The measurement of an electron's energy requires a time interval of 1.2×10^−8 s . What is the smallest possible uncertainty in the electron's energy?

Answers

Answer 1

Answer:

Answer:

$1.05* 10^(-26)J$

Explanation:

The uncertainty in energy is given by $\Delta E=(h)/(2\pi \Delta t)$

here h is plank's constant which value is $6.67* 10^(-34)$ and $\Delta t$ is the time interval which is given as $1.2* 10^(-8)sec$

So using all the parameters the smallest possible uncertainty in electrons energy is $=(6.67* 10^(-34))/(2* \pi * 1.2* 10^(-8))=1.05* 10^(-26)J$

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A dynamics cart with a friction pad is placed at the top of an inclined track and released fromrest. The cart accelerates down the incline at a rate of 0.60 m/s2. If the track is angled at 10degrees above the horizontal, determine the coefficient of kinetic friction between the cart andthe track.
For a short time the position of a roller-coaster car along its path is defined by the equations r=25 m, θ=(0.3t) rad, and z=(−8 cosθ) m, where t is measured in seconds, Determine the magnitudes of the car's velocity and acceleration when t=4s .

Two parallel wires I and II that are near each other carry currents i and 3i both in the same direction. Compare the forces that the two wires exert on each other. A. The wires exert equal magnitude attractive forces on each other. B. Wire I exerts a stronger force on wire II than II exerts on I.C. Wire II exerts a stronger force on wire I than I exerts on II. D. The wires exert equal magnitude repulsive forces on each other. E. The wires exert no forces on each other.

Answers

Answer:

A. The wires exert equal magnitude attractive forces on each other.

Explanation:

Magnetic field due to current i on current 2i

B₁ = 10⁻⁷ x 2 i / r where r is distance between the two wires

Force on wire II due to wire I per unit length

= magnetic field x current in wire II

= B₁ x 2 i

= [ 10⁻⁷ x 2 i / r ] x 2i

= 4 x 10⁻⁷ i² / r

Magnetic field due to current 2i on current i

B₂ = 10⁻⁷ x 4 i / r where r is distance between the two wires

Force on wire I due to wire II per unit length

= magnetic field x current in wire I

= B₂ x i

= [ 10⁻⁷ x 4 i / r ] x i

= 4 x 10⁻⁷ i² / r

So final forces on each wire are same .

This force will be attractive in nature . The direction of force can be known from fleming's right hand rule .

A firefighting crew uses a water cannon that shoots water at 27.0 m/s at a fixed angle of 50.0 ∘ above the horizontal. The firefighters want to direct the water at a blaze that is 12.0 m above ground level. How far from the building should they position their cannon? There are two possibilities (d1Part A:
d1=_____m
Part B:
d2=______m

Answers

Answer:

Explanation:

In projectile motion , range of projectile is given by the expressions

R = u²sin2θ / g

where u is velocity of projectile.

u = 27 m/s θ = 50

12 = 27² sin 2θ / 9.8

sin 2θ = .16

θ = 9.2 / 2

= 4.6

When we place 90- θ in place of θ , in the formula of range , we get the same value of projectile. hence at 85.4 ° , the range will be same.

What is the most important safety rule to remember during lab activities

Answers

To follow instructions

High-energy particles are observed in laboratories by photographing the tracks they leave in certain detectors; the length of the track depends on the speed of the particle and its lifetime. A particle moving at 0.993c leaves a track 1.15 mm long. What is the proper lifetime of the particle

Answers

Answer:

Lifetime = 4.928 x 10^-32 s

Explanation:

(1 / v2 – 1 / c2) x2 = T2

T2 = (1/ 297900000 – 1 / 90000000000000000) 0.0000013225

T2 = (3.357 x 10^-9 x 1.11 x 10^-17) 1.3225 x 10^-6

T2 = (3.726 x 10^-26) 1.3225 x 10^-6 = 4.928 x 10^-32 s

Final answer:

To find the proper lifetime of the particle, we can use the time dilation equation and the Lorentz factor. Plugging in the given values, we find that the proper lifetime of the particle is approximately 5.42 × 10^-9 seconds.

Explanation:

To find the proper lifetime of the particle, we can use the time dilation equation, which states that the proper time (time experienced in the frame of reference of the particle) is equal to the time observed in the laboratory frame of reference divided by the Lorentz factor. The Lorentz factor can be calculated using the equation γ = 1/√(1 - (v/c)^2), where v is the velocity of the particle and c is the speed of light. Given that the particle is moving at 0.993c, the Lorentz factor is approximately 22.82.

