A 0.12-kg metal rod carrying a current of current 4.1 A glides on two horizontal rails separation 6.3 m apart. If the coefficien
1 answer:
Answer:
The magnetic field is
Explanation:
From the question we are told that
The mass of the metal rod is
The current on the rod is
The distance of separation(equivalent to length of the rod ) is
The coefficient of kinetic friction is
The kinetic frictional force is
The constant speed is
Generally the magnetic force on the rod is mathematically represented as
For the rod to move with a constant velocity the magnetic force must be equal to the kinetic frictional force so
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Answer:
2.09 m/s
Explanation:
As the spring is stretched initially , and the mass released from rest i.e v=0. Also, The next time the speed becomes zero again is when the spring is fully compressed, and the mass is on the opposite side of the spring with respect to its equilibrium position, after a time t=0.600 s. This illustrates half oscillation of the system.
Therefore, for the period of a full oscillation of the system
T= 2t => 2(0.6)=> 1.2 s
As the frequency is the reciprocal of the period, we have
f= 1/T => 1/1.2
f= 0.833 Hz
The angular frequency'ω' is given by,
ω= 2πf => 2π x 0.833=>5.23 rad/s
For the maximum velocity of the object in a spring-mass system:
V= Aω
where A is the amplitude of the oscillation. As here, the amplitude of the motion corresponds to the initial displacement of the object (A=0.400 m)
V= 0.4 x 5.23 =>2.09 m/s
Answer:
20 K
Explanation:
It is given that,
The change in temperature is 20 C.
We need to find the change in thermodynamic temperature.
If teperauture T₁ = 0° C = 0+273 = 273 K
T₂ = 20° C = 20 + 273 = 293 K
The change in temperature,
So, the change in temperature of 20°C is equivalent to 20 K.
Complete Question
Suppose a NASCAR race car rounds one end of the Martinsville Speedway. This end of the track is a turn with a radius of approximately 57.0 m . If the track is completely flat and the race car is traveling at a constant 30.5 m/s (about 68 mph ) around the turn.
Required:
a. What is the race car's centripetal (radial) acceleration?
b. What is the force responsible for the centripetal acceleration in this case?
O normal
O gravity
O friction
O weight
Answer:
question a
question b
correct option is option 3
Explanation:
From the question we are told that
The radius is
The constant speed at which the race car is travelling is
Generally from the question we are told that the track is completely flat so the only force pulling the car to the middle is the frictional force hence the centripetal force is due to the frictional force
Generally the centripetal acceleration is mathematically represented as
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