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3 years ago

A car traveling west at 70 miles per hour for 3 hours covers a distance of

Physics

1 answer:

gtnhenbr [62]3 years ago

5 0

70x3=210 mph... its that easy

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PLEASE HELP ME I REALLY NEED HELP

allochka39001 [22]

Answer:

The answer is A.

Explanation:

The graph would show the unit rate otherwise known as a ratio

3 0

3 years ago

2) A 20 kg mass moving at a speed of 3 m/s is stopped by a constant force of 15 N. How many seconds must the force act on the ma

hammer [34]

Take the object's starting direction of motion to be the positive direction, so that a stopping force acts in the opposite direction. By Newton's second law, the object undergoes an acceleration a such that

-15 N = (20 kg) a

Solve for a :

a = - (15 N) / (20 kg) = -0.75 m/s²

The object's velocity v at time t is then given by

v = 3 m/s + (-0.75 m/s²) t

so the time it takes for the object to slow to a rest is

0 = 3 m/s + (-0.75 m/s²) t

t = (3 m/s) / (0.75 m/s²) = 4.0 s

3 0

3 years ago

Two capacitors, C1 = 25.0 μF and C2 = 31.0 μF, are connected in series, and a 6.0-V battery is connected across them.a. Find the

Brrunno [24]

Answer:

$13.83928\times 10^{-6}\ F$

$249.10704\times 10^{-6}\ J$

$137.89848\times 10^{-6}\ J$

$111.20842\times 10^{-6}\ J$

2.98273 V

Explanation:

$C_1=25\ mu F$

$C_2=31\ mu F$

V = Voltage = 6 V

Equivalent capacitance is given by

$\dfrac{1}{C}=\dfrac{1}{C_1}+\dfrac{1}{C_2}\\\Rightarrow C=\dfrac{C_1C_2}{C_1+C_2}\\\Rightarrow C=\dfrac{25\times 10^{-6}\times 31\times 10^{-6}}{(25+31)\times 10^{-6}}\\\Rightarrow C=13.83928\times 10^{-6}\ F$

Equivalent capacitance is $13.83928\times 10^{-6}\ F$

Energy stored is given by

$E=\dfrac{1}{2}CV^2\\\Rightarrow E=\dfrac{1}{2}\times 13.83928\times 10^{-6}\times 6^2\\\Rightarrow E=249.10704\times 10^{-6}\ J$

Total energy stored is $249.10704\times 10^{-6}\ J$

Charge is given by

$Q=CV\\\Rightarrow Q=13.83928\times 10^{-6}\times 6\\\Rightarrow Q=83.03568\times 10^{-6}\ C$

Voltage is given by

$V_1=\dfrac{Q}{C_1}\\\Rightarrow V_1=\dfrac{83.03568\times 10^{-6}}{25\times 10^{-6}}\\\Rightarrow V_1=3.3214272\ V$

$E_1=\dfrac{1}{2}C_1V_1^2\\\Rightarrow E_1=\dfrac{1}{2}\times 25\times 10^{-6}\times 3.3214272^2\\\Rightarrow E_1=137.89848\times 10^{-6}\ J$

Energy strored in C1 is $137.89848\times 10^{-6}\ J$

$V_2=\dfrac{Q}{C_2}\\\Rightarrow V_2=\dfrac{83.03568\times 10^{-6}}{31\times 10^{-6}}\\\Rightarrow V_2=2.67857\ V$

$E_2=\dfrac{1}{2}C_2V_2^2\\\Rightarrow E_2=\dfrac{1}{2}\times 31\times 10^{-6}\times2.67857^2\\\Rightarrow E_2=111.20842\times 10^{-6}\ J$

Energy stored in C2 is $111.20842\times 10^{-6}\ J$

$E=E_1+E_2\\\Rightarrow E=137.89848\times 10^{-6}+111.20842\times 10^{-6}\\\Rightarrow E=249.107\times 10^{-6}\ J$

So, the energy is equivalent

Equivalent capacitance

$C=C_1+C_2\\\Rightarrow C=25+31\\\Rightarrow C=56\times 10^{-6}\ F$

$E=\dfrac{1}{2}CV^2\\\Rightarrow V=\sqrt{\dfrac{2E}{C}}\\\Rightarrow V=\sqrt{\dfrac{2\times 249.107\times 10^{-6}}{56\times 10^{-6}}}\\\Rightarrow V=2.98273\ V$

The voltage would be 2.98273 V

8 0

4 years ago

15 points!

VMariaS [17]

Answer:

$4.12\times 10^{-5}\ J$ .

Explanation:

Given that,

Capacitance, $C=1.4\times 10^{-7}\ F$

Charge stored in the capacitor, $Q=3.4\times 10^{-6}\ C$

We need to find the electric potential energy stored in the capacitor. The formula for the electric potential energy stored in the capacitor is given by :

$E=\dfrac{Q^2}{2C}$

Put all the values,

$E=\dfrac{(3.4\times 10^{-6})^2}{2\times 1.4\times 10^{-7}}\\\\=4.12\times 10^{-5}\ J$

So, the required electric potential eenergy is equal to $4.12\times 10^{-5}\ J$ .

3 0

3 years ago

Suppose an endothermic reaction has a positive change in entropy greater than the heat absorbed. What can be said about the spon

pentagon [3]

If the entropy is greater than the enthalpy, it will have more spontinaity

8 0

4 years ago

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