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2 years ago

9

A particle is confined to a one-dimensional box that is 50 pm long. What is the smallest possible uncertainty in momentum for th

e particle?

1 answer:

nevsk [136]2 years ago

6 0

Answer:

The smallest possibility is 0.01E-22kgm/s

Explanation:

Using

Momentum= h/4πx

= 6.6x 10^-34Js/ 4(3.142* 50*10-12m)

= 0.01*10^-22kgm/s

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A car of 900 kg mass is moving at the velocity of 60 km/hr. It is brought into rest at 50 meter distance by applying a brake. No

kozerog [31]

Answer: $-2502N$

Explanation:

$(V_2)^2=(V_1)^2+2ad$

where;

$V_2$ = final velocity = 0

$V_1$ = initial velocity = 60 km/h = 16.67 m/s

$a$ = acceleration

$d$ = distance

First all of, because acceleration is given in m/s and not km/h, you need to convert 60km/h to m/s. Our conversion factors here are 1km = 1000m and 1h = 3600s

$60km/h(\frac{1000m}{1km} )(\frac{1h}{3600s} )=16.67m/s$

Solve for a;

$(V_2)^2=(V_1)^2+2ad$

Begin by subtracting $(V_1)^2$

$(V_2)^2-(V_1)^2=2ad$

Divide by 2d

$\frac{(V_2)^2-(V_1)^2}{2d} =a$

Now plug in your values:

$a=\frac{(0)^2-(16.67 m/s)^2}{2(50m)}$

$a=\frac{0-277.89m^2/s^2}{100m}$

$a=-2.78m/s$

If you're wondering why I calculated acceleration first is because in order to find force, we need 2 things: mass and acceleration.

$F=ma$

m = mass = 900kg

a = acceleration = -2.78m/s

$F=(900kg)(-2.78m/s)\\F=-2502N$

It's negative because the force has to be applied in the opposite direction that the car is moving.

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2 years ago