Answer:
4
Explanation:
In order for the current to continue flowing through the circuit (and for the bulbs to continue shining), there must be a closed path containing the battery where current can flow. Let's see the effect of removing each bulb on the circuit:
- 1: when removing bulb 1 only, the current can still flow through the path battery-bulb 3- bulb 4
- 2: when removing bulb 2 only, the current can still flow through the path battery-bulb 3- bulb 4
- 3: when removing bulb 3 only, the current can still flow through the path battery-bulb 1-bulb 2- bulb 4
- 4: when removing bulb 4 only, the current can no longer flow. In fact, there is no closed path that contains the battery now, so the current will not flow and all the bulbs will stop shining.
None of the choices is an appropriate response.
There's no such thing as the temperature of a molecule. Temperature and
pressure are both outside-world manifestations of the energy the molecules
have. But on the molecular level, what it is is the kinetic energy with which
they're all scurrying around.
When the fuel/air mixture is compressed during the compression stroke,
the temperature is raised to the flash point of the mixture. The work done
during the compression pumps energy into the molecules, their kinetic
energy increases, and they begin scurrying around fast enough so that
when they collide, they're able to stick together, form a new molecule,
and release some of their kinetic energy in the form of heat.
Answer:
E_Phase = 560V
Explanation:
The computation of the voltage i.e. dropped across each phase is shown below:
Given that
The delta connection line voltage is
E_line = 560 V
And, in the case of delta connection, the line voltage would be equivalent to the phase voltage
That means
E_Phase = E_Line
= 560 V
Hence, the voltage i.e. dropped across each phase is
E_Phase = 560V
Answer:
-35 m/s
Explanation:
Momentum is conserved.
Momentum before firing = momentum after firing
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
Before the bullet is fired, the bullet and rifle have no velocity, so u₁ and u₂ are 0.
0 = m₁v₁ + m₂v₂
Given m₁ = 0.7 kg, v₁ = 350 m/s, and m₂ = 7 kg:
0 = (0.7 kg) (350 m/s) + (7 kg) v
v = -35 m/s
The rifle recoils at 35 m/s in the opposite direction.
950kg. It is W=mg. therefore, mass= W/g which is W/10
=9500/10
=950