Next, we can use the equation Δx = βγcτ, where Δx is the length of the track, β is the velocity of the particle in units of the speed of light (v/c), γ is the Lorentz factor, c is the speed of light, and τ is the proper lifetime of the particle. Plugging in the given values, we have 1.15 mm = 0.993 * 22.82 * c * τ. Solving for τ, we find that the proper lifetime of the particle is approximately 5.42 × 10^-9 seconds.

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A suspended platform of negligible mass is connected to the floor below by a long vertical spring of force constant 1200 N/m. A circus performer of mass 70 kg falls from rest onto the platform from a height of 5.8 m above it. Find the maximum spring compression

Answers

Answer:

The maximum spring compression = 3.21 m

Explanation:

The height of the circus performer above the platform connected to string material = 5.8 m

Let the maximum compression of the spring from the impact of the circus performer be x.

According to the law of conservation of energy, the difference in potential energy of the circus performer between the initial height and the level at which spring is compressed to is equal to the work done on the spring to compress it by x

Workdone on the spring by the circus performer = (1/2)kx²

where k = spring constant = 1200 N/m

Workdone on the spring by the circus performer = (1/2)(1200)x² = 600x²

The change in potential energy of the circus performer = mg (5.8 + x)

m = mass of the circus performer = 70 kg

g = acceleration due to gravity = 9.8 m/s²

The change in potential energy of the circus performer = (70)(9.8)(5.8 + x) = (3978.8 + 686x)

600x² = 3978.8 + 686x

600x² - 686x - 3978.8 = 0

Solving this quadratic equation

x = 3.21 m or - 2.07 m

Since the negative answer doesn't satisfy the laws of physics, our correct answer is 3.21 m

Hope this Helps!!!

Two atoms collide while moving in a dilute gas. The larger atom has a mass M1 = 6 Daltons and a speed v1 = 200 m/s, while the smaller has a mass M2 = 1 Daltons. During the collision both atoms simply bounce off each other. They do not change their speeds, but after the collision they each change their directions, bouncing in the indicated directions. (You may express your results using the mass unit "Daltons". 1 Dalton is approximately equal to the mass of a proton or neutron and is defined as one-twelfth the mass of a single neutral carbon-12 atom in its ground state.)A. What is the magnitude of the change in the momentum, Δp1, of mass M1?
B. What is the change in the total momentum of the pair?
C. What is the magnitude of the change in the momentum Δp2, of mass M2?

Answers

Answer:

a). ΔP1=-2.4 $x10^(3) (D*m)/(s)$

b). Pp=0 F=0

c). ΔP2=2.4 $x10^(3) (D*m)/(s)$

Explanation:

Initial momentum

$P_(1)=m_(1)*v_(i1)$

Final momentum

$P_(1f)=m_(1)*v_(f1)=-m_(1)*v_(i1)$

The change of momentum m1 is:

a).

ΔP1= $P_(1f)-P_(1)$

ΔP1= $-m_(1)*v_(i1)-m_(1)*v_(i1)$

ΔP1= $-2*m_(1)*v_(i1)$

ΔP1= $-2*6 D*200(m)/(s)$

ΔP1= $-2.4x10^(3)(D*m)/(s)$

b).

The law of conservation of energy in this case there is not external forces so the momentum of the pair change is equal to zero

P=0

Fx=0

c).

ΔP1+ΔP2=0

ΔP2=-ΔP1

ΔP2=- $-2.4x10^(3)(D*m)/(s)$

ΔP2= $2.4x10^(3)(D*m)/(s)$

Final answer:

The magnitude of the change in momentum of mass M1 is 2400 Daltons*m/s. The change in the total momentum of the pair is 2000 Daltons*m/s. The magnitude of the change in momentum of mass M2 is -400 Daltons*m/s.

Explanation:

A. To find the magnitude of the change in momentum of mass M1, we use the formula Δp1 = m1 * Δv1, where m1 is the mass of M1 and Δv1 is the change in velocity of M1. Since M1 simply changes direction, its change in velocity is equal to 2 times its original velocity. Therefore, Δp1 = m1 * (2v1) = 6 * (2 * 200) = 2400 Daltons*m/s.

B. The change in the total momentum of the pair is equal to the sum of the changes in momentum of M1 and M2. Since M2 also changes direction, its change in velocity is equal to 2 times its original velocity. Therefore, the change in the total momentum is Δp1 + Δp2 = 2400 Daltons*m/s + (-400 Daltons*m/s) = 2000 Daltons*m/s.

C. To find the magnitude of the change in momentum of mass M2, we use the same formula as in part A, but with the values for M2. Δp2 = m2 * Δv2 = 1 * (2 * (-200)) = -400 Daltons*m/s.

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