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3 years ago

1. why does the sun have the greatest amount of gravity in our solar system?

2 answers:

6 0

The lighter object orbits the heavier one, and the Sun is, by far, the heaviest object in the solar system. ... Heavier objects (really, more massive ones) produce a bigger gravitational pull than lighter ones, so as the heavyweight in our solar system, the Sun exerts the strongest gravitational pull.

melomori [17]3 years ago

3 0

Answer:

it is the biggest object that is the closest to us

Explanation:

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Your house is 45.0 m from a powerline carrying 152 A of current. How much magnetic field does the current create at your house?

Sedaia [141]

This question involves the concepts of th magnetic field and current.

The magnetic field created by the current at the house is "6.75 x 10⁻⁷ T".

<h3>Magnetic Field</h3>

The magnetic field created by a current carrying wire can be given by the following formula:

$B=\frac{\mu_o I}{2\pi r}$

where,

B = magnetic field = ?
$\mu_o$ = permeabiliy of free space =4π x 10⁻⁷
I = current = 152 A
r = distance = 45 m

$B=\frac{4\pi x\ 10^{-7}(152)}{2\pi(45)}$

B = 6.75 x 10⁻⁷ T

Learn more about magnetic field here:

brainly.com/question/23096032

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1 year ago

In the deep ocean, a water wave with wavelength 95 m travels at 12 m/s. Suppose a small boat is at the crest of this wave, 1.2 m

dimaraw [331]

Answer:

the vertical position is 1.1971m

Explanation:

Recall that

$y = 1.2 * cos (\frac{2*\pi * distance travelled}{wavelenght})$

recall that $v = \frac{distance}{time}$

$thus; distance = v* t$

this implies that distance = 12 * 5.0

= 60

therefore; $y = 1.2 cos \frac{2*3.142*60}{95}$

y = 1.1971m

the

3 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,

$v_x = 28.58m/s.$