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3 years ago

A toy car moves 8 min 4 s at the constant velocity. What is the car's velocity?

Physics

1 answer:

Schach [20]3 years ago

6 0

Explanation:

Formula:

$velocity = (d \div t)$

d = distance given

t = the amount of time given

Substitute the given values into the formula for velocity:

$v = 8 \div 4$

velocity is shortened for v.

8 (distance) divided by 4 (time) equals the velocity.

Solve:

$2 = 8 \div 4$

The velocity of the toy car equals: B. 2 m/s.

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3 years ago

An object is moving along a straight line, and the uncertainty in its position is 1.90 m.

just olya [345]

Answer:

$2.78\times 10^{-35}\ \text{kg m/s}$

$6.178\times 10^{-34}\ \text{m/s}$

$0.31\times 10^{-4}\ \text{m/s}$

Explanation:

$\Delta x$ = Uncertainty in position = 1.9 m

$\Delta p$ = Uncertainty in momentum

h = Planck's constant = $6.626\times 10^{-34}\ \text{Js}$

m = Mass of object

From Heisenberg's uncertainty principle we know

$\Delta x\Delta p\geq \dfrac{h}{4\pi}\\\Rightarrow \Delta p\geq \dfrac{h}{4\pi\Delta x}\\\Rightarrow \Delta p\geq \dfrac{6.626\times 10^{-34}}{4\pi\times 1.9}\\\Rightarrow \Delta p\geq 2.78\times 10^{-35}\ \text{kg m/s}$

The minimum uncertainty in the momentum of the object is $2.78\times 10^{-35}\ \text{kg m/s}$

Golf ball minimum uncertainty in the momentum of the object

$m=0.045\ \text{kg}$

Uncertainty in velocity is given by

$\Delta p\geq m\Delta v\geq 2.78\times 10^{-35}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{m}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{0.045}\\\Rightarrow \Delta v\geq 6.178\times 10^{-34}\ \text{m/s}$

The minimum uncertainty in the object's velocity is $6.178\times 10^{-34}\ \text{m/s}$

Electron

$m=9.11\times 10^{-31}\ \text{kg}$

$\Delta v\geq \dfrac{\Delta p}{m}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{9.11\times 10^{-31}}\\\Rightarrow \Delta v\geq 0.31\times 10^{-4}\ \text{m/s}$

The minimum uncertainty in the object's velocity is $0.31\times 10^{-4}\ \text{m/s}$ .

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3 years ago

A proton and an electron are fixed in space with a separation of 859 nm. Calculate the electric potential at the midpoint betwee

makkiz [27]

Answer:

The electric potential at the midpoint between the two particles is 3.349 X 10⁻³ Volts

Explanation:

Electric potential is given as;

V = E*r

where;

E is the electric field strength, = kq/r²

V = ( kq/r²)*r

V = kq/r

k is coulomb's constant = 8.99 X 10⁹ Nm²/C²

q is the charge of the particles = 1.6 X 10⁻¹⁹ C

r is the distance between the particles = 859 nm

At midpoint, the distance = r/2 = 859nm/2 = 429.5 nm

V = (8.99 X 10⁹ * 1.6 X 10⁻¹⁹)/ (429.5 X 10⁻⁹)

V = 3.349 X 10⁻³ Volts

Therefore, the electric potential at the midpoint between the two particles is 3.349 X 10⁻³ Volts

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3 years ago