3. A 0.35 kg puck slides across the ice with an average force of friction of 0.15 N acting on it. It slides 82 m before coming t
1 answer:
Answer:
Work done is 12.3 J
Explanation:
We have,
Mass of puck, m = 0.35 kg
Force of friction acting on the puck when it slides is 0.15 N
Distance travelled by the puck is 82 m.
It is required to find the work done on the puck. Finally the puck comes to rest and the force of friction is acting on it. It means the applied force is 0.15 N. Work done is given by
The work done on the puck is 12.3 J.
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The induced electric field inside the cylindrical region is directly proportional to r
What is the electric field?
An electric field is the physical field that surrounds electrically charged particles and exerts a force on all other charged particles in the field.
- When a loop of wire is placed in a time-varying magnetic field, an induced electric field arises.
- This induced electric field allows the flow of current in the loop. There is no potential associated with this induced electric field.
Here. a magnetic field with magnitude linearly varies with time.
Radial distance = r
Area of this region, A = π*r^2
Let the induced electric field be E.
Magnetic flux:
∅(B) = B*A
Let the magnetic field be kt,
where k is some constant and t is time.
(B) = kt * (πr^2)
Using Maxwell's equation of electromagnetism,
Hence
E ∝ r
The induced electric field is directly proportional to r
So, option 2 is correct.
Learn more about Maxwell's equation here
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Answer:
La temperatura final es de aproximadamente 159,94°C
Explanation:
Los parámetros dados son;
La masa de la llave española de acero, m₁ = 200 gramos
La temperatura de la llave, T₁ = 550 ° C
La masa del recipiente de aluminio que contiene agua, m₂ = 250 gramos
La masa del agua en el recipiente de aluminio, m₃ = 220 gramos
La capacidad calorífica específica del hierro, , c₁ = 0.499 ca/(g·°C)
La capacidad calorífica específica del aluminio, , c₂ = 0.217 cal/(g·°C)
La capacidad calorífica específica del agua, , c₃= 1 cal/(g·°C)
En equilibrio térmico, tenemos;
m₁·c₁·(T₁ - T) = m₂·c₂·(T -T₂) + m₃·c₃·(T - T₂)
Conectando los valores, da;
200 × 0.499 × (550 - T) = 250 × 0.217 × (T -18) + 220 × 1 × (T - 18)
Simplificando, usando una calculadora gráfica, obtenemos;
De también encontramos 'T' al convertirlo en el tema de la ecuación anterior aún usando una calculadora gráfica;
T = 1196530/7481 °C ≈ 159.94°C
La temperatura final,T ≈ 159.94°C.