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Akimi4
2 days ago
14

A pressure cooker is a pot whose lid can be tightly sealed to prevent gas from entering or escaping. even without knowing how bi

g the pressure cooker is, or what altitude it is being used at, we can make predictions about how much force the lid will experience under different conditions.
Physics
1 answer:
Sav [1.1K]2 days ago
5 0
Even if we lack details about the size of the pressure cooker or the altitude of its operation, we can reliably assess the force on the lid based on prior knowledge because, similar to boiling water, the pressure buildup inside the cooker increases in line with the rising temperature. 
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Two resistors ( 3 ohms & 6 ohms) in a series circuit with a power supply = 12 volts. The current through resistor 6 ohms is
Ostrovityanka [942]

In a series circuit...

-- The overall resistance equals the sum of the individual resistances.

-- The current remains identical throughout the circuit.

The total resistance in this circuit is (3Ω + 6Ω )  =  9Ω

<pThe current at every point measures (V/R) = (12v / 9Ω ) = 1.33 A.

Select choice (a).

6 0
6 days ago
Use the terms "force", "weight", "mass", and "inertia" to explain why it is easier to tackle a 220 lb football player than a 288
ValentinkaMS [1149]
<span>Answer
A person who weighs 220 lb has less mass than someone who weighs 288 lb, so accelerating the 220 lb player requires less force. The heavier player therefore carries greater momentum. Because 288 lb corresponds to more weight (and mass), that player has higher inertia and is harder to stop. For these reasons it is easier to tackle a 220 lb player than a 288 lb player. 
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7 0
14 days ago
Read 2 more answers
It takes 56.5 kilojoules of energy to raise the temperature of 150 milliliters of water from 5°C to 95°C. If you
ValentinkaMS [1149]
You would gain an additional 40/60 of energy, which equals 2/3. To find the actual energy consumption, multiply 5/3 by the needed energy.
7 0
5 days ago
A ball is dropped from the top of a cliff. By the time it reaches the ground, all the energy in its gravitational potential ener
ValentinkaMS [1149]

The ball was released from a height of 20 meters

Explanation:

The scenario is as follows:

1. A ball drops from the edge of a cliff.

2. Upon reaching the ground, the energy held in its gravitational potential energy transforms entirely into kinetic energy.

   This implies K.E = P.E.

3. The ball impacts the ground at a speed of 20 m/s.

4. The gravitational field strength noted is 10 N/kg.

<pOur goal is to ascertain the height from which the ball was dropped.

<pSince the ball was dropped from a cliff, its initial velocity is 0.<p→ K.E = \frac{1}{2}m(v^{2}-v_{0}^{2})

where v is the final velocity, v_{0} is the initial velocity, and m is the mass.

<p→ v = 20 m/s and v_{0} = 0 m/s.<p→ K.E = \frac{1}{2}m(20^{2}-0^{2})

→ K.E = \frac{1}{2}m(400)

→ K.E = 200 m joules when the ball strikes the ground.

<p→ P.E = mg h

where g is the gravitational field strength, m is mass, and h signifies height.

<p→ g = 10 N/kg.<p→ P.E = m(10)(h)

→ P.E = 10m h joules.

<p→ P.E = K.E.

→ 10m h = 200 m.

Dividing through by 10m yields:

→ h = 20 meters.

The ball was released from a height of 20 meters.

Learn more

To understand more about gravitational potential energy, visit

8 0
4 days ago
A closed system of mass 10 kg undergoes a process during which there is energy transfer by work from the system of 0.147 kJ per
Softa [913]

Result: -50.005 kJ

Details:

Provided Data

mass of the system = 10 kg

work done = 0.147 kJ/kg

Elevation change (\Delta h)=-50 m

initial speed (v_1)=15 m/s

Final Speed (v_2)=30 m/s

Specific internal Energy (\Delta U)=-5 kJ/kg

according to the first Law of thermodynamics

Q-W=\Delta H

\Delta H=\Delta KE+ \Delta PE +\Delta U

where KE represents kinetic energy

PE indicates potential energy

U denotes internal Energy

\Delta H=\frac{m(v_2^2-v_1^2)}{2}+mg(\Delta h)+\Delta U

Q=W+\Delta KE+ \Delta PE +\Delta U

Q=0.147\times 10+\frac{10\cdot (30^2-15^2)}{2}+10\cdot 9.81\cdot (-5)-5\times 10

Q = 1.47 + 3.375 - 4.850 - 50

Q = -50.005 kJ

5 0
13 days ago
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