Answer:
a. The spheres will attract each other.
Explanation:
When two conducting spheres are connected by a conducting wire and a negatively charged rod is brought near it then this will induce opposite (positive) charge at the nearest point on the sphere and by the conservation of charges there will also be equal amount of negative charge on the farthest end of this conducting system this is called induced polarization.
- When the conducting wire which joins them is cut while the charged rod is still in proximity to of one of the metallic sphere then there will be physical separation of the two equal and unlike charges on the spheres which will not get any path to flow back and neutralize.
- Hence the two spheres will experience some amount of electrostatic force between them.
The heat flows into my body and I start with a warming feeling going down my throat as it starts there I get this warm feeling and I break out in a coldsweat from consuming a warm drink a cold drink will help cool me down :
Explanation:
If drink warm will make u warm drinking cool will make u cool down faster
The greatest height the ball will attain is 3.27 m
<h3>Data obtained from the question</h3>
- Initial velocity (u) = 8 m/s
- Final velocity (v) = 0 m/s (at maximum height)
- Acceleration due to gravity (g) = 9.8 m/s²
The maximum height to which the ball can attain can be obtained as follow:
v² = u² – 2gh (since the ball is going against gravity)
0² = 8² – (2 × 9.8 × h)
0 = 64 – 19.6h
Collect like terms
0 – 64 = –19.6h
–64 = –19.6h
Divide both side by –19.6
h = –64 / –19.6h
h = 3.27 m
Thus, the greatest height the ball can attain is 3.27 m
Learn more about motion under gravity:
brainly.com/question/13914606
Respuesta:
0,0560 cal / gºC.
Explicación:
Cantidad de calor; (Q)
Q = mcΔt; Δt = t2 - t1
m = masa, c = capacidad calorífica específica; Δt = cambio de temperatura
c de agua = 1 cal / gºC
c de aluminio = 0,22 cal / gºC
QTotal = Q de agua + Q de aluminio
Q de agua = 450 * 1 * (26 - 23) = 1350 cal
Q de aluminio = 60 * 0.22 * (26 - 23) = 39.6 cal
QTotal = 1350 + 39,6 = 1389,6 cal
Calor perdido = calor ganado
QTotal = calor perdido
- 1389,6 = 335,2 * c * (26 - 100)
-1389,6 = −24804,8 * c
c = 1389,6 / 24804,8
c = 0,056021 cal / gºC.
Capacidad calorífica específica de la plata = 0,0560 cal / gºC.
Answer:
a-
V= IR
9V = I ×( 12+6)
I = 9/ 18 A = 0.5 A
b
V=IR
240 = 6 A ×( 20 + R)
40 = 20 + R
R = 20 ohm
c
resultant resistance of the 2 parallel resistances= Ro
1/Ro = 1/ 5 + 1/ 20
1/Ro =( 20+5)/100
= 1/Ro = 1/4
Ro= 4 ohm
V=IR
V = 2A × ( 1+ 4 OHM)
V = 10V
d
equivalent resistance = Ro
1/Ro = 1/(2+8) + 1/(5+5)
1/Ro = 1/10 +1/10
2/10 = 1/ Ro
Ro= 10/2 = 5 ohm
V = IR
12V = I × 5Ohm
I=2.4 A
