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3 years ago

What is the impulse that a car with a mass of 6kg experiences when its velocity changes from 23m/s to 6m/s?

1 answer:

8 0

A.
Impulse is calculated through the equation,

I = mΔv

where I is impulse, m is mass and Δv is change is velocity. Substituting the known values,

I = (6 kg)(23 m/s - 6 m/s)
I = 102 kg m/s

Answer: 102 kg m/s

B.
Impulse in this item is calculated by multiplying the force by the time.

I = Force / time = (29 kg m/s²)(7 s) = 203 kg m/s

Answer: 203 kg m/s

C.
To answer this item, we have to multiply the mass and the velocity to get the impulse and divide the answer by the time to get the force.

F = m(v)/t

Substituting,

F = (0.1 kg)(-5.69 m/s)/0.64 s
F = -0.89 N

Answer: -0.89 N

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A point charge with a charge q1 = 4.00 μC is held stationary at the origin. A second point charge with a charge q2 = -4.10 μC mo

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Answer:

W = -0.480 J

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q₁ = 4 μC

q₂ = -4.10 μC

$W = kq_1q_2(\dfrac{1}{a}+\dfrac{1}{b})$

$b = \sqrt{(0.27-0)^2+(0.27-0)^2}$

b = 0.381

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3 years ago

The uncertainty in the position of an electron along an x axis is given as 68 pm. What is the least uncertainty in any simultane

umka21 [38]

Answer:

$\Delta p_x=7.75\times 10^{-25}\ kg-m/s$

Explanation:

Given that,

The uncertainty in the position of an electron along the x-axis is, $\Delta x=68\ pm=68\times 10^{-12}\ m$

We need to find the east uncertainty in any simultaneous measurement of the momentum component of this electron.

We know that the Heisenberg's uncertainty principle gives the relation between the uncertainty in position and the momentum of electron as :

$\Delta p_x{\cdot}\Delta x\ge \dfrac{h}{4\pi }$

Putting all the values, we get :

$\Delta p_x{\cdot}\ge \dfrac{h}{4\pi \Delta x}\\\\\Delta p_x \ge \dfrac{6.63\times 10^{-34}}{4\pi \times 68\times 10^{-12}}\\\\\Delta p_x\ge 7.75\times 10^{-25}\ kg-m/s$

So, the momentum component of this electrons is greater than $7.75\times 10^{-25}\ kg-m/s$ .

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4 years ago