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3 years ago

Describe a situation where your body can act as a lightning conductor.

1 answer:

7 0

When someone is struct by lightning, the electricity passes through the body, into the earth. Here, our body acts as a lightning conductor to complete the earthing process.

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Please answer B, C, E, and D

Natasha_Volkova [10]

B- the biosphere needs the hydrosphere (water) to survive. They drink the water.
C- the biosphere use the géosphère as land/ home.
D-The géosphère depends on the hydrosphere for water. Without the water, the géosphère would eventually turn dry and nothing will be able to grow
E- The animals in the hydrosphere need air to survive. The gills on sea animals are used to filter out the water from the oxygen and use the oxygen to breathe. As the fish opens its mouth, water runs over the gills, and blood in the capillaries picks up oxygen that's dissolved in the water.

3 0

2 years ago

Read 2 more answers

A) A car accelerates uniformly at 3.7 m s⁻² as it passes a stationary road train. The initial speed of the car is 30 m s⁻¹, and

Tanya [424]

(a) The length of the train is 54.6 m

(b) The braking distance of the road train is 194.44 m

The given parameters:

acceleration of the car, a = 3.7 m/s²

initial velocity of the car, u = 30 m/s

final velocity of the car, v = 130 km/h = 36.11 m/s

To find:

the length of the train

The length of the train is the distance travelled by the car

The distance traveled by the car is calculated as:

$v^2 = u^2 + 2as\\\\2as = v^2 - u^2\\\\s = \frac{ v^2 - u^2}{2a} \\\\where;\\\\s \ is \ the \ distance \ travelled \ by \ the \ car\\\\s = \frac{(36.11)^2 - (30)^2}{2\times 3.7} \\\\s = 54.585 \ m\\\\s \approx 54.6 \ m$

Thus, the length of the train is 54.6 m

(b) The braking distance of a road train travelling at 25 m s⁻¹

$a = \frac{v_1^2}{2s_1} = \frac{v_2^2}{2s_2} \\\\Given;\\\\s_1 = 70 \ m\\\\v_1 = 15 \ m/s\\\\v_2 = 25 \ m/s\\\\s_2 = ?\\\\2s_2v_1^2 = 2s_1v_2^2\\\\s_2 = \frac{ 2s_1v_2^2}{2v_1^2} \\\\s_2 = \frac{s_1v_2^2}{v_1^2} \\\\s_2 = \frac{70\times 25^2}{15^2} \\\\s_2 = 194.44 \ m$

Thus, the braking distance of the road train is 194.44 m

Learn more here: brainly.com/question/19572178

6 0

3 years ago

A worker on the roof of a house drops his 0.58 kg hammer, which slides down the roof at constant speed of 6.69 m/s. The roof mak

shusha [124]

Answer:

17.3 m

Explanation:

Given that,

Mass of a hammer is 0.58 kg

Velocity with which the hammer slides is 6.69 m/s at constant speed.

The roof makes an angle of 18 ◦ with the horizontal, and its lowest point is 18.2 m from the ground. We need to find the horizontal distance traveled by the hammer between the time is leaves the roof of the house and the time it hits the ground. Firstly, we will find the time taken by the hammer when it reaches ground in vertical direction.

$y=y_0+v_ot +\dfrac{1}{2}gt^2$

Putting all the values,

$0=18.2+6.69t-\dfrac{1}{2}\times 9.8t^2\\\\-4.9t^2+6.69t+18.2=0\\\\t=\dfrac{-6.69+\sqrt{6.69^{2}-4\cdot-4.9\cdot18.2}}{2\cdot-4.9}, \dfrac{-6.69-\sqrt{6.69^{2}-4\cdot-4.9\cdot18.2}}{2\cdot-4.9}\\\\t=-1.36\ s\ \text{and}\ t=2.72\ s$

Neglecting negative value,

To find horizontal distance, multiply 2.72 s with the horizontal component of velocity.

$d=2.72\times 6.69\times \cos(18)\\\\d=17.3\ m$

3 0

3 years ago

Find the cube roots of 27(cos 327° + i sin 327° ). Write the answer in trigonometric form.

Sati [7]

Answer:

$z^{\frac{1}{3} }= -0.978 + i\cdot 2.836$ , $z^{\frac{1}{3} }= -1.967 - i\cdot 2.265$ , $z^{\frac{1}{3} }= 2.945 - i\cdot 0.571$

Explanation:

The cube root of the complex number can determined by the following De Moivre's Formula:

$z^{\frac{1}{n} } = r^{\frac{1}{n} }\cdot \left[\cos\left(\frac{x + 2\pi\cdot k}{n} \right) + i\cdot \sin\left(\frac{x+2\pi\cdot k}{n} \right)\right]$

Where angles are measured in radians and k represents an integer between $0$ and $n - 1$ .

The magnitude of the complex number is $27$ and the equivalent angular value is $1.817\pi$ . The set of cubic roots are, respectively:

k = 0

$z^{\frac{1}{3} } = 3\cdot \left[\cos \left(\frac{1.817\pi}{3} \right)+i\cdot \sin\left(\frac{1.817\pi}{3} \right)]$

$z^{\frac{1}{3} }= -0.978 + i\cdot 2.836$

k = 1

$z^{\frac{1}{3} } = 3\cdot \left[\cos \left(\frac{3.817\pi}{3} \right)+i\cdot \sin\left(\frac{3.817\pi}{3} \right)]$

$z^{\frac{1}{3} }= -1.967 - i\cdot 2.265$

k = 2

$z^{\frac{1}{3} } = 3\cdot \left[\cos \left(\frac{5.817\pi}{3} \right)+i\cdot \sin\left(\frac{5.817\pi}{3} \right)]$

$z^{\frac{1}{3} }= 2.945 - i\cdot 0.571$

5 0

3 years ago

Which of the following is the flow of electrons through a wire or a conductor

Hatshy [7]

Wires or silver and copper

8 0

3 years ago