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4 years ago

PLEASE HELP

2 answers:

4 0

Don't over think it!

For example, if you do 5 laps in 5 minutes then you do one lap per minute. If you use that example and you apply it to your problem you'll notice that the answer is simply C. 1km/min.

Best wishes!

Grace [21]4 years ago

4 0

C. 1km/min because the car covers a distance of 5km over a period of 5 minutes, so the car travels 1 kilometer each minute.

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An electron in an atom has an uncertainty of 0.2 nm. If it is doubled to 0.4 nm by what factor does the uncertainty in momentum

fenix001 [56]

Answer:

The uncertainty in momentum changes by a factor of 1/2.

Explanation:

By Heisenberg's uncertainty principle, ΔpΔx ≥ h/2π where Δp = uncertainty in momentum and Δx = uncertainty in position = 0.2 nm. The uncertainty in momentum is thus Δp ≥ h/2πΔx. If the uncertainty in position is doubled, that is Δx₁ = 2Δx = 0.4 nm, the uncertainty in momentum Δp₁ now becomes Δp₁ ≥ h/2πΔx₁ = h/2π(2Δx) = (h/2πΔx)/2 = Δp/2.

So, the uncertainty in momentum changes by a factor of 1/2.

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4 years ago

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3 years ago

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Consider a box with the same mass and velocity instead sliding without friction toward the incline. How does this object's kinet

Umnica [9.8K]

Answer:

Object's kinetic energy is high on frictionless medium.

Explanation:

There is a great difference in object's kinetic energy which is present between sliding with friction and sliding without friction because friction decreases the speed of an object so if an object moves on the frictionless medium, it moves and covers more distance and we can say that the object has more kinetic energy while on the other hand, if a body moves on the medium and experience friction so it moves with little speed so we can say that the body has low kinetic energy.

8 0

3 years ago

Two wires are stretched between two fixed supports and have the same length. One wire A there is a second-harmonic standing wave

lina2011 [118]

(a) Greater

The frequency of the nth-harmonic on a string is an integer multiple of the fundamental frequency, $f_1$ :

$f_n = n f_1$

So we have:

- On wire A, the second-harmonic has frequency of $f_2 = 660 Hz$ , so the fundamental frequency is:

$f_1 = \frac{f_2}{2}=\frac{660 Hz}{2}=330 Hz$

- On wire B, the third-harmonic has frequency of $f_3 = 660 Hz$ , so the fundamental frequency is

$f_1 = \frac{f_3}{3}=\frac{660 Hz}{3}=220 Hz$

So, the fundamental frequency of wire A is greater than the fundamental frequency of wire B.

(b) $f_1 = \frac{v}{2L}$

For standing waves on a string, the fundamental frequency is given by the formula:

$f_1 = \frac{v}{2L}$

where

v is the speed at which the waves travel back and forth on the wire

L is the length of the string

(c) Greater speed on wire A

We can solve the formula of the fundamental frequency for v, the speed of the wave:

$v=2Lf_1$

We know that the two wires have same length L. For wire A, $f_1 = 330 Hz$ , while for wave B, $f_B = 220 Hz$ , so we can write the ratio between the speeds of the waves in the two wires:

$\frac{v_A}{v_B}=\frac{2L(330 Hz)}{2L(220 Hz)}=\frac{3}{2}$

So, the waves travel faster on wire A.

7 0

3 years ago

Olaf is standing on a sheet of ice that covers the football stadium parking lot in Buffalo, New York; there is negligible fricti

Bas_tet [7]

Answer:

v = 0.059 m/s

Explanation:

To find the final speed of Olaf and the ball you use the conservation momentum law. The momentum of Olaf and the ball before catches the ball is the same of the momentum of Olaf and the ball after. Then, you have:

$mv_{1i}+Mv_{2i}=(m+M)v$ (1)

m: mass of the ball = 0.400kg

M: mass of Olaf = 75.0 kg

v1i: initial velocity of the ball = 11.3m/s

v2i: initial velocity of Olaf = 0m/s

v: final velocity of Olaf and the ball

You solve the equation (1) for v and replace the values of all variables:

$v=\frac{mv_{1i}}{m+M}=\frac{(0.400kg)(11.3m/s)}{0.400kg+75.0kg}=0.059\frac{m}{s}$

Hence, after Olaf catches the ball, the velocity of Olaf and the ball is 0.059m/s

3 0

3 years ago