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irga5000 [103]
1 year ago
15

You drop a 30 g pebble down a well. You hear a splash 2.7 s later. What was the pebble’s speed just as it reaches the water? Ass

ume g = 9.8 m/s2
Physics
1 answer:
Flauer [41]1 year ago
7 0

vf=vi+at

No need for rearranging because it is already set up for Vf (final velocity)

a= -9.8m/s² (because it is falling)

vi= 0

t= 2.7

0 + -9.8(2.7) = vf

vf = -26.5 (-26.46)

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Large and small numbers are often best entered in exponential notation. For example, the mass of the earth is 5.98x1024 kg and t
Doss [256]

Answer:

The charge to mass ratio is -1.76\times10^{11}

Explanation:

Mass\ of\ electron=m_e=9.11\times10^{-31} kg\\ Charge\ of\ the\ electron=q_e=-1.60\times10^{-19} C\\

We need to find how much charge is contained  in the electron per unit of mass, to do this we divide the charge in an electron and the mass of an electron:

\frac{Charge\ of\ electron}{Mass\ of\ electron}=\frac{q_e}{m_e}\frac{C}{kg}=\frac{-1.60\times10^{-19} C}{9.11\times10^{-31} kg}=-1.76\times10^{11} \frac{C}{kg}

4 0
2 years ago
Read 2 more answers
A piston-cylinder chamber contains 0.1 m3 of 10 kg R-134a in a saturated liquid-vapor mixture state at 10 °C. It is heated at co
vaieri [72.5K]

Answer:

(A) 10132.5Pa

(B)531kJ of energy

Explanation:

This is an isothermal process. Assuming ideal gas behaviour then the relation P1V1 = P2V2 holds.

Given

m = 10kg = 10000g, V1 = 0.1m³, V2 = 1.0m³

P1 = 101325Pa. M = 102.03g/mol

P2 = P1 × V1 /V2 = 101325 × 0.1 / 1 = 10132.5Pa

(B) Energy is transfered by the r134a in the form of thw work done in in expansion

W = nRTIn(V2/V1)

n = m / M = 10000/102.03 = 98.01mols

W = 98.01 × 8.314 × 283 ×ln(1.0/0.1)

= 531kJ.

6 0
2 years ago
Two astronauts, A and B, both with mass of 60Kg, are moving along a straight line in the same direction in a weightless spaceshi
sveta [45]

Answer:

V=14m/s

Explanation:

From the Question we are told that

Mass of A and B is 60kg

Speed of A=2m/s

Speed of B=1m/s

Mass of bag =5kg

Generally the momentum of the astronaut  A and bag is mathematically given as

  M_A=(60+5)*2

   M_A=130kgm/s

Generally to avoid collision the speed of astronaut be should be less than or equal to that of astronaut A with the bag

Therefore for minimum requirement speed of astronaut A should be given by astronaut B's speed which is equal to 1

Therefore

  130=(60*1)=(5*v)

   V=14m/s

7 0
1 year ago
A bug starts at point A, crawls 8.0 cm east, then 5.0 cm south, 3.0 west, and 4.0 cm north to point B.
Sholpan [36]

Answer:

5cm east& 1cm west from A

Explanation:

https://brainly.ph/question/2753392

7 0
1 year ago
A race car driver must average 200km/hr for four laps to qualify for a race. Because of engine trouble, the car averages only 17
vampirchik [111]
The average speed would have to be 260 km/hr due to the driver originally going 30 km/hr too slow the first two laps
5 0
2 years ago
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