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3 years ago

Which of the following is a luminous object? the Moon Jupiter a comet the Sun

2 answers:

6 0

The Sun is the only luminous object on that list ... it has its own light.

Jupiter, the Moon, and comets are only visible when the Sun shines on them.

babunello [35]3 years ago

3 0

The Sun is a luminous object.

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Bonus: (It's not that hard, you just have to pay attention to units.) The Saturn V rocket first stage

agasfer [191]

$v = 2.45×10^3\:\text{m/s}$

Explanation:

Newton's 2nd Law can be expressed in terms of the object's momentum, in this case the expelled exhaust gases, as

$F = \dfrac{d{p}}{d{t}}$ (1)

Assuming that the velocity remains constant then

$F = \dfrac{d}{dt}(mv) = v\dfrac{dm}{dt}$

Solving for $v,$ we get

$v = \dfrac{F}{\left(\frac{dm}{dt}\right)}\;\;\;\;\;\;\;(2)$

Before we plug in the given values, we need to convert them first to their appropriate units:

The thrust F is

$F = 7.5×10^6\:\text{lbs}×\dfrac{4.45\:\text{N}}{1\:\text{lb}} = 3.34×10^7\:\text{N}$

The exhaust rate dm/dt is

$\dfrac{dm}{dt} = 15\dfrac{T}{s}×\dfrac{2000\:\text{lbs}}{1\:\text{T}}×\dfrac{1\:\text{kg}}{2.2\:\text{lbs}}$

$\;\;\;\;\;= 1.36×10^4\:\text{kg/s}$

Therefore, the velocity at which the exhaust gases exit the engines is

$v = \dfrac{F}{\left(\frac{dm}{dt}\right)} = \dfrac{3.34×10^7\:\text{N}}{1.36×10^4\:\text{kg/s}}$

$\;\;\;= 2.45×10^3\:\text{m/s}$

6 0

1 year ago

A hockey puck with a mass of 0.175 kg slides over the ice. The puck initially slides with a speed of 5.25 m/s, but it comes to a

Neko [114]

Answer:

1.70 J

Explanation:

The heat dissipated is the difference in the kinetic energies.

This is given by

$E = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$

$v_i$ and $v_f$ are the initial and final velocities.

With m = 0.175 kg,

$E = \frac{1}{2}\times0.175(2.85^2 - 5.25^2) = -1.701\text{ J}$

The negative sign appears because energy is lost.

7 0

2 years ago

You need to pull up the cart up this ramp. How would you change this ramp so it is easier to pull the cart up?

natali 33 [55]

I would make the ramp flatter. In doing so the ramp would have to be longer.

6 0

3 years ago

PE=30J, m=?, g=10m/s2, h=10m

OleMash [197]

Based on the given, this is probably a gravitational potential energy problem (PEgrav). The formula for PEgrav is:

PEgrav = mgh

Where:
m = mass (kg)
g = acceleration due to gravity
h = height (m)

With this formula you can derive the formula for your unknown, which is mass. First put in what you know and then solve for what you do not know.

$PEgrav=mgh$
$30J=m(10)(10[tex] \frac{30}{100} =m$ )[/tex]

Do operations that you can with what is given first.

$30J=m(100m)$

Transpose the 100 to the other side of the equation. Do not forget that when you transpose, you do the opposite operation.

$\frac{30}{100} =m$

m = 0.30kg

5 0

2 years ago

A 0.300 kg block is pressed against a spring with a spring constant of 8050 N/m until the spring is compressed by 6.00 cm. When

natita [175]

Answer:

a) $\mu_{k} = 0.704$ , b) $R = 0.312\,m$