All categories

3 years ago

If the accuracy in measuring the position of a particle increases, the accuracy in measuring its velocity will

Physics

1 answer:

astraxan [27]3 years ago

7 0

Answer:

Therefore the answer is the precision in the speed DECREASES

Explanation:

In quantum mechanics, we have the uncertainty principle that establishes that when the accuracy of the position increases the accuracy the speed decreases, being related by the expression

Δx Δv ≥ h'/ 2

h' = h/2π

Therefore the answer is the precision in the speed DECREASES

You might be interested in

Help please I'll mark as brainliest for the first correct answer

Alenkinab [10]

Answer:

first one i think is this. work = 1/2 kx^2 = 1/2 Fx

2nd, is 0.08 J

Explanation:

EE = ½ kx²

EE = ½ (400 N/m) (0.02 m) ²

EE = 0.08.

THIRD, Velocity of the stone is 4 m/s when it leaves catapult.

3 0

2 years ago

If the net force acting on an object is 100 N to the left, Are the forces unbalanced or balanced?

hodyreva [135]

Answer:

unbalanced

Explanation:

none

8 0

2 years ago

Priscilla is driving her car on a busy street and Harvey passes her on his motorcycle. What will happen to the sound from his mo

Evgesh-ka [11]

The answer to this question is A

8 0

2 years ago

Jorge wishes to observe the progress of his vacuum pump in evacuating a tall bell jar for the purpose of demonstrating that soun

natka813 [3]

Answer:

1.147 %

Explanation:

$\rho$ = Density of mercury = $1.36\times 10^4\ kg/m^3$

g = Acceleration due to gravity = 9.81 m/s²

h = Height of mercury = 8.69 m

Standard atmospheric pressure = $1.01\times 10^5\ Pa$

6.55 mmHg converting to Pa

$\rho gh\\ =1.36\times 10^4\times 9.81\times 8.69\times 10^{-3}\\ =1159.38504\ Pa$

Dividing air pressure by the above value

$\frac{1159.38504}{1.01\times 10^5}\times 100=1.147\ \%$

The fraction of atmospheric pressure is 1.147 %

8 0

2 years ago

A car is stationary. It accelerates at 0.8 ms^2

lara31 [8.8K]

Answer:

the car’s final displacement is 60 m

Explanation:

Given;

initail velocity of the car, u = 0

acceleration of the car, a = 0.8 m/s²

time of motion, t = 10 s

The first displacement of the car:

$x_1 = ut + \frac{1}{2} at^2\\\\x_1 = 0 + \frac{1}{2} (0.8)(10)^2\\\\x_1 = 40 \ m$

The second displacement of the car;

acceleration, a = 0.4 m/s², time of motion, t = 10 s

$x_2 = ut + \frac{1}{2} at^2\\\\x_2 = 0 + \frac{1}{2} (0.4)(10)^2\\\\x_2 = 20 \ m$

The final displacement of the car;

x = x₁ + x₂

x = 40 m + 20 m

x = 60 m

Therefore, the car’s final displacement is 60 m

8 0

2 years ago