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3 years ago

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A solenoid has 100 turns per cm and a total resistance of 60 Ω center of the solenoid when it is connected to a 12 volt battery

Determine the magnetic field in the 21 A flat c

1 answer:

charle [14.2K]3 years ago

8 0

Answer:

2.5 x 10^-3 T

Explanation:

n = 100 turns / cm = 10000 turns / m, R = 60 ohm, V = 12 V

i = V / R = 12 / 60 = 0.2 A

Magnetic field due to solenoid is

B = μ0 x n x i

B = 4 x 3.14 x 10^-7 x 10000 x 0.2

B = 2.5 x 10^-3 T

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A 6 kilogram block in outer space is moving at -100 m/s (to the left). It suddenly experiences three forces as shown below (in t

Blizzard [7]

Answer:

x = 1474.9 [m]

Explanation:

To solve this problem we must use Newton's second law, which tells us that the sum of forces must be equal to the product of mass by acceleration.

We must understand that when forces are applied on the body, they tend to slow the body down to stop it.

So as the body continues to move to the left, it is slowing down. Therefore we must calculate this deceleration value using Newton's second law. We must perform a sum of forces on the x-axis equal to the product of mass by acceleration. With leftward movement as negative and rightward forces as positive.

ΣF = m*a

$10 +12*sin(60)= - 6*a\\a = - 3.39[m/s^{2}]$

Now using the following equation of kinematics, we can calculate the distance of the block, before stopping completely. The initial speed must be 100 [m/s].

$v_{f}^{2} =v_{o}^{2}-2*a*x$

where:

Vf = final velocity = 0 (the block stops)

Vo = initial velocity = 100 [m/s]

a = - 3.39 [m/s²]

x = displacement [m]

$0 = 100^{2}-2*3.39*x\\x=\frac{10000}{2*3.39}\\x=1474.9[m]$

4 0

3 years ago

Notice that in each conversion factor the numerator equals the denominator when units are taken into account. A common error in

navik [9.2K]

Answer:

he factor for the temporal part 1.296 107 s² = h²

m / s² = 12960 km / h²

Explanation:

This is a unit conversion exercise.

In the unit conversion, the size of the object is not changed, only the value with respect to which it is measured is changed, for this reason in the conversion the amount that is in parentheses must be worth one.

In this case, it is requested to convert a measure km/h²

Unfortunately, it is not clearly indicated what measure it is, but the most used unit in physics is m / s² , which is a measure of acceleration. Let's cut this down

the factor for the distance is 1000 m = 1 km

the factor for time is 3600 s = 1 h

let's make the conversion

m / s² (1km / 1000 m) (3600 s / 1h)²

note that as time is squared the conversion factor is also squared

m / s² = 12960 km / h²

the factor for the temporal part 1.29 107 s² = h²

6 0

3 years ago

Which statement best describes why an island's food web could be considered a closed system?

garik1379 [7]

I would say Option B) because Option C) is wrong since matter cannot be created. A closed system does not exchange matter so it's not Option D). Since an island is an isolated area, Option A) is wrong.

3 0

3 years ago

Read 2 more answers

U1=20 m/s turn it to km/h

notka56 [123]

It is 72 km/h
I hope it helps

7 0

3 years ago

A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the

Maurinko [17]

Answer:

(a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

Explanation:

Given that,

Activity $R_{0}=10\ mCi$

Time $t_{1}=4\ hours$

Activity R= 8 mCi

(a). We need to calculate the decay constant

Using formula of activity

$R=R_{0}e^{-\lambda t}$

$\lambda=\dfrac{1}{t}ln(\dfrac{R_{0}}{R})$

Put the value into the formula

$\lambda=\dfrac{1}{4\times3600}ln(\dfrac{10}{8})$

$\lambda=0.0000154\ s^{-1}$

$\lambda=1.55\times10^{-5}\ s^{-1}$

We need to calculate the half life

Using formula of half life

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}$

Put the value into the formula

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}$

$T_{\dfrac{1}{2}}=44.719\times10^{3}\ s$

$T_{\dfrac{1}{2}}=11.3\ hr$

(b). We need to calculate the value of N₀

Using formula of $N_{0}$

$N_{0}=\dfrac{3.70\times10^{6}}{\lambda}$

Put the value into the formula

$N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}$

$N_{0}=2.38\times10^{11}\ nuclei$

(c). We need to calculate the sample's activity

Using formula of activity

$R=R_{0}e^{-\lambda\times t}$

Put the value intyo the formula

$R=10e^{-(1.55\times10^{-5}\times30\times3600)}$

$R=1.87\ mCi$

Hence, (a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

4 0

3 years ago