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kondor19780726 [428]

2 years ago

8

What is the force per meter on a lightning bolt at the equator that carries 20,000 A perpendicular to Earth’s 3.0e−5 T field? (b

) What is the direction of the force if the current is straight up and Earth’s field direction is due north, parallel to the ground?

1 answer:

jonny [76]2 years ago

8 0

To solve this problem it is necessary to apply the concepts related to Magnetic Force.

The magnetic force as magnitude of a vector is described by

$F= IlBsin\theta$

Where,

$\theta =$ Angle between the wire and the magnetic field

L = Length

B = Magnetic Field

I = Current

Since the lightning bolt is perpendicular to the earth then the angle is 90°, moreover we need find the force per meter, then

$\frac{F}{l} = IBsin\theta$

Replacing with our values

$\frac{F}{l} = (20000)(3*10^{-5})sin(90)$

$\frac{F}{l} = 0.6N/m$

Therefore the force per meter on lightning volt is 0.6N/m

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pochemuha

Answer:

(i) 208 cm from the pivot

(ii) Move further from the pivot

Explanation:

(i) Sum of the moments about the pivot of the seesaw is zero.

∑τ = Iα

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Kenny should sit 208 cm on the other side of the pivot.

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2 years ago

Onur drops a basketball from a height of 10\,\text{m}10m10, start text, m, end text on Mars, where the acceleration due to gravi

Doss [256]

Answer:

Explanation:

Given that,

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H=10m

It is dropped on planet mass

And the acceleration due to gravity on Mars is given as

g= 3.7m/s²

Time taken for the ball to reach the ground

Initial velocity of the body is zero

u=0m/s

Using equation of motion: free fall

H = ut + ½gt²

10 = 0•t + ½ × 3.7 ×t²

10 = 0 + 1.85t²

10 = 1.85t²

Then, t² =10/1.85

t² = 5.405

t = √ 5.405

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2 years ago

Evaporation of sweat requires energy and thus take excess heat away from the body. Some of the water that you drink may eventual

kotegsom [21]

Answer:

The amount of heat required is $H_t = 1.37 *10^{6} \ J$

Explanation:

From the question we are told that

The mass of water is $m_w = 20 \ ounce = 20 * 28.3495 = 5.7 *10^2 g$

The temperature of the water before drinking is $T_w = 3.8 ^oC$

The temperature of the body is $T_b = 36.6^oC$

Generally the amount of heat required to move the water from its former temperature to the body temperature is

$H= m_w * c_w * \Delta T$

Here $c_w$ is the specific heat of water with value $c_w = 4.18 J/g^oC$

So

$H= 5.7 *10^2 * 4.18 * (36.6 - 3.8)$

=> $H= 7.8 *10^{4} \ J$

Generally the no of mole of sweat present mass of water is

$n = \frac{m_w}{Z_s}$

Here $Z_w$ is the molar mass of sweat with value

$Z_w = 18.015 g/mol$

=> $n = \frac{5.7 *10^2}{18.015}$

=> $n = 31.6 \ moles$

Generally the heat required to vaporize the number of moles of the sweat is mathematically represented as

$H_v = n * L_v$

Here $L_v$ is the latent heat of vaporization with value $L_v = 7 *10^{3} J/mol$

=> $H_v = 31.6 * 7 *10^{3}$

=> $H_v = 1.29 *10^{6} \ J$

Generally the overall amount of heat energy required is

$H_t = H + H_v$

=> $H_t = 7.8 *10^{4} + 1.29 *10^{6}$

=> $H_t = 1.37 *10^{6} \ J$

4 0

2 years ago

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Answer:

B. 4 m/s

Explanation:

v=d/t

Running for 300 m at 3 m/s takes 100 seconds and running at 300 m at 6 m/s takes 50 seconds. 100 s + 50 s = 150 s (total time). Total distance is 600 m, so 600 m/ 150 s = 4 m/s.

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Answer is given below

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