initial velocity

v is its final velocity

For the car in this problem, we have

m = 900 kg

u = 30 m/s

v = 80 m/s

Therefore, the change in momentum is:

$\Delta p = (900)(80-30)=45,000 kg m/s$

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

#LearnwithBrainly

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3 years ago

A 3 kg drone starts at 5 meter above the ground and reached 25 meter above the

docker41 [41]

Recall that work is the amount of energy transferred to an object when it experiences a displacement and is acted upon by an external force. It is given a symbol of W and is measured in joules (J).

W=\vec{F}\cdot \Delta \vec{d}

We can use this formula to determine the work done by very specific forces, generating specific types of energy. We will examine three types of energy in this activity: gravitational potential, kinetic, and thermal. Before we start deriving equations for gravitational potential energy and kinetic energy, we should note that since work is the transfer and/or transformation of energy, we can also write its symbol as \Delta E.

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2 years ago