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Salsk061 [2.6K]
4 years ago
14

Which symbol represents a type of electromagnetic radiation released during radioactive decay?

Physics
1 answer:
Tanya [424]4 years ago
8 0
The answer for <span>electromagnetic radiation released during radioactive decay i</span>s C. He
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A cell of internal resistance 2 ohms supplies current to a 6 ohms resistor. What is the efficiency of the cell?. ​
Margaret [11]

75% is the efficiency of the cell.

Explanation:

Internal resistance is the resistance generated due to the devices present in the circuit. Most preferably, it is the resistance of the battery connected to the circuit.

As the current, resistance of the resistor and the internal resistance of the battery is given as 6 A, 6 ohms and 2 ohms.

Then the efficiency of the cell is E = \frac{IR}{I (R+r)} =\frac{R}{r+R}

So, the efficiency of the cell is E=\frac{R}{r+R}

Then, E = \frac{6}{2+6}=\frac{6}{8}

So, the efficiency is 75%.

75% is the efficiency of the cell.

4 0
3 years ago
During a circus act, an elderly performer thrills the crowd by catching a cannon ball shot at him. The cannon ball has a mass of
Ipatiy [6.2K]

Answer:

m v1 = (m + M) v2

v2 = m v1 / (m + M)

v2 = 7 * 74 / (74 + 65)

3.73 m/s

74 kg is too heavy for the cannonball (over 150 lbs)

4 0
3 years ago
Planet geos in orbit a distance of 1
nadya68 [22]

Planet Geos in orbit a distance of 1 A.U. (astronomical unit) from the star Astra has an orbital period of 1 "year." If planet Logos is 4 A.U. from Astra, how long does Logos require for a complete orbit?

TB = <span>8</span> years

4 0
3 years ago
Read 2 more answers
Which planet formed near the sun where the solar system temperatures were very hugh
Gnom [1K]

mercury is the closest planet from the sun

8 0
3 years ago
Read 2 more answers
Energy of a SpacecraftVery far from earth(at R=\infty), a spacecraft has run out of fuel and its kineticenergy is zero. If only
Firdavs [7]

Answer:

s_e=\sqrt{\frac{2GM_e}{R_e^2}}

Explanation:

In this case mechanical energy is conserved, which means that the sum of the initial kinetic energy and initial potential gravitational energy will be equal to the sum of the final kinetic energy and final potential gravitational energy:

K_i+U_i=K_f+U_f

Which in our case will be:

\frac{mv_i^2}{2}+\frac{-GM_em}{r_i^2}=\frac{mv_f^2}{2}+\frac{-GM_em}{r_f^2}

Which, since v_i=0m/s, r_i=infinity, r_f=R_e, v_f=s_e and canceling <em>m</em> means that:

\frac{s_f^2}{2}=\frac{GM_e}{R_e^2}

Solving for the final velocity we get:

s_e=\sqrt{\frac{2GM_e}{R_e^2}}

6 0
3 years ago
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