All categories

3 years ago

A lightning rod effectively protects a building by transferring the electrical current to the ____________.

Physics

1 answer:

maria [59]3 years ago

6 0

It transfers the electrical current from the point of contact on the lightning rod into the ground.

You might be interested in

A 1.2 L weather balloon on the ground has a temperature of 25°C and is at atmospheric pressure (1.0 atm). When it rises to an el

Irina-Kira [14]

Answer:

71.19 C

Explanation:

25C = 25 + 273 = 298 K

Applying the ideal gas equation we have

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

where P, V and T are the pressure, volume and temperature of the gas at 1st and 2nd stage, respectively. We can solve for the temperature and the 2nd stage:

$T_2 = T_1\frac{P_2V_2}{P_1V_1} = 298\frac{0.77*1.8}{1.2*1} = 298*1.155 = 344.19 K = 344.19 - 273 = 71.19 C$

4 0

3 years ago

With a diameter that's 11 times larger than Earth's, _______ is the largest planet.

katovenus [111]

With a diameter that's 11 times larger than Earth's, Jupiter is the largest planet

6 0

3 years ago

sunlight, wind, and running water are essentially "free." Yet renewable energy sources are a very small part of our energy consu

dimulka [17.4K]

The heat and energy of coal burning is more easier and more efficient to turn into electricity than any method using renewable resources.

7 0

3 years ago

Read 2 more answers

A rocket blasts off from the Earth's surface. During the initial phase of flight, the engine of the rocket burns fuel at a rate

Katen [24]

Answer:

αβ = Ma

Explanation:

By Newton's 2nd Law, the equation governing the motion of the rocket while the rocket is burning fuel is

αβ = Ma where α = rocket's fuel burning rate, β = relative to the velocity of the rocket, M = instantaneous mass of the rocket and a = acceleration of rocket.

5 0

2 years ago

1. Do alto de uma plataforma com 15m de altura, é lançado horizontalmente um projéctil. Pretende-se atingir um alvo localizado n

sveta [45]

Answer:

(a). The initial velocity is 28.58m/s

(b). The speed when touching the ground is 33.3m/s.

Explanation:

The equations governing the position of the projectile are

$(1).\: x =v_0t$

$(2).\: y= 15m-\dfrac{1}{2}gt^2$

where $v_0$ is the initial velocity.

(a).

When the projectile hits the 50m mark, $y=0$ ; therefore,

$0=15-\dfrac{1}{2}gt^2$

solving for $t$ we get:

$t= 1.75s.$

Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that

$50m = v_0(1.75s)$

which gives

$\boxed{v_0 = 28.58m/s.}$

(b).

The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,

$v_x = 28.58m/s.$

the vertical component of the velocity is

$v_y = gt \\v_y = (9.8m/s^2)(1.75s)\\\\{v_y = 17.15m/s.$

which gives a speed $v$ of

$v = \sqrt{v_x^2+v_y^2}$

$\boxed{v =33.3m/s.}$

4 0

3 years ago