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2 years ago

A 2.72 A current flows through a wire for

Physics

1 answer:

agasfer [191]2 years ago

8 0

Current=2.72A=I
Time=t=9.88s
Charge=Q

$\\ \rm\longmapsto I=\dfrac{Q}{t}$

$\\ \rm\longmapsto Q=It$

$\\ \rm\longmapsto Q=2.72\times 9.88$

$\\ \rm\longmapsto Q=26.87C$

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How much energy is stored in a 2.80-cm-diameter, 14.0-cm-long solenoid that has 200 turns of wire and carries a current of 0.800

ozzi

Answer:

The energy stored in the solenoid is 7.078 x 10⁻⁵ J

Explanation:

Given;

diameter of the solenoid, d = 2.80 cm

radius of the solenoid, r = d/2 = 1.4 cm

length of the solenoid, L = 14 cm = 0.14 m

number of turns, N = 200 turns

current in the solenoid, I = 0.8 A

The cross sectional area of the solenoid is given as;

$A = \pi r^2\\\\A = \pi (0.014)^2\\\\A = 6.16*10^{-4} \ m^2$

The inductance of the solenoid is given by;

$L = \frac{\mu_0 N^2A}{l} \\\\L = \frac{(4\pi*10^{-7})(200^2)(6.16*10^{-4})}{0.14}\\\\L = 2.212*10^{-4} \ H$

The energy stored in the solenoid is given by;

E = ¹/₂LI²

E = ¹/₂(2.212 x 10⁻⁴)(0.8)²

E = 7.078 x 10⁻⁵ J

Therefore, the energy stored in the solenoid is 7.078 x 10⁻⁵ J

8 0

3 years ago

LOTS OF POINTSA rocket of mass 40 000 kg takes off and flies to a height of 2.5 km as its engines produce 500 000 N of thrust.

Svetradugi [14.3K]

Answer:

i) E = 269 [MJ] ii)v = 116 [m/s]

Explanation:

This is a problem that encompasses the work and principle of energy conservation.

In this way, we establish the equation for the principle of conservation and energy.

$E_{k1}+W_{1-2}=E_{k2}\\where:\\E_{k1}= kinetic energy at moment 1\\W_{1-2}= work between moments 1 and 2.\\E_{k2}= kinetic energy at moment 2.$

$W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]$

At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.

Er = 269*10^6[J]

ii ) With the energy calculated at the previous point, we can calculate the speed developed.

$E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]$

8 0

3 years ago

Students hypothesized that by running an electric current through the wire of the apparatus shown here, they could cause a non-m

astra-53 [7]

The answer is d i think

5 0

3 years ago

Read 2 more answers

What is the magnitude of the electric field at a distance of 89 cm from a 27 μC charge, in units of N/C?

GuDViN [60]

Answer:

306500 N/C

Explanation:

The magnitude of an electric field around a single charge is calculated with this equation:

$E(r) = \frac{1}{4 \pi *\epsilon 0} \frac{q}{r^2}$

With ε0 = 8.85*10^-12 C^2/(N*m^2)

Then:

$E(0.89) = \frac{1}{4 \pi *8.85*10^-12} \frac{27*10^-6}{0.89^2}$

E(0.89) = 306500 N/C

3 0

3 years ago

What is the empirical formula of metaldehyde, C8H16O4?

inessss [21]

Answer:

C8H16O4

Explanation:

ok??

7 0

2 years ago