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2 years ago

E= (500.0lm) 4 (100)

Physics

1 answer:

drek231 [11]2 years ago

7 0

Answer:

didn't understand your question

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Calculate frequency if wavelength is 35 m and wave speed is 546 m/s.

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2 years ago

A) 1.2-kg ball is hanging from the end of a rope. The rope hangs at an angle 20° from the vertical when a 19 m/s horizontal wind

Marat540 [252]

Answer:

Part a)

$F_v = 4.28 N$

Part B)

$L = 1.02 m$

Part C)

$v = 1.25 m/s$

Explanation:

Part A)

As we know that ball is hanging from the top and its angle with the vertical is 20 degree

so we will have

$Tcos\theta = mg$

$T sin\theta = F_v$

$\frac{F_v}{mg} = tan\theta$

$F_v = mg tan\theta$

$F_v = 1.2\times 9.81 (tan20)$

$F_v = 4.28 N$

Part B)

Here we can use energy theorem to find the distance that it will move

$-\mu mg cos\theta L + mg sin\theta L = -\frac{1}{2}mv^2$

$(-(0.37)m(9.81) cos15 + m(9.81) sin15)L = - \frac{1}{2}m(1.4)^2$

$(-3.5 + 2.54)L = - 0.98$

$L = 1.02 m$

Part C)

At terminal speed condition we know that

$F_v = mg$

$bv^2 = mg$

$2.5 v^2 = 3.9$

$v = 1.25 m/s$

7 0

2 years ago

A long, straight, horizontal wire carries a left-to-right current of 40 A. If the wire is placed in a uniform magnetic field of

sladkih [1.3K]

Answer:

The magnitude of the resultant of the magnetic field is $4.11\times10^{-5}\ T$

Explanation:

Given that,

Current = 40 A

Magnetic field $B=3.7\times10^{-5}\ T$

Distance = 22 cm

We need to calculate the magnetic field

Using formula of magnetic field

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

Where, r = distance

I = current

Put the value into the formula

$B'=\dfrac{4\pi\times10^{-7}\times20}{2\pi\times0.22}$

$B'=1.8\times10^{-5}\ T$

We need to calculate the magnitude of the resultant of the magnetic field

Using formula of resultant

$B''=\sqrt{B^2+B'^2}$

Put the value into the formula

$B''=\sqrt{(3.7\times10^{-5})^2+(1.8\times10^{-5})^2}$

$B''=4.11\times10^{-5}\ T$

Hence, The magnitude of the resultant of the magnetic field is $4.11\times10^{-5}\ T$

6 0

2 years ago