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4 years ago

6

How is an electric current carried?

1 answer:

jok3333 [9.3K]4 years ago

4 0

By moving electrons in a wire

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Lightning flashes one mile (1609 m) away from you. How much time does it take the light to travel that distance?

hichkok12 [17]

Answer:

$t=5.36\times 10^{-6}\ s$

Explanation:

Given that,

Lightning flashes one mile (1609 m) away from you.

We need to find the time it take the light to travel that distance. Let the time be t. We know that,

speed = distance/time

$t=\dfrac{d}{v}\\\\t=\dfrac{1609}{3\cdot10^{8}}\\\\=5.36\times 10^{-6}\ s$

So, the required time is $5.36\times 10^{-6}\ s$

7 0

3 years ago

At a circus, a human cannonball is shot from a cannon at 15m/s at an angle of 40° above horizontal. She

Citrus2011 [14]

Answer:

a

The height is $H = 6.74 \ m$

b

The horizontal distance is $D = 23.74 \ m$

Explanation:

From the question we are told that

The speed is $v = 15 \ m/s$

The angle is $\theta = 40^o$

The height of the cannon from the ground is h = 2 m

The distance of the net from the ground is k = 1 m

Generally the maximum height she reaches is mathematically represented as

$H = \frac{v^2 sin^2 \theta }{2 * g } + h$

=> $H = \frac{(15)^2 [sin (40)]^2 }{2 * 9.8} + 2$

=> $H = 6.74 \ m$

Generally from kinematic equation

$s = ut + \frac{1}{2} at^2$

Here s is the displacement which is mathematically represented as

s = [-(h-k)]

=> s = -(2-1)

=> s = -1 m

There reason why s = -1 m is because upward motion canceled the downward motion remaining only the distance of the net from the ground which was covered during the first half but not covered during the second half

a = -g = -9.8

$u = v sin (\theta)$

So

$-1 = (vsin 40 )t + \frac{1}{2} * (-9.8) t^2$

=> $-4.9t^2 + 9.6418t + 1 = 0$

using quadratic formula to solve the equation we have

$t = 2.07 \ s$

Generally distance covered along the horizontal is

$D = v cos (40) * 2.07$

=> $D = 15 cos (40) * 2.07$

=> $D = 23.74 \ m$

7 0

3 years ago

What are the benefits of physical exercises?

makvit [3.9K]

Answer:

Manage your weight

Have lower blood pressure

Lower your risk of falls

it reduces your risk of heart attack

7 0

3 years ago

How deep can an object with 6360N hitting on a sponge get ???

borishaifa [10]

Answer:

The sponge must go $\dfrac{636}{m}\ \text{meter}$ deep

Explanation:

If F = 6360 N, then it is required to find how deep can an object with this force hitting on a sponge get.

We know that, F = mgh

m is mass

g is acceleration due to gravity

$h=\dfrac{F}{mg}\\\\h=\dfrac{6360}{10m}\\\\h=\dfrac{636}{m}\ \text{meter}$

So, the sponge must go $\dfrac{636}{m}\ \text{meter}$ deep.

5 0

4 years ago

You push on the top edge of a 1.8 m tall solid circular cylinder of iron which is 4.00 cm in diameter. You push with a horizonta

Citrus2011 [14]

Answer:

$\triangle x=3.2*10^-^5 m$

Explanation:

From the question we are told that

Height of circular cylinder is $h= 1.8m$

Diameter of cylinder $D=4cm=>0.04m$

Horizontal Force $F=900N$

$Y = 10.0 *10^1^0 N/m^2 \\B = 9.0 * 10^1^0 N/m^2\\S = 4.0 * 10^1^0 N/m^2\\$

Generally the formula for shear modulus is mathematically represented by

$G=\frac{\tau}{\gamma}$

Where

$G=Shear modulus\\\tau= shear\ stress\\\gamma=shear strain$

$G=\frac{F/A}{\triangle x/L}$

$G=\frac{900/\pi r^2}{\triangle x/1.8}$

$4.0*10^1^0=\frac{900/\pi r^2}{\triangle x/1.8}$

$4.0*10^1^0=\frac{900}{\pi r^2} *\frac{1.8}{\triangle x}$

${\triangle x} =\frac{1620}{\pi r^2* 4.0*10^1^0}$

$\triangle x=3.2*10^-^5 m$

5 0

3 years ago