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3 years ago

Jax went to the vet and the scale reads 9 pounds. what does this mean?

Physics

1 answer:

makkiz [27]3 years ago

8 0

if jax is an animal it means it is under weight. and if its a human there severly anorexic

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A machine is rolling a metal cylinder under pressure. The radius of the cylinder is decreasing at a constant rate of 0.05 inches

Luden [163]

Yes, the volume of the cylinder will remain constant. As the radius decreases, the height will increase to make sure that the volume is kept the same.
We have been given a value of dr/dt and are required to find dh/dt
Because the volume is constant, we can plug it into the formula for the volume of the cylinder and rearrange it to make h the subject:
128 = πr²h
h = 128/πr²
Now we differentiate both sides:
dh/dr = -256/πr³
Applying the chain rule:
dh/dt = dh/dr x dr/dt
dh/dt = (-256/πr³) x -0.05
dh/dt = 64/5πr³; substituting the value of r
dh/dt = 64/5π(1.5)³
dh/dt = 1.21 in/sec

4 0

4 years ago

A locomotive approaches its next stop and accelerates at -0.12 m/s^2, coming to a complete stop in 30 seconds. This motion could

Masja [62]

Answer:

Answer: positive velocity & negative acceleration

Explanation:

Accelerated Motion

Both the velocity and acceleration are vectors because they have magnitude and direction. When the motion is restricted to one dimension, i.e. left-right or up-down, the direction is marked with the sign according to some preset reference.

The locomotive is moving at a certain speed with a (so far) unknown sign but the acceleration has a negative sign. Since the locomotive comes to a complete stop it means the velocity and the acceleration are of opposite signs.

Thus the velocity is positive.

Answer: positive velocity & negative acceleration

4 0

3 years ago

A soccer player kicks a soccer ball with a force of 1.8 N. If the mass of the ball is .43 kg. How fast will the ball accelerate?

Vesnalui [34]

Answer:

The acceleration of the ball is 4.18 [m/s^2]

Explanation:

By Newton's second law we can find the acceleration of the ball

$F = m*a\\where:\\F = force applied [N] or [kg*m/s^2]\\m = mass of the ball [kg]\\a = acceleration [m/s^s]$

Now we have:

$a = F/m\\a = \frac{1.8 [kg*m/s^s]}{0.43[kg]} \\a = 4.18 [kg]$

4 0

3 years ago

Read 2 more answers

Compare the circular velocity of a particle orbiting in the Encke Division, whose distance from Saturn 133,370 km, to a particle

Ket [755]

Answer:

The particle in the D ring is 1399 times faster than the particle in the Encke Division.

Explanation:

The circular velocity is define as:

$v = \frac{2 \pi r}{T}$

Where r is the radius of the trajectory and T is the orbital period

To determine the circular velocity of both particles it is necessary to know the orbital period of each one. That can be done by means of the Kepler’s third law:

$T^{2} = r^{3}$

Where T is orbital period and r is the radius of the trajectory.

Case for the particle in the Encke Division:

$T^{2} = r^{3}$

$T = \sqrt{(133370 Km)^{3}}$

$T = \sqrt{(2.372x10^{15} Km)}$

$T = 4.870x10^{7} Km$

It is necessary to pass from kilometers to astronomical unit (AU), where 1 AU is equivalent to 150.000.000 Km ( $1.50x10^{8} Km$ )

1 AU is defined as the distance between the earth and the sun.

$\frac{4.870x10^{7} Km}{1.50x10^{8}Km} . 1AU$

$T = 0.324 AU$

But 1 year is equivalent to 1 AU according with Kepler’s third law, since 1 year is the orbital period of the earth.

$T = \frac{0.324 AU}{1 AU} . 1 year$

$T = 0.324 year$

That can be expressed in units of days

$T = \frac{0.324 year}{1 year} . 365.25 days$

$T = 118.60 days$

Circular velocity for the particle in the Encke Division:

$v = \frac{2 \pi r}{T}$

$v = \frac{2 \pi (133370 Km)}{(118.60 days)}$

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

$118.60 days .\frac{86400 s}{1 day}$ ⇒ 10247040 s

$133370 Km .\frac{1000 m}{1 Km}$ ⇒ 133370000 m

$v = \frac{2 \pi (133370000 m)}{(10247040 s)}$

$v = 81.778 m/s$

Case for the particle in the D Ring:

For the case of the particle in the D Ring, the same approach used above can be followed

$T^{2} = r^{3}$

$T = \sqrt{(69000 Km)^{3}}$

$T = \sqrt{(3.285x10^{14} Km)}$

$T = 1.812x10^{7} Km$

$\frac{1.812x10^{7} Km}{1.50x10^{8}Km} . 1 AU$

$T = 0.120 AU$

$T = \frac{0.120 AU}{1 AU} . 1 year$

$T = 0.120 year$

$T = \frac{0.120 year}{1 year} . 365.25 days$

$T = 43.83 days$

Circular velocity for the particle in D Ring:

$v = \frac{2 \pi r}{T}$

$v = \frac{2 \pi (69000 Km)}{(43.83 days)}$

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

$43.83 days . \frac{86400 s}{1 day}$ ⇒ 3786912 s

$69000 Km . \frac{1000 m}{ 1 Km}$ ⇒ 69000000 m

$v = \frac{2 \pi (69000000 m)}{(3786912 s)}$

$v = 114.483 m/s$

$\frac{114.483 m/s}{81.778 m/s} = 1.399$

The particle in the D ring is 1399 times faster than the particle in the Encke Division.

7 0

3 years ago

1. In the laboratory, the life time of a particle moving with speed 2.8 x 10^10 cm\s is found to be 2.5 x 10^-7.

timofeeve [1]

Answer: n the laboratory, the life time of a particle moving with speed 2.8 x 10^10 cm\s is found to be 2.5 x 10^-7. Calculate the proper life of the ...

Explanation:

5 0

3 years ago