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3 years ago

La velocidad de la luz en el vacío es c= 3000.000 km\s la luz del sol tarda en llegar a la tierra 8 minutos y 14 segundos

Physics

1 answer:

ehidna [41]3 years ago

3 0

La velocidad correcta de la luz en el vacío es 300.000 km/s .

La distancia = (velocidad) x (duración de tiempo)

Duración de tiempo = 494 segundos, porque cada minuto = 60 segundos

La distancia = (300.000 km/s) x (494 s)

<em>La distancia = 148.200.000 km</em>

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What is the magnitude of the force, directed parallel to the ramp, that he needs to exert on the crate to get it to start moving

lions [1.4K]

Complete question is;

Jason works for a moving company. A 75 kg wooden crate is sitting on the wooden ramp of his truck; the ramp is angled at 11°.

What is the magnitude of the force, directed parallel to the ramp, that he needs to exert on the crate to get it to start moving UP the ramp?

Answer:

F = 501.5 N

Explanation:

We are given;

Mass of wooden crate; m = 75 kg

Angle of ramp; θ = 11°

Now, for the wooden crate to slide upwards, it means that the force of friction would be acting in an opposite to the slide along the inclined plane. Thus, the force will be given by;

F = mgsin θ + μmg cos θ

From online values, coefficient of friction between wooden surfaces is μ = 0.5

Thus;

F = (75 × 9.81 × sin 11) + (0.5 × 75 × 9.81 × cos 11)

F = 501.5 N

3 0

3 years ago

A straightforward method of finding the density of an object is to measure its mass and then measure its volume by submerging it

Veseljchak [2.6K]

Incomplete question as the unit of volume is not written correctly.So the complete question is here:

A straightforward method of finding the density of an object is to measure its mass and then measure its volume by submerging it in a graduated cylinder. What is the density of a 240-g rock that displaces 89.0 cm³?

Answer:

$d_{Density}=2.7g/cm^{3}$

Explanation:

Given data

Mass m=240g

Volume V=89.0 cm³

To find

Density d

Solution

If rock displaces 89.0 cm³ of water means volume of rock is also 89cm³

$d_{Density}=\frac{mass}{volume}\\d_{Density}=\frac{240g}{89.0cm^{3} } \\d_{Density}=2.7g/cm^{3}$

5 0

3 years ago

A 6.99-g bullet is moving horizontally with a velocity of +341 m/s, where the sign + indicates that it is moving to the right (s

Ratling [72]

Answer:

a). 1.218 m/s

b). R=2.8 $^{-3}$

Explanation:

$m_{bullet}=6.99g*\frac{1kg}{1000g}=6.99x10^{-3}kg$

$v_{bullet}=341\frac{m}{s}$

Momentum of the motion the first part of the motion have a momentum that is:

$P_{1}=m_{bullet}*v_{bullet}$

$P_{1}=6.99x10^{-3}kg*341\frac{m}{s} \\P_{1}=2.3529$

The final momentum is the motion before the action so:

a).

$P_{2}=m_{b1}*v_{fbullet}+(m_{b2}+m_{bullet})*v_{f}}$

$P_{2}=1.202 kg*0.554\frac{m}{s}+(1.523kg+6.99x10^{-3}kg)*v_{f}$

$P_{1}=P_{2}$

$2.529=0.665+(1.5299)*v_{f}\\v_{f}=\frac{1.864}{1.5299}\\v_{f}=1.218 \frac{m}{s}$

b).

kinetic energy

$K=\frac{1}{2}*m*(v)^{2}$

Kinetic energy after

$Ka=\frac{1}{2}*1.202*(0.554)^{2}+\frac{1}{2}*1.523*(1.218)^{2}\\Ka=1.142 J$

Kinetic energy before

$Kb=\frac{1}{2}*mb*(vf)^{2}\\Kb=\frac{1}{2}*6.99x10^{-3}kg*(341)^{2}\\Kb=406.4J$

Ratio = $\frac{Ka}{Kb}$

$R=\frac{1.14}{406.4}\\R=2.8x10^{-3}$

3 0

3 years ago

As the proton approaches the uranium nucleus, the repulsive force slows down the proton until it comes momentarily to rest, afte

Klio2033 [76]

Answer:

$r^2 = \frac{ 2 k \ Ze^2}{ 2m}$

Explanation:

For this exercise we must use the principle of conservation of energy

starting point. The proton very far from the nucleus

Em₀ = K = ½ m v²

final point. The point where the proton is stopped (v = 0)

Em_f = U = q V

where the potential is

V = k Ze / r²

Let us consider that all the charge of the nucleus is in the center, therefore r is the distance from this point to the proton that is approaching

Energy is conserved

Em₀ = Em_f

½ m v² = e ( $k \frac{Ze}{r^2}$ )

$r^2 = \frac{ 2 k \ Ze^2}{ 2m}$

with this expression we can find the closest approach distance (r)

3 0

3 years ago

A 782.10 kg car is brought from 7.60 m/s to 3.61 m/s over a time period of 4.23 seconds. What is the average force the car exper

alexandr402 [8]

Answer:

–735.17 N

The negative sign indicate that the force is acting in opposition direction to the car.

Explanation:

The following data were obtained from the question:

Mass (m) of car = 782.10 kg

Initial velocity (u) = 7.60 m/s

Final velocity (v) = 3.61 m/s

Time (t) = 4.23 s

Force (F) =?

Next, we shall determine the acceleration of the car. This can be obtained as follow:

Initial velocity (u) = 7.60 m/s

Final velocity (v) = 3.61 m/s

Time (t) = 4.23 s

Acceleration (a) =?

a = (v – u) / t

a = (3.61 – 7.60) / 4.23

a = –3.99 / 4.23

a = –0.94 m/s²

Finally, we shall determine the force experienced by the car as shown below:

Mass (m) of car = 782.10 kg

Acceleration (a) = –0.94 m/s²

Force (F) =?

F = ma

F = 782.10 × –0.94

F = –735.17 N

The negative sign indicate that the force is acting in opposition direction to the car.

4 0

3 years ago