Amplitude is a Measure of a Waves

A car starts from 0 m along a road and accelerates at 0.5 m/s^2 to the right. A second car starts from 1000 m along the road and

Brrunno [24]

Answer:

e) 31.6 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

Equation of motion

$s=ut+\frac{1}{2}at^2\\\Rightarrow s_1=0\times t+\frac{1}{2}\times 0.5\times t^2\\\Rightarrow s_1=\frac{1}{2}0.5t^2\ m$

$s=ut+\frac{1}{2}at^2\\\Rightarrow s_2=0\times t+\frac{1}{2}\times 0.5\times t^2\\\Rightarrow s_2=\frac{1}{2}1.5t^2\ m$

$s_1+s_2=1000$

$\\\Rightarrow 1000=\frac{1}{2}0.5t^2+\frac{1}{2}1.5t^2\\\Rightarrow 1000=\frac{0.5t^2+1.5t^2}{2}\\\Rightarrow 1000=\frac{2t^2}{2}\\\Rightarrow 1000=t^2\\\Rightarrow t=\sqrt{1000}\\\Rightarrow t=31.6\ s$

Time taken by the cars to meet 31.6 seconds.

5 0

3 years ago

Compute the velocity of an object orbiting at height 2Re above the surface of earth

I am Lyosha [343]

Answer:

The orbital speed can be found using v = SQRT(G*M/R). The R value (radius of orbit) is the earth's radius plus the height above the earth - in this case, 6.59 x 106 m.

3 0

3 years ago

There are two parallel conductive plates separated by a distance d and zero potential. Calculate the potential and electric fiel

taurus [48]

Answer:

The total electric potential at mid way due to 'q' is $\frac{q}{4\pi\epsilon_{o}d}$

The net Electric field at midway due to 'q' is 0.

Solution:

According to the question, the separation between two parallel plates, plate A and plate B (say) = d

The electric potential at a distance d due to 'Q' is:

$V = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d}$

Now, for the Electric potential for the two plates A and B at midway between the plates due to 'q':

For plate A,

$V_{A} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}$

Similar is the case with plate B:

$V_{B} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}$

Since the electric potential is a scalar quantity, the net or total potential is given as the sum of the potential for the two plates:

$V_{total} = V_{A} + V_{B} = \frac{1}{4\pi\epsilon_{o}}.q(\frac{1}{\frac{d}{2}} + \frac{1}{\frac{d}{2}}$

$V_{total} = \frac{q}{4\pi\epsilon_{o}d}$

Now,

The Electric field due to charge Q at a distance is given by:

$\vec{E} = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d^{2}}$

Now, if the charge q is mid way between the field, then distance is $\frac{d}{2}$ .

Electric Field at plate A, $\vec{E_{A}}$ at midway due to charge q:

$\vec{E_{A}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}$

Similarly, for plate B:

$\vec{E_{B}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}$

Both the fields for plate A and B are due to charge 'q' and as such will be equal in magnitude with direction of fields opposite to each other and hence cancels out making net Electric field zero.

3 0

2 years ago

As a substance melts, heat energy is used to break the connections between molecules, and temperature __________ until all melti

mario62 [17]

Temperature stays the same

4 0

2 years ago

What total mass must be converted into energy

Eduardwww [97]

This question apparently wants you to get comfortable
with E = m c² . But I must say, this question is a lame
way to do it.

c = 3 x 10⁸ m/s
E = m c²

   1.03 x 10⁻¹³ joule = (m) (3 x 10⁸ m/s)²

Divide each side by (3 x 10⁸ m/s)²:

   Mass = (1.03 x 10⁻¹³ joule) / (9 x 10¹⁶ m²/s²)

   = (1.03 / 9) x (10⁻¹³ ⁻ ¹⁶) (kg)

   = 1.144 x 10⁻³⁰ kg . (choice-1)

This is roughly the mass of (1 and 1/4) electrons, so it seems
that it could never happen in nature. The question is just an
exercise in arithmetic, and not a particularly interesting one.
______________________________________

Something like this could have been much more impressive:

The Braidwood Nuclear Power Generating Station in northeastern
Ilinois USA serves Chicago and northern Illinois with electricity.
<span>The station has two pressurized water reactors, which can generate
a net total of 2,242 megawatts at full capacity, making it the largest
nuclear plant in the state.
If the Braidwood plant were able to completely convert mass
to energy, how much mass would it need to convert in order
to provide the total electrical energy that it generates in a year,
operating at full capacity ?

Energy = (2,242 x 10⁶ joule/sec) x (86,400 sec/day) x (365 da/yr)

   = (2,242 x 10⁶ x 86,400 x 365) joules

   = 7.0704 x 10¹⁶ joules .

How much converted mass is that ?

   E = m c²

Divide each side by c² : Mass = E / c² .
c = 3 x 10⁸ m/s

    Mass = (7.0704 x 10¹⁶ joules) / (9 x 10¹⁶ m²/s²)

      = 0.786 kilogram ! ! !

THAT should impress us ! If I've done the arithmetic correctly,
then roughly (1 pound 11.7 ounces) of mass, if completely
converted to energy, would provide all the energy generated
by the largest nuclear power plant in Illinois, operating at max
capacity for a year !

</span>

7 0

2 years ago

Read 2 more answers