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MariettaO [177]
3 years ago
13

How many atoms are in one mole of uranium?

Physics
2 answers:
sammy [17]3 years ago
8 0

Answer:

C. 6. 02 x 10^24

Explanation:

Mashcka [7]3 years ago
6 0

Answer:

A

Explanation:

6.02×10^23 hope this helps

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Optical tweezers use light from a laser to move single atoms and molecules around. Suppose the intensity of light from the tweez
WITCHER [35]

Answer:

Explanation:

a )

If it is totally absorbed pressure is calculated as follows .

Pressure = I / c where I is intensity of light falling .

= 1000 / 3 x 10⁸

= 3.33 x 10⁻⁶ N / m²

b ) weight of tritium atom

=  3 x 1.67 x 10⁻²⁷ kg

acceleration = force / mass

=    3.33x 10⁻⁶ / 3 x 1.67 x 10⁻²⁷

= .6646 x 10²¹ m /s²

= 66.46 x 10¹⁹ m / s²

7 0
3 years ago
A railroad car of mass 2.50*10^4 kg is moving with a speed of 4.00 m/s. It collides and couples with three other coupled railroa
NemiM [27]

Answer:

(a) 2.5 m/s

(b) 37.5 KJ

Explanation:

(a)

From the law of conservation of momentum, Initial momentum=Final momentum

mV_1+3mV_2=(m+3m)V_f=4mV_f

V_1+3V_2=4V_f and making V_f the subject then

V_f=\frac {V_1+3V_2}{4} and since V_1 is initial velocity of car, value given as 4 m/s, V_2 is the initial velocity of the three cars stuck together, value given as 2 m/s and v_f is the final velocity which is unknown. By substitution

V_f=\frac {4+(3\times2)}{4}=2.5 m/s

(b)

Initial kinetic energy is given by

\frac {mV_1^{2}}{2}+\frac {3mV_2^{2}}{2}=\frac {m(V_1^{2}+3V_2^{2}}{2}=\frac {2.5\times 10^{4}(4^{2}+3(2^{2}))}{2}=350\times10^{3} J= 350 KJ

Final kinetic energy is given by

\frac {4mV_f^{2}}{2}=\frac {4\times 2.5\times 10^{4}\times 2.5^{2}}{2}=312.5\times 10^{3} J=312.5 KJ

The energy lost is given by subtracting the final kinetic energy from the initial kinetic energy hence

Energy lost=350-312.5=37.5 KJ

6 0
4 years ago
Read 2 more answers
PLEASE HURRYYY
babymother [125]

your answer is make up artist

4 0
4 years ago
in which type of wave are vibrations at right angles to the direction in which the wave is travelling​
Pani-rosa [81]

Answer:

longitudinal waves have those properties

7 0
3 years ago
. A vertical electric field is set up in space to compensate for the gravitational force on a point charge. What is the required
Delicious77 [7]

(a) The required magnitude of the electric field when the point charge is an electron is 5.57 x 10⁻¹¹ N/C.

(b) The required magnitude of the electric field when the point charge is an proton is 1.02 x 10⁻⁷ N/C.

<h3>Magnitude of electric field </h3>

The magnitude of electric field is given by the following equation.

F = qE

But F = mg

mg = qE

E = mg/q

where;

  • E is the electric field
  • m is mass of the particle
  • g is acceleration due to gravity
  • q is charge of the particle
<h3>For an electron</h3>

E = (9.11 x 10⁻³¹ x 9.8)/(1.602 x 10⁻¹⁹)

E = 5.57 x 10⁻¹¹ N/C

<h3>For proton</h3>

E = (1.67 x 10⁻²⁷ x 9.8)/(1.602 x 10⁻¹⁹)

E = 1.02 x 10⁻⁷ N/C

Thus, the required vertical electric field is greater when the charge is proton.

Learn more about electric field here: brainly.com/question/14372859

#SPJ1

4 0
2 years ago
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