A volleyball that has an initial momentum of

Suppose you are an astronaut on a spacewalk, far from any source of gravity. You find yourself floating alongside your spacecraf

erik [133]

Answer:

c. Throw the items away from the spaceship.

Explanation:

By the Principle of action and reaction yu can get back to your spacecraft throwing the items away from the spaceship.

7 0

4 years ago

Read 2 more answers

A velocity selector is built with a magnetic field of magnitude 5.2 T and an electric field of strength 4.6 × 10 ^ 4 N/C. The di

nydimaria [60]

Explanation :

Magnetic field, B = 5.2 T

Electric field, $E=4.6\times 10^4\ N/C$

The directions of the two fields are perpendicular to each other. Hence the force due to each field will equate each other.

Electrostatic force, F =qE.........(1)

Magnetic force, F = qvB........(2)

From equation (1) and (2)

E = v B

$v=\dfrac{E}{B}$

$v=\dfrac{4.6\times 10^4\ N/C}{5.2\ T}$

$v=0.88\times 10^4\ m/s$

$v=8.8\times 10^3\ m/s$

Hence, the correct option is (a).

(2) A permanent magnet always has two poles as the North pole and south pole. The magnetic field lines start from north poles and terminate at the south pole of the magnet.

Hence, the correct option is (C).

5 0

3 years ago

PERIOD OF THE LEG The period of the leg can be approximated by treating the leg as a physical pendulum, with a period of Equatio

Lesechka [4]

Answer:

Explanation:

We can see from the question that

$T 2\pi\sqrt{\frac{I}{mgh} }$

and $I = \frac{ml^2}{3}$

$m = 16$ % of $67kg$

=> $m =10kg$

from the question $l =48$ % of $1.83m$

Substituting this into the equation

$I = \frac{10.72 * (0.8784)^2}{3}$

=> $I = 2.7571 \ kg m^2$

$h = 0.5 * L$

$h = 0.5 * 0.8784$

$h = 0.4392m$

From the equation above

$T = 2 \pi \sqrt{\frac{2.7571}{10 .72 * 9.81 * 0.4392} }$

$T = 1,534\ sec$

4 0

3 years ago

If light of wavelength 700 nm strikes such a photocathode, what will be the maximum kinetic energy, in eV , of the emitted elect

Oksana_A [137]

If the light of wavelength 700 nm strikes such a photocathode the maximum kinetic energy, in eV, of the emitted electrons is 0.558 eV.

so - $KE_{max} = hc/lembda} work

threshold when KE = 0

hc/lambda = work = 1240/900=1.38 eV

b) Kemax = hc/lambda - work = 1240/640 -1.38=0.558 eV

What is photocathode?

A photocathode electrolyte interface can be used in a photoelectrolysis cell as the primary light-harvesting junction (in conjunction with an appropriate electrochemical anode) or as an optically complementary photoactive half-cell in a tandem photoelectrode photoelectrolysis cell (Hamnett, 1982; Kocha et al, 1994).
In the case of the former, the electrode should ideally harvest photon energy across the majority of the solar spectrum in order to achieve the highest energy conversion efficiency possible.
In the latter case, however, the photocathode may only be active in a specific band of the solar spectrum in order to generate a cathodic photocurrent sufficient to match the current generated in the photoanodic half-cell.

To learn more about Photocathode from the given link:

brainly.com/question/9861585

#SPJ4

3 0

2 years ago

A cell supplies current of 0.6A and 0.2A through 1ohms and 4.0ohms resistor respectively. Calculate the internal resistance of t

Vlad [161]

<h2>Answer:</h2>

0.5Ω

<h2>Explanation:</h2>

Since different currents are passing through the resistors, then the resistors are most probably connected in parallel. This also means that the same voltage will pass across them.

Using Ohm's law, the voltage across a resistor in a circuit is given by;

V = I(R + r) -----------(i)

<em>For the 1ohm resistor, the voltage across it is given by;</em>

<em>Where;</em>

I = current passing through the 1 ohm resistor = 0.6A

R = resistance of the 1 ohm resistor = 1Ω

r = internal resistance of the cell = r

Substitute these values into equation (i) as follows;

V = 0.6(1 + r) -------------------(ii)

<em>For the 4.0ohm resistor, the voltage across it is given by;</em>

<em>Where;</em>

I = current passing through the 4.0 ohms resistor = 0.2A

R = resistance of the 4.0 ohms resistor = 4.0Ω

r = internal resistance of the cell = r

Substitute these values into equation (i) as follows;

V = 0.2(4.0 + r) -------------------(iii)

<em>Now solve equations (ii) and (iii) simultaneously;</em>

V = 0.6(1 + r)

V = 0.2(4.0 + r)

Substitute the value of V in equation (ii) into equation (iii). Therefore, we have;

0.6(1 + r) = 0.2(4.0 + r)

<em>Solve for r</em>

0.6 + 0.6r = 0.8 + 0.2r

0.6r - 0.2r = 0.8 - 0.6

0.4r = 0.2

r = $\frac{0.2}{0.4}$

r = 0.5

Therefore, the internal resistance of the cell is 0.5Ω

4 0

3 years ago