Answers
Answer:
Ohms law
Explanation:
Which states that the current flowing through any cross-section of the conductor is directly proportional to the potential differenceapplied across its end, provided physical conditions like temperature and pressure remain constant.
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Classical mechanics is an extremely well tested model. Hundreds of years worth of experiments, as well as most feats of engineering, have verified its validity. If special relativity gave very different predictions than classical physics in everyday situations, it would be directly contradicted by this mountain of evidence. In this problem, you will see how some of the usual laws of classical mechanics can be obtained from special relativity by simply assuming that the speeds involved are small compared to the speed of light.Two of the most surprising results of special relativity are time dilation and length contraction, namely, that measured intervals in time and space are not absolute quantities but instead appear differently to different observers. The equations for time dilation and length contraction can be written t=?t0 and l=l0/?, where?=11?u2c2?.Part AFind the first two terms of the binomial expansion for ?.Express your answer in terms of u and c.Hints? = 1+12(uc)2 … SubmitMy AnswersGive UpCorrectYou can see that ??1 if u?c, as is the case in most situations. If you set ?=1 in the equations for time dilation and length contraction you recover the equations of classical physics, which state essentially that there is no time dilation or length contraction. Therefore, we don't see any appreciable length contraction or time dilation in everyday life.Part BConsider a case involving a speed that is fast compared to those encountered in our everyday life: a spy plane moving at 1500m/s. Find the deviation from classical physics (??1) that relativity predicts at this speed. Use only the first two terms of the binomial expansion, as your calculator may not be able to handle the necessary number of digits otherwise.Express your answer to four significant figures.??1 = 1.250×10?11SubmitMy AnswersGive UpCorrectIf you lived for 70 years in such a spy plane moving at 1500m/s, this would amount to about 28ms of cumulative time difference between you and people who lived at rest relative to the earth when you finally landed. Thus, it is not surprising that relativistic effects are not observed in everyday life, or even at the fringes of everyday life. By using atomic clocks, which can measure time accurately to one part in 1013 or better, the time dilation at the normal speed for an airliner has been verified.Part CNow, consider the relativistic velocity addition formula:speed=v+u1+vuc2.If v=u=0.01c=1% of c, what is the relativistic sum of the two speeds?Express your answer as a percentage of the speed of light to five significant figures.
A ship is traveling at 154 m/s and accelerates at a rate of 1.80 m/s^2 for 1 minute. What will its speed be after that minute? Calculate the answer in both meters per second and kilometers per hour.
Answers
Answer:
178.75 N
Explanation:
The force necessary to start moving the crate must be equal to or more than the frictional force (resistive force) acting on the crate but moving in an opposite direction to the frictional force.
So, we find the frictional force, Fr:
Fr = -μmg
Where μ = coefficient of friction
m = mass
g = acceleration due to gravity
The frictional force is negative because it acts against the direction of motion of the crate.
Fr = -0.57 * 32 * 9.8
Fr = - 178.75 N
Hence, the force necessary to move the crate must be at least equal to but opposite in direction to this frictional force.
Therefore, this force is 178.75 N
Swinging a tennis racket against a ball is an example of a third class lever.
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Please select the best answer from the choices provided.
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Answers
Final answer:
Swinging a tennis racket against a ball as a third class lever in physics.
Explanation:
Swinging a Tennis Racket as a Third Class Lever
A tennis racket swinging against a ball is indeed an example of a third class lever in physics. In a third class lever, the effort is situated between the fulcrum and the load. In this case, the effort is provided by the player's hand gripping the racket handle, the fulcrum is the wrist joint, and the load is the ball being struck by the racket.
When a player swings the racket, the force applied by the player's hand exerts an effort on the handle of the racket. This causes the racket to rotate about the wrist joint acting as the fulcrum. The ball serves as the load, receiving the force and accelerating in the opposite direction.
Learn more about Third class lever here:
Answers
Answer:
The spacing is 5.15 μm.
