20 POINTSSS!!!!!!!!

Which of the following statements is true about a current-carrying wire in a magnetic field? A. Reversing the current direction

Zielflug [23.3K]

B. Reversing the current direction will cause the force deflecting the wire to be perpendicular to the magnetic field but in the opposite direction.

6 0

3 years ago

Electricity is distributed from electrical substations to neighborhoods at 13000 V. This is a 60 Hz oscillating (AC) voltage. Ne

Levart [38]

Answer:

the number of turns in the primary coil is 13000

Explanation:

Given the data in the question;

V₁ = 13000 V

V₂ = 120 V

N₁ = ?

N₂ = 120 turns

the relation between the voltages and the number of turns in the primary and secondary coils can be expressed as;

V₁/V₂ = N₁/N₂

V₁N₂ = V₂N₁

N₁ = V₁N₂ / V₂

so we substitute

N₁ = (13000 V × 120 turns) / 120 V

N₁ = 1560000 V-turns / 120 V

N₁ = 13000 turns

Therefore, the number of turns in the primary coil is 13000

8 0

2 years ago

Near the top of the Citigroup Center building in New York City, there is an object with mass of 4.00 ✕ 105 kg on springs that ha

jasenka [17]

Explanation:

It is given that,

Mass of an object, $m=4\times 10^5\ kg$

(a) Time period of oscillation, T = 2.4 s

The formula for the time period of spring is given by :

$T=2\pi \sqrt{\dfrac{m}{k}}$

Where

k is the force constant

$k=\dfrac{4\pi ^2 m}{T^2}$

$k=\dfrac{4\pi ^2 \times 4\times 10^5}{(2.4)^2}$

$k=2.74\times 10^6\ N/m$

(b) Displacement in the spring, x = 2.2 m

Energy stored in the spring is given by :

$U=\dfrac{1}{2}kx^2$

$U=\dfrac{1}{2}\times 2.74\times 10^6\ N/m\times (2.2\ m)^2$

$U=6.63\times 10^6\ J$

Hence, this is the required solution.

7 0

3 years ago

The amplitude of the voltage across an inductor can be greater than the amplitude of the generator EMF in an AC RLC circuit. Con

Degger [83]

Answer:

VL=2107.6v

Explanation:

at resonance xc=xL

1/wc = wL

z=R

because sqr(R^2+(xL-xc))

^

(xL=xc)

V/R=I

90/28.9=3.1142A

w=1/sqr(1.31×(2.86×10^-6))=516.63

xL= wL

xL= 516.63×1.31=676.785

VL=3.1142×676.785

VL=2107.6v

4 0

3 years ago

We timed how long it took for the ball to travel 1 meter several times, so we could calculate an “average” time to use in the ve

damaskus [11]

We need to find the average speed of the ball during the motion of 1 m

In order to find that we took several reading and found following times to cover the distance of 1 m

t1 = 2.26 s

t2 = 2.38 s

t3 = 3.02 s

t4 = 2.26 s

t5 = 2.31 s

Now in order to find the average time we can write

$T_{mean} = \frac{t_1 + t_2 + t_3 + t_4 + t_5}{5}$

$T_{mean} = \frac{2.26 + 2.38 + 3.02 + 2.26 + 2.31}{5}$

$T_{mean} = 2.45 s$

So average time to cover the distance of 1 m by ball will be 2.45 s

here 3.02 s is not the average time but we can say it is the median of the readings of all possible values which we can not use in our calculation as average time

3 0

3 years ago