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expeople1 [14]
4 years ago
5

4

Physics
1 answer:
Blizzard [7]4 years ago
6 0

Answer:

0.2 J

Explanation:

The pendulum forms a right triangle, with hypotenuse of 50 cm and base of 30 cm.  The height of this triangle can be found with Pythagorean theorem:

c² = a² + b²

(50 cm)² = a² + (30 cm)²

a = 40 cm

The height of the triangle is 40 cm.  The height of the pendulum when it is at the bottom is 50 cm.  So the end of the pendulum is lifted by 10 cm.  Assuming the mass is concentrated at the end of the pendulum, the potential energy is:

PE = mgh

PE = (0.200 kg) (9.8 N/kg) (0.10 m)

PE = 0.196 J

Rounding to one significant figure, the potential energy is 0.2 J.

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A small current element carrying a current of I = 1.00 A is placed at the origin given by d → l = 4.00 m m ^ j Find the magnetic
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Answer:

the magnitude and direction of d → B on the x ‑axis at x = 2.50 m is -6.4 × 10⁻¹¹T(Along z direction)

the magnitude and direction of d → B on the z ‑axis at z = 5.00 m is 1.6 × 10⁻¹¹T(Along x direction)

Explanation:

Use Biot, Savart, the magnetic field

d\bar{B}=\frac{U}{4\pi } \frac{i(d\bar{l}\times r)}{r^2}

Given that,

i = 1.00A

d → l = 4.00 m m ^ j

r = 2.5m

Displacement vector is

\bar{r}=x\hat i+y\hat j+z \hat k\\

\bar{r}= (2.5m) \hat i +(0m)^2 + (0m)^2

 =2.5m

on the axis of x at x = 2.5

r = \sqrt{(2.5)^2 + (0)^2 + (0)^2}

r = 2.5m

And unit vector

\hat r =\frac{\bar{r}}{r}

= \frac{2.5 \hat i}{2.5}\\\\= 1\hat i

Therefore, the magnetic field is as follow

d\bar{B}=\frac{U}{4\pi } \frac{i(d\bar{l}\times r)}{r^2}

d\bar{B} = \frac{(10^-^7)(1)(4\times10^-^3j\times i}{(2.50)^2} \\\\d\bar{B} = -6.4\times10^{-11} T

(Along z direction)

B)r = 5.00m

Displacement vector is

\bar{r}=x\hat i+y\hat j+z \hat k\\

\bar{r}= (5.00m) \hat i +(0m)^2 + (0m)^2

 =5.00m

on the axis of x at x = 5.0

r = \sqrt{(5.00)^2 + (0)^2 + (0)^2}

r = 5.00m

And unit vector

\hat r =\frac{\bar{r}}{r}

= \frac{5.00 \hat i}{5.00}\\\\= 1\hat i\\

Therefore, the magnetic field is as follow

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d\bar{B} = \frac{(10^-^7)(1)(4\times10^-^3j\times i}{(5.00)^2} \\\\d\bar{B} = 1.6\times10^{-11} T

(Along x direction)

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