Ask question

Ask question

All categories

3 years ago

10

In an orbiting space shuttle, you are handed two identical boxes, one filled with sand and the other filled with feathers. How c

an you determine which is which without opening the boxes?

1 answer:

Aleks04 [339]3 years ago

3 0

Explanation:

Simply, shake the boxes. The box that is difficult to shake will be the sand and the box that will be easier will be the feathers.

You might be interested in

A small truck has a mass of 2145 kg. How much work is required to decrease the speed of the vehicle from 25.0 m/s to 12.0 m/s on

MAXImum [283]

Answer:

The work required is -515,872.5 J

Explanation:

Work is defined in physics as the force that is applied to a body to move it from one point to another.

The total work W done on an object to move from one position A to another B is equal to the change in the kinetic energy of the object. That is, work is also defined as the change in the kinetic energy of an object.

Kinetic energy (Ec) depends on the mass and speed of the body. This energy is calculated by the expression:

$Ec=\frac{1}{2} *m*v^{2}$

where kinetic energy is measured in Joules (J), mass in kilograms (kg), and velocity in meters per second (m/s).

The work (W) of this force is equal to the difference between the final value and the initial value of the kinetic energy of the particle:

$W=\frac{1}{2} *m*v2^{2}-\frac{1}{2} *m*v1^{2}$

$W=\frac{1}{2} *m*(v2^{2}-v1^{2})$

In this case:

W=?
m= 2,145 kg
v2= 12 $\frac{m}{s}$

v1= 25 $\frac{m}{s}$

Replacing:

$W=\frac{1}{2} *2145 kg*((12\frac{m}{s} )^{2}-(25\frac{m}{s} )^{2})$

W= -515,872.5 J

<u><em>The work required is -515,872.5 J</em></u>

3 0

3 years ago

When objects are heated they tend to expand because

densk [106]

Hello!
===
When objects are heated, their molecules tend to vibrate fast. As they vibrate, the space between each atom increases. This keeps on happening, and the object expands until it has cooled down.
===
Hope this helps! :)

5 0

3 years ago

Tennis balls traveling at greater than 100 mph routinely bounce off tennis rackets. At some sufficiently high speed, however, th

Kipish [7]

Answer:

Probability of tunneling is $10^{- 1.17\times 10^{32}}$

Solution:

As per the question:

Velocity of the tennis ball, v = 120 mph = 54 m/s

Mass of the tennis ball, m = 100 g = 0.1 kg

Thickness of the tennis ball, t = 2.0 mm = $2.0\times 10^{- 3}\ m$

Max velocity of the tennis ball, $v_{m} = 200\ mph$ = 89 m/s

Now,

The maximum kinetic energy of the tennis ball is given by:

$KE = \frac{1}{2}mv_{m}^{2} = \frac{1}{2}\times 0.1\times 89^{2} = 396.05\ J$

Kinetic energy of the tennis ball, KE' = $\frac{1}{2}mv^{2} = 0.5\times 0.1\times 54^{2} = 154.8\ m/s$

Now, the distance the ball can penetrate to is given by:

$\eta = \frac{\bar{h}}{\sqrt{2m(KE - KE')}}$

$\bar{h} = \frac{h}{2\pi} = \frac{6.626\times 10^{- 34}}{2\pi} = 1.0545\times 10^{- 34}\ Js$

Thus

$\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}$

$\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}$

$\eta = 1.52\times 10^{-35}\ m$

Now,

We can calculate the tunneling probability as:

$P(t) = e^{\frac{- 2t}{\eta}}$

$P(t) = e^{\frac{- 2\times 2.0\times 10^{- 3}}{1.52\times 10^{-35}}} = e^{-2.63\times 10^{32}}$

$P(t) = e^{-2.63\times 10^{32}}$

Taking log on both the sides:

$logP(t) = -2.63\times 10^{32} loge$

$P(t) = 10^{- 1.17\times 10^{32}}$

6 0

3 years ago

To measure the strength of an earthquake, you can use either a _____ scale or _____ scale.

allsm [11]

To measure the strength of an earthquake, you can use either a Richter scale or Mercalli scale. Richter scale uses the amplitued of the wave and the distance from the source. Mercalli scale uses observations of people and is not considered to be scientific as Richter scale.

8 0

4 years ago

Read 2 more answers

The magnetic field inside a superconducting solenoid is 4.00 T. The solenoid has an inner diameter of 6.20 cm and a length of 26

Delvig [45]

Answer:

(a) The magnetic energy density in the field is 6.366 J/m³

(b) The energy stored in the magnetic field within the solenoid is 5 kJ

Explanation:

magnitude of magnetic field inside solenoid, B = 4 T

inner diameter of solenoid, d = 6.2 cm

inner radius of the solenoid, r = 3.1 cm = 0.031 m

length of solenoid, L = 26 cm = 0.26 m

(a) The magnetic energy density in the field is given by;

$u _B = \frac{B^2}{2\mu_o} \\\\u _B = \frac{(4)^2}{2(4\pi*10^{-7})}\\\\u_B = 6.366*10^6 \ J/m^3$

(b) The energy stored in the magnetic field within the solenoid

$U_B = u_B V\\\\U_B = u_B AL$

$U_B = u_B(A)(L)\\\\U_B = 6.366*10^6(\pi * 0.031^2)(0.26) \\\\U_B = 4997.69 J\\\\U_B = 5 \ KJ\\$

6 0

3 years ago