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BartSMP [9]
2 years ago
14

An electric current is created in a long thin wire. How will increasing the current and changing the direction of the current ef

fect the direction that a compass placed under the wire will point?
a. It compass will spin continuously.
b. The compass will flip exactly 180 degrees.
c. The compass will spin exactly 360 degrees.
d. The compass will not react and still point north.
Physics
2 answers:
PilotLPTM [1.2K]2 years ago
8 0
The answer is : B. The compass will spin exactly 360 Degrees

Compass will point out to the earth's magnetic field. If we change the direction of the current , we created an effect which make the compass detect the magnetic field in the exact opposite direction, that is why it will flip exactly 180 degrees
dexar [7]2 years ago
6 0

On USATestprep it is B 180 degrease

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If a point has 40 J of energy and the electric potential is 8 V, what must be the charge?
Alekssandra [29.7K]

If a point has 40 J of energy and the electric potential is 8 V, the charge must be: A. 5 C

<u>Given the following the details;</u>

  • Energy = 40 Joules
  • Electric potential = 8 Volts

To find the quantity of charge;

Mathematically, the quantity of charge with respect to electric potential is given by the formula;

Quantity \; of \; charge = \frac{Energy}{Electric \; potential}

Substituting the values into the formula, we have;

Quantity \; of \; charge = \frac{40}{8}

<em>Quantity of charge = 5 Coulombs</em>

Therefore, the quantity of charge must be <em>5 Coulombs.</em>

Find more information: brainly.com/question/21808222

8 0
3 years ago
Read 2 more answers
A 790kg car moving at 7 m/s takes a turn around a circle with a radius of 20 m. Determined the net force (in Newton’s) acting up
prisoha [69]

1935.5 N is the "net force" acting on a car.

<u>Explanation</u>:

Given that,  

Mass of the car is 790 kg.

Velocity of the car is 7 m/s. (v)

It turned around with 20 m. (r)  

We know that, Net force = m × a

\text { Here, acceleration of the car is radial acceleration } a_{\mathrm{rad}}=\frac{v^{2}}{r}

\mathrm{a}_{\mathrm{rad}}=\frac{7^{2}}{20}

\mathrm{a}_{\mathrm{rad}}=\frac{49}{20}

a_{\text {rad }}=2.45 \mathrm{m} / \mathrm{s}^{2}

Now, Net force = m × a

Net force = 790 × 2.45

Net force = 1935.5 N

4 0
3 years ago
A coin is placed on a large disk which rotates uniformly at a rate of 1 rot/s. The coefficient of friction between the coin and
s2008m [1.1K]

Answer:

r = 0.02 m

Explanation:

from the question we have :

speed = 1 rps = 1x 60 = 60 rpm

coefficient of friction (μ) = 0.1

acceleration due to gravity (g) = 9.8 m/s^{2}

maximum distance without falling off (r) = ?

to get how far from the center of the disk the coin can be placed without having to slip off we equate the formula for the centrifugal force with the frictional force on the turntable force

mv^2 / r =  m x g x μ

v^2 / r =  g x μ  .......equation 1

where

velocity (v) = angular speed (rads/seconds) x radius

angular speed (rads/seconds) = (\frac{2π}{60} ) x rpm

angular speed (rads/seconds) = (\frac{2 x π}{60} ) x 60 = 6.28 rads/ seconds

now

velocity = 6.28 x r = 6.28 r

now substituting the value of velocity into equation 1

v^2 / r =  g x μ

(6.28r)^2 / r = 9.8 x 0.1

39.5 x r = 0.98

r = 0.02 m

6 0
2 years ago
A series circuit has a capacitor of 0.25 × 10−6 F, a resistor of 5 × 103 Ω, and an inductor of 1 H. The initial charge on the ca
svetoff [14.1K]

Answer:

q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

Explanation:

Given that L=1H, R=5000\Omega, \ C=0.25\times10^{-6}F, \ \ E(t)=12V, we use Kirchhoff's 2nd Law to determine the sum of voltage drop as:

E(t)=\sum{Voltage \ Drop}\\\\L\frac{d^2q}{dt^2}+R\frac{dq}{dt}+\frac{1}{C}q=E(t)\\\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+\frac{1}{0.25\times10^{-6}}q=12\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+4000000q=12\\\\m^2+5000m+4000000=0\\\\(m+4000)(m+1000)=0\\\\m=-4000  \ or \ m=-1000\\\\q_c=c_1e^{-4000t}+c_2e^{-1000t}

#To find the particular solution:

Q(t)=A,\ Q\prime(t)=0,Q\prime \prime(t)=0\\\\0+0+4000000A=12\\\\A=3\times10^{-6}\\\\Q(t)=3\times10^{-6},\\\\q=q_c+Q(t)\\\\q=c_1e^{-4000t}+c_2e^{-1000t}+3\times10^{-6}\\\\q\prime=-4000c_1e^{-4000t}-1000c_2e^{-1000t}\\q\prime(0)=0\\\\-4000c_1-1000c_2=0\\c_1+c_2+3\times10^{-6}=0\\\\#solving \ simultaneously\\\\c_1=10^{-6},c_2=-4\times10^{-6}\\\\q=10^{-6}e^{-4000t}-4\times10^{-6}e^{-1000t}+3\times10^{-6}\\\\q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

Hence the charge at any time, t is q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

6 0
3 years ago
A 120 kg box is on the verge of slipping down an inclined plane with an angle of inclination of 47º. What is the coefficient of
Alex_Xolod [135]

Given :

A 120 kg box is on the verge of slipping down an inclined plane with an angle of inclination of 47º.

To Find :

The coefficient of static friction between the box and the plane.

Solution :

Vertical component of force :

mg\ sin\ \theta =  120\times 10 \times sin\ 47^\circ{}=877.62 \ N

Horizontal component of force(Normal reaction) :

mg\ cos\ \theta =  120\times 10 \times cos\ 47^\circ{}=818.40 \ N

Since, box is on the verge of slipping :

mg\ sin\ \theta= \mu(mg \ cos\ \theta)\\\\\mu = tan \ \theta\\\\\mu = tan\ 47^o\\\\\mu = 1.07

Therefore, the coefficient of static friction between the box and the plane is 1.07.

Hence, this is the required solution.

7 0
3 years ago
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