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3 years ago

Based on the data table, find the acceleration and then predict the velocity at the time 3.5 seconds.

Physics

2 answers:

kumpel [21]3 years ago

4 0

Answer:B

Explanation:

Gala2k [10]3 years ago

3 0

The answer is B. Idk how to explain. Choose b.

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Answer:

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3 years ago

What are some predicted affects of climate change linked to global warming

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Decrease in atmospheric ozone, which results in lower protection from various stellar particles

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3 years ago

1. A heat engine operates between two reservoirs at T2 = 600 K and T1 = 350 K. It takes in 1.00 x 103 J of energy from the highe

Alex73 [517]

Answer:

$\Delta S_u=2.1429\ J.K^{-1}$

$W_c=416.67\ J$

Explanation:

Given:

temperature of source reservoir, $T_H=600\ K$

temperature of sink reservoir, $T_L=350\ K$

energy absorbed from the source, $Q_{in}=1000\ J$

work done, $W=250\ J$

Now change in entropy of the surrounding:

$\Delta S_u=\frac{dQ_L}{T_L}$

Since heat engine is a device that absorbs heat from a high temperature reservoir and does some work giving out heat in the universe as the byproduct.

$\Delta S_u=\frac{Q_H-W}{T_L}$

$\Delta S_u=\frac{1000-250}{350}$

$\Delta S_u=2.1429\ J.K^{-1}$

We know Carnot efficiency is given as:

$\eta_c=1-\frac{T_L}{T_H}$

$\eta_c=1-\frac{350}{600}$

$\eta_c=0.4167$

Now the Carnot work done:

$W_c=Q_H\times \eta_c$

$W_c=1000\times 0.4167$

$W_c=416.67\ J$ .......................(1)

From eq. (1) we have the Carnot work, so the difference:

$\Delta W=W_c-W$

$\Delta W=416.67-250$

$\Delta W=166.67\ J$

Now, we find:

$T_L.\Delta S_u=350\times 2.1429$

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3 years ago

Two small plastic spheres are given positive electrical charges. When they are 15.0 cm apart, the repulsive force between them h

Tems11 [23]

Answer:

a) q_1=q_2= 7.42*10^-7 C

b) q_2= 3.7102*10^-7 C , q_1 = 14.8*10^-7 C

Explanation:

Given:

F_e = 0.220 N

separation between spheres r = 0.15 m

Electrostatic constant k = 8.99*10^9

Find: charge on each sphere

part a

q_1 = q_2

Using coulomb's law:

F_e = k*q_1*q_2 / r^2

q_1^2 = F_e*r^2/k

q_1=q_2= sqrt (F_e*r^2/k)

Plug in the values and evaluate:

q_1=q_2= sqrt (0.22*0.15^2/8.99*10^9)

q_1=q_2= 7.42*10^-7 C

part b

q_1 = 4*q_2

Using coulomb's law:

F_e = k*q_1*q_2 / r^2

q_2^2 = F_e*r^2/4*k

q_2= sqrt (F_e*r^2/4*k)

Plug in the values and evaluate:

q_2= sqrt (0.22*0.15^2/4*8.99*10^9)

q_2= 3.7102*10^-7 C

q_1 = 14.8*10^-7 C

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3 years ago

Read 2 more answers

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