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lianna [129]
3 years ago
15

A net force of 6.8 N accelerates a 31 kg scooter across a level parking lot. What is the magnitude of the scooter’s acceleration

?a) 0.22 m/s2b) 0.69 m/s2c) 3.2 m/s2d) 4.6 m/s2
Physics
1 answer:
mrs_skeptik [129]3 years ago
3 0

Answer:

a)0.22 m/s².

Explanation:

Given that

Net force ,F= 6.8 N

mass ,m = 31 kg

From the second law of Newton's

F = m a   ---------------1

Where

F=Net force ,m=mass

a=Acceleration

Now by putting the values in the equation 1

F = m a

6.8 = 31 x a

a=\dfrac{6.8}{31}\ m/s^2

a = 0.219\ m/s^2

a = 0.22\ m/s^2

Therefore the acceleration of the scooter will be 0.22 m/s².

The answer will be "a".

a)0.22 m/s².

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Carbon-14 is used to determine the time an organism was living. The amount of carbon-14 an organism has is constant with the atm
lutik1710 [3]

Answer:

The age of the organism is approximately 11460 years.

Explanation:

The amount of carbon-14 decays exponentially in time and is defined by the following equation:

\frac{n(t)}{n_{o}} = e^{-\frac{t}{\tau} } (1)

Where:

n_{o} - Initial amount of carbon-14.

n(t) - Current amount of carbon-14.

t - Time, measured in years.

\tau - Time constant, measured in years.

Then, we clear the time within the formula:

t = -\tau \cdot \ln \frac{n(t)}{n_{o}} (2)

In addition, time constant can be calculated by means of half-life of carbon-14 (t_{1/2}), measured in years:

\tau = \frac{t_{1/2}}{\ln 2}

If we know that \frac{n(t)}{n_{o}} = 0.25 and t_{1/2} = 5730\,yr, then the age of the organism is:

\tau = \frac{5730\,yr}{\ln 2}

\tau \approx 8266.643\,yr

t = -(8266.643\,yr)\cdot \ln 0.25

t \approx 11460.001\,yr

The age of the organism is approximately 11460 years.

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A snowball is rolling down a hill at 4.5 m/s and accumulating snow as it goes. Its diameter begins at 0.50 m and ends at the bot
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To find the change in centripetal acceleration, you should first look for the centripetal acceleration at the top of the hill and at the bottom of the hill.

The formula for centripetal acceleration is:
Centripetal Acceleration = v squared divided by r

where:
v = velocity, m/s
r= radium, m

assuming the velocity does not change:

at the top of the hill:
centripetal acceleration = (4.5 m/s^2) divided by 0.25 m
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at the bottom of the hill:
centripetal acceleration = (4.5 m/s^2) divided by 1.25 m
                                      = 16.2 m/s^2

to find the change in centripetal acceleration, take the difference of the two.
change in centripetal acceleration = centripetal acceleration at the top of the hill - centripetal acceleration at the bottom of the hill

= 81 m/s^2 - 16.2 m/s^2
= 64.8 m/s^2 or 65 m/s^2
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r was thirsty and decided to mix up a pitcher of lemonade. She put lemon juice, water, and sugar into a pitcher and stirred it t
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The amount of work done to bring an electron (q = −e) from right side of hydrogen nucleus to left side in the k shell is________
maksim [4K]

Answer:

d) 2Fr

Explanation:

We know that the work done in moving the charge from the right side to the left side in the k shell is W = ∫Fdr from r = +r to -r. F = force of attraction between nucleus and electron on k shell. F = qq'/4πε₀r² where q =charge on electron in k shell  -e and q' = charge on nucleus = +e. So, F = -e × +e/4πε₀r² = -e²/4πε₀r².

We now evaluate the integral from r = +r to -r

W = ∫Fdr

= ∫(-e²/4πε₀r²)dr

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= -e²/4πε₀∫dr/r²

= -e²/4πε₀ × -[1/r] from r = +r to -r

W = e²/4πε₀[1/-r - 1/+r] = e²/4πε₀[-2/r} = -2e²/4πε₀r.

Since F = -e²/4πε₀r², Fr = = -e²/4πε₀r² × r = = -e²/4πε₀r and 2Fr = -2e²/4πε₀r.

So W = -2e²/4πε₀r = 2Fr.

So, the amount of work done to bring an electron (q = −e) from right side of hydrogen nucleus to left side in the k shell is W = 2Fr

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