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Semmy [17]
3 years ago
5

An experiment on the earth's magnetic field is being carried out 1.00 m from an electric cable. What is the maximum allowable cu

rrent int eh cable if the experiment is to be accurate?
Physics
1 answer:
Whitepunk [10]3 years ago
4 0

Answer:

The allowed current in the cable is 1.15 A.

Explanation:

Given that,

Distance = 1.00 m

Suppose the magnetic field is 2.3\times10^{-5}\ T and  if the experiment is to be accurate to 1.0 %

We need to calculate the current

Using formula of magnetic field

B=\dfrac{\mu_{0}I}{2\pi r}

I=\dfrac{B\times2\pi r}{\mu_{0}}

Put the value into the formula

I=\dfrac{2.3\times10^{-5}\times2\times\pi\times1.00}{4\pi\times10^{-7}}

I=115\ A

If the experiment is to be accurate to 1.0%

Then,

We need to calculate the allowed current in the cable

I'=(1.0%)\times I

I'=\dfrac{1.0}{100}\times115

I'=1.15\ A

Hence, The allowed current in the cable is 1.15 A.

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A cake is removed from a 350◦F oven and placed on a cooling rack in a 70◦F room. After 30 minutes the cake is 200◦F. When will i
galben [10]

Answer:

350 F to 100 F it take approx 87.33 min  

Explanation:

given data

oven = 350◦F

cooling rack = 70◦F

time = 30 min

cake = 200◦F

solution

we apply here Newtons law of cooling  

\frac{dT}{dt} = -k(T-Ta)

\frac{dy}{dt} = \frac{d}{dt} (T(t) -Ta)

= \frac{dT}{dt} -\frac{dTa}{dt} =\frac{dT}{dt} = -k(T-Ta)

-ky \frac{dy}{dt} = -ky

T(t) -Ta = (To -Ta) e^{-kt} T(t) = Ta+ (To -Ta)  e^{-kt}

put her value for time 30 min and T(t) = 200◦F and To =350◦F  and Ta = 70◦F

so here

200 = 70 + ( 350 - 70 ) e^{-k30}

k = 0.025575

so here for  T(t) = 100F

100 = 70 + ( 350 - 70 ) e^{-0.025575*t}

time = 87.33 min

so here 350 F to 100 F it take approx 87.33 min  

5 0
3 years ago
Peristalsis is the action caused by contractions of muscles in this structure in the digestive system...
ollegr [7]

Answer:

I believe the answer is esophagus

8 0
3 years ago
A rectangular coil of 65 turns, dimensions 0.100 m by 0.200 m, and total resistance 10.0 ? rotates with angular speed 29.5 rad/s
vovikov84 [41]

Answer:

Explanation:

N = 65

Area, A = 0.1 x 0.2 = 0.02 m^2

R = 10 ohm

ω = 29.5 rad/s

B = 1 T

(a) at t = 0

e = N x B x A x ω

e = 65 x 1 x 0.02 x 29.5

e = 38.35 V

(b) The maximum rate of change of magnetic flux is equal to the maximum value of induced emf.

Ф = 38.35 Wb/s

(c) e = NBAω Sinωt

e = 65 x 1 x 0.02 x 29.5 x Sin (29.5 x 0.05)

e = 38.174 V

(d) Maximum torque

τ = M B Sin 90

τ = N i A B

τ = N e A B / R

τ = 65 x 38.35 x 0.02 x 1 / 10

τ = 5 Nm

8 0
3 years ago
A car, moving along a straight stretch of highway, begins to accelerate at 0.0323 m/s 2 . It takes the car 63.3 s to cover 1 km.
rosijanka [135]
X=ut+0.5at²
1000 = 63.3u + 0.5 * 0.0323 * 63.3²
u = (1000 - 0.1615*63.3²)/63.3
u = 5.57 m/s

i don't even know if this is correct
7 0
4 years ago
In chemistry, potential energy is usually considered to be the energy _____________.
Lunna [17]

Answer:

The answer is option D. Stored in the chemical bonds.

Hope it helps............

4 0
3 years ago
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