Explanation:
Given that,
Electron with energy = 25 eV
Wave length = 0.25 nm
Separation d= 0.16 mm
Distance D=3.3 m
We need to calculate the spacing
Using formula of width
Put the value into the formula
Hence, The spacing is 5.15 μm.
Final answer:
To calculate the spacing between maxima in a double slit interference pattern, we use the formula x = L * λ / d. Converting the given units to meters and plugging the values into the formula, we find that the spacing between maxima on the screen is approximately 5.14 micro meters.
Explanation:
To calculate the spacing between maxima, we can utilize the formula for double slit interference, θ = λ/d where λ represents the wavelength of the electron, d is the distance between the two slits, and θ is the angle of diffraction. Considering the small angle approximation for tan θ ≈ θ, we get x = L * λ / d, where x is the distance between maxima on the screen, and L is the distance from the slits to the screen.
Firstly, the electron's wavelength needs to be converted from nm to m, resulting in λ = 0.25 * 10^-9 m. Similarly, the slit separation d should be converted from mm to m, giving d = 0.16 * 10^-3 m. Inserting these values into the formula along with L = 3.3 m, we can solve for x.
x = (3.3 m * 0.25 * 10^-9 m) / 0.16 * 10^-3 m =~ 5.14 μm
So, the spacing between maxima on the screen is approximately 5.14 micrometers.
Learn more about Double Slit Interference here:
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Answers
Answers
Answer:
560 watts
The rate of energy dissipation is 560 W
Explanation:
Rate of energy dissipation is the rate of energy consumption in the x-ray tube.
P = VI
Given;
Voltage V = 80 kV = 80,000
Current I = 7mA = 0.007 A
Substituting the given values;
P = 80,000 V × 0.007A
P = 560 Joules per second
P = 560 watts
The rate of energy dissipation is 560 W
Answers
Answer:
Initial velocity of ball 2 = 38.21 m/s = 125.36 ft/s
At the time of collision velocity of ball one is descending.
Explanation:
Velocity of ball 1 = 146 ft/sec = 44.50m/s
The balls are to collide at an altitude of 234 ft
H = 234 ft = 71.32 m
We have equation of motion
v² = u² + 2as
v² = 44.50² + 2 x (-9.81) x 71.32
v = ±24.10 m/s.
Time for each velocity can be calculated using equation of motion
v = u + at
24.10 = 44.50 - 9.81 t , t = 2.07 s
-24.10 = 44.50 - 9.81 t , t = 6.99 s
Since the second ball throws after 2.3 seconds we can avoid case with t = 2.07 s.
So at the time of collision velocity of ball one is descending.
The collision occurs at t = 6.99 s.
Time of flight of ball 2 = 6.99 - 2.3 = 4.69 seconds.
Height traveled by ball 2 = 71.32 m
We need to find velocity
We have
s = ut + 0.5 at²
71.32 = u x 4.69 - 0.5 x 9.81 x 4.69²
u = 38.21 m/s = 125.36 ft/s
Initial velocity of ball 2 = 38.21 m/s = 125.36 ft/s
Answer:
v2=139 ft
Explanation:
First we just look at the motion of the first particle. It is moving vertically in a gravitational field so is decelerating with rate g = 9.81 m/s^2 = 32.18 ft/s^2. We can write it's vertical position as a function of time.
h_1=v_1*t-(a*t/2)
We set this equal to 234 ft to find when the body is passing that point, a solve the quadratic equation for t.
t_1,2=v_1±(√v_1^2-4*a/2*h_1)/a=2.57 s, 7.44 s
Since we know the second ball was launched after 2.3 seconds, we know that the time we are looking for is the second one, when the first ball is descending. The second ball will have 2.3 seconds less so the time we further use is t_c = 7.44 - 2.3 = 5.14 s. With this the speed of the second ball needed for collision at given height, can be found.
Solving a similar equation, but this time for v2 to obtain the result.
h_2=234 ft=v2*t_c-(a*t_c^2/2)--->v2=139 ft