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4 years ago

Mercury was named after the roman god of speed why is it an appropriate name for the planet

1 answer:

5 0

Because it's the planet in our solar system with the shortest,
fastest orbit around the sun ... only 88 Earth days.

The people who named it didn't know that ... they still thought that
the sun and all the planets revolve around the Earth. But they did
see it zip from one side of the sun to the other, faster than any other
planet ... the result of having the shortest, fastest orbit of any planet.

You might be interested in

Which type of rays have a wavelength shorter than that of visible light?

aniked [119]

Answer:

Explanation:

wavelength shorter means energy is higher

the wavelength

radio waves>microwave>infrared rays>gamma rays

6 0

2 years ago

Particle A and particle B are held together with a compressed spring between them. When they are released, the spring pushes the

yaroslaw [1]

Answer:

K_a = 8,111 J

Explanation:

This is a collision exercise, let's define the system as formed by the two particles A and B, in this way the forces during the collision are internal and the moment is conserved

initial instant. Just before dropping the particles

p₀ = 0

final moment

p_f = m_a v_a + m_b v_b

p₀ = p_f

0 = m_a v_a + m_b v_b

tells us that

m_a = 8 m_b

0 = 8 m_b v_a + m_b v_b

v_b = - 8 v_a (1)

indicate that the transfer is complete, therefore the kinematic energy is conserved

starting point

Em₀ = K₀ = 73 J

final point. After separating the body

Em_f = K_f = ½ m_a v_a² + ½ m_b v_b²

K₀ = K_f

73 = ½ m_a (v_a² + v_b² / 8)

we substitute equation 1

73 = ½ m_a (v_a² + 8² v_a² / 8)

73 = ½ m_a (9 v_a²)

73/9 = ½ m_a (v_a²) = K_a

K_a = 8,111 J

3 0

3 years ago

If a satellite weighs 321 lb. on the earth's surface (R = 4,000 miles), how much does it weigh 12,000 miles above the surface? (

Sever21 [200]

The gravitational force between the Earth and the satellite (its "weight") is inversely proportional to the distance between the centers of both objects.

On the surface, their centers are separated by 1 Earth radius.

12,000 miles above the surface, they're separated by 4 Earth radiii.

(4/1) = 4

So after the move, the satellite's weight is (1/4²) = 1/16 of its surface weight.

(321 lb) / (16) = (20 and a hair) lb

The correct choice from the given list is " <em>>20 lb "</em> .

3 0

3 years ago

An object of mass kg is released from rest m above the ground and allowed to fall under the influence of gravity. Assuming the f

IgorLugansk [536]

Answer:

Explanation:

From, the given information: we are not given any value for the mass, the proportionality constant and the distance

Assuming that:

the mass = 5 kg and the proportionality constant = 50 kg

the distance of the mass above the ground x(t) = 1000 m

Let's recall that:

$v(t) = \dfrac{mg}{b}+ (v_o - \dfrac{mg}{b})^e^{-bt/m}$

Similarly, The equation of mption:

$x(t) = \dfrac{mg}{b}t+\dfrac{m}{b} (v_o - \dfrac{mg}{b}) (1-e^{-bt/m})$

replacing our assumed values:

where $v_=0 \ and \ g= 9.81$

$x(t) = \dfrac{5 \times 9.81}{50}t+\dfrac{5}{50} (0 - \dfrac{(5)(9.81)}{50}) (1-e^{-(50)t/5})$

$x(t) = 0.981t+0.1 (0 - 0.981) (1-e^{-(10)t}) \ m$

$\mathbf{x(t) = 0.981t-0.981(1-e^{-(10)t}) \ m}$

So, when the object hits the ground when x(t) = 1000

Then from above derived equation:

$\mathbf{x(t) = 0.981t-0.981(1-e^{-(10)t}) \ m}$

$1000= 0.981t-0.981(1-e^{-(10)t}) \ m$

By diregarding $e^{-(10)t} \ m$

$1000= 0.981t-0.981$

1000 + 0.981 = 0.981 t

1000.981 = 0.981 t

t = 1000.981/0.981

t = 1020.36 sec

7 0

3 years ago

A physics major is working to pay her college tuition by performing in a traveling carnival. She rides a motorcycle inside a hol

il63 [147K]

Answer:

v = 12.1 m/s

Explanation:

When at the top of the circle, there are two forces acting on the combined mass of the rider and the motorcycle.
These are the force of gravity (downward) and the normal force, which is directed from the surface away from it, perpendicular to the surface.
In this case, as the motorcycle runs in the interior of the circle, at the top point this force is completely vertical, and is also downward.
Since the motorcycle is moving in a vertical circle, there must be a force, keeping the object moving around a circle.
This force is the centripetal force, aims towards the center of the circle, and is just the net force aiming in this direction at any point.
At the top point, this force is just the sum of the normal force and the weight of the mass of the rider and the motorcycle combined, as follows (we take the direction towards the center as positive):

$F_{c} = N + m*g (1)$

Now, we know that the centripetal force is related with the tangential speed at this point and the radius of the circle as follows:

$F_{c} = m*\frac{v^{2}}{r} (2)$

Since the normal force takes any value as needed to make (1) equal to (2), if the speed diminishes, it will be needed less force to keep the equality valid.
In the limit, when the motorcyvle tires barely touch the surface, this normal force becomes zero.
In this condition, from (1) and (2), we can find the minimum possible value of the speed that still keeps the motorcycle touching the surface, as follows:
$v_{min} =\sqrt{r*g} =\sqrt{15.0m*9.8m/s2} = 12.1 m/s (3)$

6 0

3 years ago