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Serga [27]
3 years ago
6

9. Use the following information to answer parts a, b, c and d

Physics
1 answer:
N76 [4]3 years ago
7 0

Answer:

iittdhusludepeotlke is the only person who has ever seen a long distance friend from which he is the 3PM and 3AM and I

Explanation:

gt6e6ofkhithodzipfrhFfu0sa6eo5ostiatkstoss ragers the following 3AM is not the only one sentences you have to say about it is not an 2475232347 email from a substance abuse treatment center for a medical professional who is a substance addict or a physician in my life to be 367inches 4 to the end 6th

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ASAP ASAP ASAP
mylen [45]

The experiments will involve two billiard balls of known masses, m₁ and m₂, and velocities u₁ and u₂. The two are allowed to collide and the velocities of the balls after the collision v₁ and v₂ are recorded.

The momentum before and after the collision is then calculated as follows:

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

<h3>What is the statement of the law of conservation of momentum?</h3>

The law of the conservation of momentum states that the momentum before and after collision in a system of colliding bodies is conserved

The momentum of a body is calculated using the formula below:

Momentum = mass * velocity.

Hence, for the two billiard balls, the momentum before and after the collision is conserved.

Learn more about momentum at: brainly.com/question/1042017

#SPJ1

3 0
1 year ago
Help in physics please :(((​
Arlecino [84]

Answer:

I am sorry I can't draw graphical ok how to draw the graph where what is your position the displacement of time and work 7 kilometres east in 2 hours and what will happen to the time and 72 in 1 hour what is the displacement you after take the displacement formula that is total time taken divided by the distance travelled ok displacement and distance travelled is different about its terms ok

8 0
3 years ago
A robin flies a distance of 45963 cm. How far has it flown in kilometers?
alina1380 [7]

Answer:

0.46km

Explanation:

45963cm/100cm=459.63m/1000m=0.45963 or 0.46km

6 0
2 years ago
A tree falls in a forest. How many years must pass before the 14C activity in 1.03 g of the tree's carbon drops to 1.02 decay pe
Illusion [34]

Answer:

t = 5.59x10⁴ y

Explanation:

To calculate the time for the ¹⁴C drops to 1.02 decays/h, we need to use the next equation:

A_{t} = A_{0}\cdot e^{- \lambda t}    (1)

<em>where A_{t}: is the number of decays with time, A₀: is the initial activity, λ: is the decay constant and t: is the time.</em>

To find A₀ we can use the following equation:  

A_{0} = N_{0} \lambda   (2)

<em>where N₀: is the initial number of particles of ¹⁴C in the 1.03g of the trees carbon </em>

From equation (2), the N₀ of the ¹⁴C in the trees carbon can be calculated as follows:        

N_{0} = \frac{m_{T} \cdot N_{A} \cdot abundance}{m_{^{12}C}}

<em>where m_{T}: is the tree's carbon mass, N_{A}: is the Avogadro's number and m_{^{12}C}: is the ¹²C mass.  </em>

N_{0} = \frac{1.03g \cdot 6.022\cdot 10^{23} \cdot 1.3\cdot 10^{-12}}{12} = 6.72 \cdot 10^{10} atoms ^{14}C    

Similarly, from equation (2) λ is:

\lambda = \frac{Ln(2)}{t_{1/2}}

<em>where t 1/2: is the half-life of ¹⁴C= 5700 years </em>

\lambda = \frac{Ln(2)}{5700y} = 1.22 \cdot 10^{-4} y^{-1}

So, the initial activity A₀ is:  

A_{0} = 6.72 \cdot 10^{10} \cdot 1.22 \cdot 10^{-4} = 8.20 \cdot 10^{6} decays/y    

Finally, we can calculate the time from equation (1):

t = - \frac{Ln(A_{t}/A_{0})}{\lambda} = - \frac {Ln(\frac{1.02decays \cdot 24h \cdot 365d}{1h\cdot 1d \cdot 1y \cdot 8.20 \cdot 10^{6} decays/y})}{1.22 \cdot 10^{-4} y^{-1}} = 5.59 \cdot 10^{4} y              

I hope it helps you!

4 0
3 years ago
A sheet of metal is 2mm wide 10cm tall and 15cm long. it was 4g. what is the density? <br>​
Hoochie [10]

Answer:

Ro = 133 [kg/m³]

Explanation:

In order to solve this problem, we must apply the definition of density, which is defined as the relationship between mass and volume.

Ro = m/V

where:

m = mass [kg]

V = volume [m³]

We will convert the units of length to meters and the mass to kilograms.

L = 15 [cm] = 0.15 [m]

t = 2 [mm] = 0.002 [m]

w = 10 [cm] = 0.1 [m]

Now we can find the volume.

V = 0.15*0.002*0.1\\V = 0.00003 [m^{3} ]

And the mass m = 4 [gramm] = 0.004 [kg]

Ro = 0.004/0.00003\\Ro = 133 [kg/m^{3}]

3 0
2 years ago
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