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3 years ago

Define time in related with gravity.

Physics

1 answer:

lakkis [162]3 years ago

5 0

Answer:

The gravitational field is really a curving of space and time. The stronger the gravity, the more spacetime curves, and the slower time itself proceeds.Time itself is slowing down and speeding up because of the relativistic way in which mass warps space and time.

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Two parallel metal plates are at a distance of 8.00 m apart.The electric field between the plates is uniform directed towards th

dezoksy [38]

<h2>The K.E of the charge is 1.02 x 10⁻¹⁷ J</h2>

Explanation:

When the charge of 2e is placed in between the plates .

The force applied on this charge by plates is = q E

here q is the magnitude of charge = 2 e = 2 x 1.6 x 10⁻¹⁹ C

and E is the magnitude of electric field intensity

The work done = Force x displacement

Thus W = q E x S

here S is displacement

Therefore W = 2 x 1.6 x 10⁻¹⁹ x 4 x 8

= 1.02 x 10⁻¹⁷ J

This work will be converted into the kinetic energy of charge .

Thus K.E = 1.02 x 10⁻¹⁷ J

4 0

3 years ago

Use dimensional analysis to determine how the linear acceleration a in m/s2 of a particle traveling in a circle depends on some,

astraxan [27]

I have to say, i love this kind of problems.

So, we got the linear acceleration, as we all know, linear the acceleration its, in dimensional units:

$[a]=[\frac{distance}{time^2}]$ .

Now, we got the radius

$[r] = [distance]$

the angular frequency

$[\omega] = \frac{1}{s}$

and the mass

$[m]=[mass]$ .

Now, the acceleration doesn't have units of mass, so it can't depend on the mass of the particle.

The distance in the acceleration has exponent 1, and so does in the radius. As the radius its the only parameter that has units of distance, this means that the radius must appear with exponent 1. Lets write

$a \propto r$ .

The time in the acceleration has exponent -2 As the angular frequency its the only parameter that has units of time, this means that the angular frequency must appear, but, the angular frequency has an exponent of -1, this means it must be squared

$a \propto r \omega^2$ .

We are almost there. If this were any other problem, we would write:

$a = A r \omega^2$

where A its an dimensionless constant. Its common for this constants to appears if we need an conversion factor. If we wanted the acceleration in cm/s^2, for example. Luckily for us, the problem states that there is no dimensionless constant involved, so:

$a = r \omega^2$

5 0

3 years ago

A ball is dropped from a rooftop 60m high. <br> How long is the ball in the air?

alina1380 [7]

Answer: 3.49 s

Explanation:

We can solve this problem with the following equation of motion:

$y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}$ (1)

Where:

$y=0 m$ is the final height of the ball

$y_{o}=60 m$ is the initial height of the ball

$V_{o}=0 m/s$ is the initial velocity (the ball was dropped)

$g=9.8 m/s^{2}$ is the acceleratio due gravity

$t$ is the time

Isolating $t$ :

$t=\sqrt{\frac{2 y_{o}}{g}}$ (2)

$t=\sqrt{\frac{2 (60 m)}{9.8 m/s^{2}}}$ (3)

Finally we find the time the ball is in the air:

$t=3.49 s$ (4)

7 0

4 years ago

An automobile travels on a straight road for 40 km at 30 km/h. It then continues in the same direction for another 40 km at 60 k

Butoxors [25]

Answer:

The average velocity is 40km/h.

Explanation:

The average velocity is $\bar{v}=\frac{\Delta x }{\Delta t}$ , where $\Delta x$ is the distance traveled and $\Delta t$ the time elapsed.

The distance traveled is clearly 80km since it's all done in the same direction, we only need to know the time elapsed. For this we calculate the time elapsed on the first part, and add it to the time elapsed on the second part using always the formula $\Delta t=\frac{\Delta x }{v}$ , where v is the velocity on each part, which is constant.

The time elapsed for the first part is $\Delta t_1=\frac{40 km}{30km/h}=\frac{4}{3}h$ , and the time elapsed for the second part is $\Delta t_2=\frac{40 km}{60km/h}=\frac{2}{3}h$ , giving us a total time of $\Delta t_1+\Delta t_2=\frac{4}{3}h+\frac{2}{3}h$ =2h.