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brilliants [131]
3 years ago
11

TIMED! URGENT! REALLY APPRECIATE HELP!! TYSM!!!!!!

Physics
2 answers:
JulsSmile [24]3 years ago
8 0

Answer:

<em>d = 27.03 m</em>

Explanation:

<u>Constant Speed Motion </u>

An object travels at constant speed if the ratio of the distance traveled by the time taken is constant.

Expressed in a simple equation, we have:

\displaystyle v=\frac{d}{t}

Where  

v = Speed of the object

d = Distance traveled

t = Time taken to travel d.

From the equation above, we can solve for d:

d = v . t

It's required to find the distance traveled by someone walking at v=1.7 m/s for t=15.9 s. Substituting in the last equation:

d = 1.7 m/s * 15.9 s

d = 27.03 m

Diano4ka-milaya [45]3 years ago
3 0
The answer is 27.03 I just multiplied the two numbers
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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
3 years ago
What is most likely to happen to light that hits an opaque object?
ddd [48]
B. is the answer.

C is not correct because the light is actually reflected off of an opaque object.
4 0
3 years ago
A 50-g cube of ice, initially at 0.0°C, is dropped into 200 g of water in an 80-g aluminum container, both initially at 30°C.
MakcuM [25]

Answer:

b. 9.5°C

Explanation:

m_i = Mass of ice = 50 g

T_i = Initial temperature of water and Aluminum = 30°C

L_f = Latent heat of fusion = 3.33\times 10^5\ J/kg^{\circ}C

m_w = Mass of water = 200 g

c_w = Specific heat of water = 4186 J/kg⋅°C

m_{Al} = Mass of Aluminum = 80 g

c_{Al} = Specific heat of Aluminum = 900 J/kg⋅°C

The equation of the system's heat exchange is given by

m_i(L_f+c_wT)+m_wc_w(T-T_i)+m_{Al}c_{Al}=0\\\Rightarrow 0.05\times (3.33\times 10^5+4186\times T)+0.2\times 4186(T-30)+0.08\times 900(T-30)=0\\\Rightarrow 1118.5T-10626=0\\\Rightarrow T=\dfrac{10626}{1118.5}\\\Rightarrow T=9.50022\ ^{\circ}C

The final equilibrium temperature is 9.50022°C

4 0
3 years ago
3. During a tug-of-war, Team A pulls with a
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Answer:

8000 - 5000 =3000

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8 0
2 years ago
Read 2 more answers
A 51-kg woman runs up a vertical flight of stairs in 5.0 s. Her net upward displacement is 5.0 m. Approximately, what average po
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Answer:

The average power the woman exerts is 0.5 kW

Explanation:

We note that power, P = The rate at which work is done = Work/Time

Work = Energy

The total work done is the potential energy gained which is the energy due to vertical displacement

Given that the vertical displacement = 5.0 m, we have

Total work done = Potential energy gained = Mass, m × Acceleration due to gravity, g × Vertical height, h

m = 51 kg

g = Constant = 9.81 m/s²

h = 5.0 m

Also, time, t = 5.0 s

Total work done = 51 kg × 9.81 m/s²× 5 m = 2501.55 kg·m²/s² = 2501.55 J

P = 2501.55 J/(5 s) = 500.31 J/s = 500.31 W ≈ 500 W = 0.5 kW.

6 0
3 years ago
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