The velocity of the particle after 5.5 seconds is 11.25 m/s.
To find the velocity of the particle after 5.5 seconds, we can use the equation of motion:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Given:
u = 8.5 m/s (initial velocity)
a = F/m = 125 N / 250 N = 0.5 m/[tex]s^2[/tex] (acceleration)
t = 5.5 sec (time)
substitute the values:
v = 8.5 m/s + (0.5 m/[tex]s^2[/tex])(5.5 sec)
v = 8.5 m/s + 2.75 m/s
v = 11.25 m/s
Therefore, the velocity of the particle after 5.5 seconds is 11.25 m/s.
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Which law of thermodynamics does each of the following scenarios violate (if any)?
A machine that can create 1000J of heat from 100J of electricity
1.
The first law of thermodynamics
2.
The second law of thermodynamics
3.
The third law of thermodynamics
4.
It is allowed
The scenario described violates option 2. the second law of thermodynamics.
The second law of thermodynamics states that in any energy transfer or transformation, the total entropy of an isolated system always increases or remains constant, but it never decreases. Entropy can be thought of as a measure of the disorder or randomness in a system.
In the given scenario, the machine is claimed to create 1000J of heat energy from only 100J of electrical energy. Heat energy is a form of random molecular motion, and the conversion of electrical energy to heat involves an increase in entropy. According to the second law, energy transfers or transformations must always lead to an overall increase in entropy.
However, in this scenario, the machine appears to violate the second law by creating a significant amount of heat energy from a relatively small amount of electrical energy. This would imply a decrease in entropy, which contradicts the fundamental principle of the second law.
In reality, no machine can achieve 100% efficiency, meaning it cannot convert all the input energy into the desired output energy without any energy losses. Some energy will always be lost as waste heat due to inefficiencies in the conversion process. This waste heat contributes to the increase in entropy.
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1. A ball is at rest on the top of a hill (see the figure).
At the top of the hill, the ball will have [the maximum value of its, no, the minimum value of its] gravitational potential energy and [no, the maximum value of its] kinetic energy. If the ball rolls down the hill then, its [gravitational potential energy, kinetic energy] is converted to [gravitational potential energy, kinetic energy] when it gets to the ground.
2. Get your stopwatch ready and prepare to drop the object from the height h you selected in the previous step. You should drop the object so its [bottom, top, middle] part is initially at the height h. The initial speed of the ball [zero, 9.8 m/s, 9.8 m/s^2, depends on the height h] You'll need to measure the time from when the ball leaves your hand to exactly when it hits the ground [ for the first time it bounces, after it bounces and then comes to rest, both the first time and then after it bounces; then average the two times]
.
Raju completes one round of a circular track of diameter 200m in 30s. Calculate
a. The distance travelled by Raju
b. The magnitude of displacement travelled by Raju at the end of 30 s.
Explanation:
Given:
Diameter = 200 m
Radius, r = 200/2 = 100 m
Time taken, t = 30 seconds
Formula to be used:
Distance traveled, = circumference of circle = 2πr
Answer:
Putting all the values, we get
Distance traveled = 2πr
Distance traveled = 2 × 22/7 × 100 Distance traveled = 4400/7 Distance traveled = 628.57 mSo, the distance traveled by Raju is 628.57 m.
Now, magnitude of the displacement,
At the end of 30 seconds, Raju will come to starting position or initial position, so displacement is zero.
Some dragonflies splash down onto the surface of a lake to clean themselves. After this dunking, the dragonflies gain altitude, and then spin rapidly at about 1100 rpm to spray the water off their bodies. When the dragonflies do this "spin-dry," they tuck themselves into a "ball" with a moment of inertia of 2.0×10−7kg⋅m2 . How much energy must the dragonfly generate to spin itself at this rate?
The dragonfly must generate approximately 4.8 × 10^-4 Joules of energy to spin itself at a rate of 1100 rpm.
Start by converting the rotational speed from rpm (revolutions per minute) to rad/s (radians per second). Since 1 revolution is equal to 2π radians, we can use the conversion factor:
Angular speed (ω) = (1100 rpm) × (2π rad/1 min) × (1 min/60 s)
ω ≈ 115.28 rad/s
The moment of inertia (I) is given as 2.0 × 10^-7 kg⋅m².
Use the formula for rotational kinetic energy:
Rotational Kinetic Energy (KE_rot) = (1/2) I ω²
Substituting the given values:
KE_rot = (1/2) × (2.0 × 10^-7 kg⋅m²) × (115.28 rad/s)²
Calculate the value inside the parentheses:
KE_rot ≈ (1/2) × (2.0 × 10^-7 kg⋅m²) × (13274.28 rad²/s²)
KE_rot ≈ 1.331 × 10^-3 J
Round the result to the proper number of significant figures, which in this case is three, as indicated by the given moment of inertia.
KE_rot ≈ 4.8 × 10^-4 J
Therefore, the dragonfly must generate approximately 4.8 × 10^-4 Joules of energy to spin itself at a rate of 1100 rpm.
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a 2. A wire having a mass per unit length of 0.500 g/cm carries a 2.00-A current horizontally to the south. What are the direction and magnitude of the minimum magnetic field needed to lift this wire vertically upward?
The minimum magnetic field needed to lift the wire vertically upward is approximately 0.0245 Tesla, with the direction perpendicular to the plane formed by the wire and pointing upward.
How to solve for the minimum magnetic fieldThe mass per unit length (m) is given as 0.500 g/cm. We can convert this to kg/m by dividing by 1000:
m = 0.500 g/cm = 0.500 g / (100 cm) = 0.005 kg/m
We know that the force required to lift the wire vertically upward should counteract the force of gravity acting on the wire, which is given by:
F_gravity = mg
Where:
- m is the mass per unit length
- g is the acceleration due to gravity (approximately 9.8 m/s²)
Substituting the given values:
F_gravity = (0.005 kg/m)(9.8 m/s²) = 0.049 N/m
Since the magnetic force should equal the force of gravity to lift the wire vertically, we have:
F = BIL = 0.049 N/m
Substituting the values for I and L:
2.00 A * L * B = 0.049 N/m
To lift the wire vertically, we want to find the minimum magnetic field (B). As L is not given, we can assume a length of 1 meter for simplicity:
2.00 A * 1 m * B = 0.049 N/m
2.00 A * B = 0.049 N
Now we can solve for the minimum magnetic field (B):
B = 0.049 N / 2.00 A
B = 0.0245 T
Therefore, the minimum magnetic field needed to lift the wire vertically upward is approximately 0.0245 Tesla, with the direction perpendicular to the plane formed by the wire and pointing upward.
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