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3 years ago

Which of the following is an example of a material that allows electrons to easily move through it?

1 answer:

6 0

Conductors allow for charge transfer through the free movement of electrons

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Find the force between 2C and -1C separated by a distance 1m. Indicate if it is an attractive force (negative) or a repulsive fo

telo118 [61]

Answer:

Explanation:

Its definitely an Attractive force since the two charges are Unlike.

From Coulombs Law

F=kq1q2/R²

Given

K=9x10^9

R=1m

q1=2C

q2=-1C

F=(9x10^9 x 2 x -1)/1²

F= - 1.8x10^10N. (Attractive).

6 0

3 years ago

If the current in this circuit is 3 A, what must be the value of R3? It's 5 Ω

Svetllana [295]

Explanation :

It is given that,

In the given figure all the three resistors are in series.

Current flowing in the circuit, $I=3\ A$

Voltage, $V=30\ V$

We know that in series circuit the current flowing in all resistors is same.

Using Ohm's law, we get:

$V=IR$

$30\ V=3\ A(2\ \Omega+3\ \Omega+R_3)$

$R_3=5\ \Omega$

Hence, this is the required solution.

5 0

3 years ago

Read 2 more answers

A 5-m steel beam is lowered by means of two cables unwinding at the same speed from overhead cranes. As the beam approaches the

ArbitrLikvidat [17]

Answer:

a) The angular acceleration of the beam is 0.5 rad/s²CW (direction clockwise due the tangential acceleration is positive)

b) The acceleration of point A is 3.25 m/s²

The acceleration of point E is 0.75 m/s²

Explanation:

a) The relative acceleration of B with respect to D is equal:

$a_{B} =a_{D} +(a_{B/D} )_{n} +(a_{B/D} )_{t}$

Where

aB = absolute acceleration of point B = 2.5 j (m/s²)

aD = absolute acceleration of point D = 1.5 j (m/s²)

(aB/D)n = relative acceleration of point B respect to D (normal direction BD) = 0, no angular velocity of the beam

(aB/D)t = relative acceleration of point B respect to D (tangential direction BD)

$a_{B} =a_{D} +(a_{B/D} )_{t}$

$2.5j=1.5j +(a_{B/D} )_{t}\\(a_{B/D} )_{t}=j=1m/s^{2}$

We have that

(aB/D)t = BDα

Where α = acceleration of the beam

BDα = 1 m/s²

Where

BD = 2

$2\alpha =1\\\alpha =0.5rad/s^{2}CW$

b) The acceleration of point A is:

$a_{A} =a_{D} +(a_{A/D} )_{t}$

(aA/D)t = ADαj

$a_{A} =a_{D} +AD\alpha j\\a_{A}=1.5j+(3.5*0.5)j\\a_{A}=3.25jm/s^{2}$

The acceleration of point E is:

(aE/D)t = -EDαj

$a_{E} =a_{D} -ED\alpha j\\a_{E}=1.5j-(1.5*0.5)j\\a_{E}=0.75jm/s^{2}$

7 0

3 years ago

A fair ground ride spins its occupants inside a flying saucer-shaped container. if the horizontal circular path the riders follo

Licemer1 [7]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
Below is the solution:

centripetal accel = 1.5*g
ω²r = 1.5*9.8m/s²
ω² * 8m = 14.7 m/s²
ω = 1.36 rad/s * 1rev/2πrads * 60s/min = 12.9 rpm

6 0

3 years ago

The flash unit in a camera uses a special circuit to "step up" the 3.0 V from the batteries to 360 V , which charges a capacitor

aksik [14]

Answer:

The capacitance of the capacitor is $1.85*10^{-5}F$