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2 years ago

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I have a 3 cc piece of aluminum with a density of steel with a mass of 24.0 g. I cut it into 2 equal pieces. How has the density

of the steel pieces changed

1 answer:

tresset_1 [31]2 years ago

3 0

Answer:

What happens to the density of an object if the object is cut in half? ... The density remains the same because cutting the object in half will divide the mass & volume by the same amount. Also, the density of a substance remains the same no matter what size it is.

Explanation:

What happens to the density of an object if the object is cut in half? ... The density remains the same because cutting the object in half will divide the mass & volume by the same amount. Also, the density of a substance remains the same no matter what size it is.

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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

2 years ago

A cyclist going downhill is accelerating at 1. 2 m/s2. If the final velocity of the cyclist is 16 m/s after 10 seconds, what is

mel-nik [20]

Answer:

$\boxed {\boxed {\sf v_i= 4 \ m/s}}$

Explanation:

We are asked to find the cyclist's initial velocity. We are given the acceleration, final velocity, and time, so we will use the following kinematic equation.

$v_f= v_i + at$

The cyclist is acceleration at 1.2 meters per second squared. After 10 seconds, the velocity is 16 meters per second.

$v_f$ = 16 m/s
a= 1.2 m/s²
t= 10 s

Substitute the values into the formula.

$16 \ m/s = v_i + (1.2 \ m/s^2)(10 \ s)$

Multiply.

$16 \ m/s = v_i + (1.2 \ m/s^2 * 10 \ s)$

$16 \ m/s = v_i + 12 \ m/s$

We are solving for the initial velocity, so we must isolate the variable $v_i$ . Subtract 12 meters per second from both sides of the equation.

$16 \ m/s - 12 \ m/s = v_i + 12 \ m/s -12 \ m/s$

$4 \ m/s = v_i$

The cyclist's initial velocity is <u>4 meters per second.</u>

6 0

2 years ago

When a force is acting at axis of rotation,why torque is zero?

Vikki [24]

Explanation:

see, torque=force × perpendicular distance

...that perpendicular distance is between axis of rotation and the point where force acts... so in above's case perpendicular distance is zero... so the torque is zero!

4 0

2 years ago

Please !!! I really need help !!! How do I understand these ?!!!!

trasher [3.6K]

Answer

The answer for the first one I think is false.

The second one would be true i think. I hope i got it right and have a wonderful day

8 0

2 years ago

Read 2 more answers

Two moles of an ideal monatomic gas are at a temperature of 345 K. Then, 2250 J of heat is added to the gas, and 870 J of work i

Lapatulllka [165]

Answer: The final temperature is 470K

Explanation: Using the relation;

Q= ΔU +W

Given, n = 2mol

Initial temperature T1= 345K

Heat =Q= 2250J

Workdone=W=-870J(work is done on gas)

T2 =Final temperature =?

ΔU =3/2nR(T2-T1)

ΔU=3/2 × 2 ×8.314 (T2 - 345)

ΔU=24.942(T2-345)

Therefore Q = 24.942(T2-345)+ (-870)

2250=24.942(T2-345)+ (-870)

125.09=(T2-345)

T2 =470K

Therfore the final temperature is 470K

5 0

2 years ago