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3 years ago

I need this right ASAP!!! plz plz plz

Physics

2 answers:

UkoKoshka [18]3 years ago

8 0

Answer:

The answer is D

Nucleus is neutral

KiRa [710]3 years ago

5 0

The answer is D
electrically neutral region at the center of the atom.

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The magnetic flux through each turn of a 110-turn coil is given by ΦB = 9.75 ✕ 10−3 sin(ωt), where ω is the angular speed of the

Xelga [282]

Answer:

Explanation:

Given that a coil has a turns of

N = 110 turns

And the flux is given as function of t

ΦB = 9.75 ✕ 10^-3 sin(ωt),

Given that, at an instant the angular velocity is 8.70 ✕ 10² rev/min

ω = 8.70 ✕ 10² rev/min

Converting this to rad/sec

1 rev = 2πrad

Then,

ω = 8.7 × 10² × 2π / 60

ω = 91.11 rad/s

Now, we want to find the induced EMF as a function of time

EMF is given as

ε = —NdΦB/dt

ΦB = 9.75 ✕ 10^-3 sin(ωt),

dΦB/dt = 9.75 × 10^-3•ω Cos(ωt)

So,

ε = —NdΦB/dt

ε = —110 × 9.75 × 10^-3•ω Cos(ωt)

Since ω = 91.11 rad/s

ε = —110 × 9.75 × 10^-3 ×91.11 Cos(91.11t)

ε = —97.71 Cos(91.11t)

The EMF as a function of time is

ε = —97.71 Cos(91.11t)

Extra

The maximum EMF will be when Cos(91.11t) = -1

Then, maximum emf = 97.71V

8 0

3 years ago

PLEASE HELP IM UNSURE OF WHAT THE ANSWER IS

katovenus [111]

Answer:

The horizontal forces (Cols 2 & 3) balance for W and Y.

The vertical forces (Cols 1 & 4) also balance for W and Y.

This is not true for either X (unbalanced horizontal forces) or

Z (also unbalnced horizontal forces)

6 0

3 years ago

What is the electric potential of a 2.2 µC charge at a distance of 6.3 m from the charge? Recall that Coulomb’s constant is k =

iren2701 [21]

1) The electric potential at a distance r from a single point charge is given by
$V(r) = k \frac{q}{r}$
where k is the Coulomb's constant, q is the charge and r is the distance from the charge.

The charge in this problem is
$q=2.2 \mu C =2.2 \cdot 10^{-6} C$
So the potential at distance $r=6.3 m$ is
$V=k \frac{q}{r}=(8.99 \cdot 10^9 Nm^2 C^{-2}) \frac{2.2 \cdot 10^{-6} C}{6.3 m}=3139 V$

2) By using the same formula as before, we can find the electric potential at distance r=99 m from the charge:
$V=k \frac{q}{r}=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{2.2 \cdot 10^{-6} C}{99 m}=198 V$

3 0

3 years ago

Read 2 more answers

I will give more points <br>for this thank you I u helped<br>me

scZoUnD [109]

Answer:

weight = mass×gravity

15×9.8
147N

<h2>stay safe healthy and happy.</h2>

6 0

3 years ago

A ramp 20 m long and 4 m high is used to lift a heavy box. A pulley system with 4 rope sections supporting the load is used to l

kobusy [5.1K]

The ramp has dimensions of 20 m length and 4 m height. The angle made by the ramp on the horizontal ground is
θ = sin⁻¹ (4/20)
After getting the angle, the force required to move the box can then be solved given that the mass of the box is known and also the coefficient of friction of the ramp

5 0

3 years ago