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3 years ago

Conceptual question about relativity: Imagine two events--

1 answer:

6 0

They are not the same event in that they occur in different places and times in most frames of reference. In the photon's frame they are not separated in either space nor time because photons don't experience time and at least mathematically all points on the spacetime manifold are the same point to a photon. What the zero spacetime interval can tell us though, is that the events are connected by a light beam (light-like separation). There is as much time between the events as there is space and one event can conceptually cause the other. They are on the cusp between time-like and space-like events.

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Which statement is true about the reaction shown by this chemical equation?

KatRina [158]

Answer:

it's C endothermic

Explanation:

5 0

2 years ago

The distance between two stations is 180 km. A train takes 2 hours to cover this distance. The speed of the train in m/sec is...

Mamont248 [21]

Answer:

Av = 25 [m/s]

Explanation:

To solve this problem we must use the definition of speed, which is defined as the relationship between distance over time. for this case we have.

$Av=\frac{distance}{time}$

where:

Av = speed [km/h] or [m/s]

distance = 180 [km]

time = 2 [hr]

Therefore the speed is equal to:

$Av = \frac{180}{2} \\Av = 90 [km/h]$

Now we must convert from kilometers per hour to meters per second

$90[\frac{km}{h}]*1000[\frac{m}{1km}]*1[\frac{h}{3600s} ]= 25 [m/s]$

4 0

2 years ago

A parallel-plate capacitor with plates of area 360 cm2 is charged to a potential difference V and is then disconnected from the

Softa [21]

Answer:

$Q=3.9825\times 10^{-9} C$

Explanation:

We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.

1 m =100 cm

Surface area =S= $\frac{360}{10000}=0.036 m^2$

$\Delta d=0.8 cm=0.008 m$

$\Delta V=100 V$

We have to find the charge Q on the positive plates of the capacitor.

V=Initial voltage between plates

d=Initial distance between plates

Initial Capacitance of capacitor

$C=\frac{\epsilon_0 S}{d}$

Capacitance of capacitor after moving plates

$C_1=\frac{\epsilon_0 S}{(d+\Delta d)}$

$V=\frac{Q}{C}$

Potential difference between plates after moving

$V=\frac{Q}{C_1}$

$V+\Delta V=\frac{Q}{C_1}$

$\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}$

$\frac{Q(d+\Delta d)}{\epsilon_0 S}-\frac{Qd}{\epsilon_0S}=100$

$\frac{Q\Delta d}{\epsilon_0 S}=100$

$\epsilon_0=8.85\times 10^{-12}$

$Q=\frac{100\times 8.85\times 10^{-12}\times 0.036}{0.008}$

$Q=3.9825\times 10^{-9} C$

Hence, the charge on positive plate of capacitor= $Q=3.9825\times 10^{-9} C$

6 0

3 years ago

A charge q1 of -5.00 x 10^-9 C and a charge q2 of -2.00x 10^-9 C are separated by a distance of 40.0 cm. Find the equilibrium po

aleksandrvk [35]

Answer:

Explanation:

Let the equilibrium position of third charge be x distance from q₁.

Force on third charge due to q₁

= 9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x²

Force on third charge due to q₂

= 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²

Both the force will act in opposite direction and for balancing , they should be equal.

9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x² = 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²

5 / x² = 2 / ( .4 - x )²

Taking square root on both sides

2.236 / x = 1.414 / .4 - x

2.236 ( .4 - x ) = 1.414 x

.8944 - 2.236 x = 1.414 x

.8944 = 3.65 x

x = .245 m

24.5 cm

So the third charge should be at a distance of 24.5 cm from q₁ .

4 0

2 years ago

Please answer help me

Andrews [41]

I believe the answer is x

7 0

3 years ago