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olganol [36]
3 years ago
8

Helpppppppppp meeeeeeeeeee plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz

Physics
1 answer:
borishaifa [10]3 years ago
3 0
It’s either A or C but most likely A
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German physicist Werner Heisenberg related the uncertainty of an object's position ( Δ x ) to the uncertainty in its velocity (
timama [110]

According to the information given, the Heisenberg uncertainty principle would be given by the relationship

\Delta x \Delta v \geq \frac{h}{4\pi m}

Here,

h = Planck's constant

\Delta v = Uncertainty in velocity of object

\Delta x = Uncertainty in position of object

m = Mass of object

Rearranging to find the position

\Delta x \geq \frac{h}{4\pi m\Delta v}

Replacing with our values we have,

\Delta x \geq \frac{6.625*10^{-34}m^2\cdot kg/s}{4\pi (9.1*10^{-31}kg)(0.01*10^6m/s)}

\Delta x \geq 5.79*10^{-9}m

Therefore the uncertainty in position of electron is 5.79*10^{-9}m

8 0
3 years ago
Hydraulic brakes use Pascal’s principle. The driver exerts a force of 100 N on the brake pedal. This force is increased by the s
Rus_ich [418]

Answer:

Explanation:

Pascal's law states that when there is an increase in pressure at any point in a confined fluid, the pressure is equally distributed at every other point in the container.

The formulas that relate to this are given below:

P1 = P2 (since the pressures are equal throughout).

P2 is the pressure transmitted in the hydraulic system

N.B : P= Force/ Area

Therefore we have  

F1/A1 = P2

F1= the force exerted by the driver = 100N  

P2= 100/A1 N/m² or 100/A1 pa or 100/(101325× A1) atm

3 0
3 years ago
WHAT IS THE POTENIAL ENERGY OF A 3 KILOGRAM BALL THAT IS ON THE GROUND
Ratling [72]

Answer:

147.15 Joules.

Explanation:

A 3 kilogram mass at a height of 5 meters, while acted on by Earth's gravity would have 147.15 Joules of potential energy, PE = 3kg * 9.81 m/s2 * 5m = 147.15 J.

5 0
2 years ago
A trolley at rest is pushed to accelerate at a
laila [671]
V^2 = 2a S
V = square root ( 2 x 0.5 . 4)
V = 2 m/s
Answer is B
6 0
3 years ago
What is the minimum horizontal force needed to make the box start moving? The coefficients of kinetic and static friction betwee
aleksandr82 [10.1K]

Answer:

minimum horizontal force required to start the block motion is F = 0.41 mg

Explanation:

As we know that box is placed on the horizontal floor

so here the normal force on the box is counter balanced by the weight of the box

so we will have

F_n = mg

now in order to displace the box from its position we need to apply the external force which will just exceed the limiting static friction force

So here we will have

F_s = \mu_s F_n

here we know that

\mu_s = 0.41

F_s = 0.41(mg)

so minimum horizontal force required to start the block motion is F = 0.41 mg

5 0
3 years ago
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