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3 years ago

Determine the quantity of work done when a crane lifts a 100-N block from 2m above the ground to 6m above the ground

Physics

1 answer:

Sedaia [141]3 years ago

7 0

You can use the formula:
Work = Force × Distance
W = 100 * (6 - 2)
W = 100 * 4
W = 400 J

hope this helps :)

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According to Einstein's special theory of relativity, in which frames of reference will all laws of physics apply?

kenny6666 [7]

Answer: Within any frame of reference that is accelerating

Special relativity was proposed on 1905 by Einstein, who developed his theory based on the following two postulates:

1. The laws of physics are the same in all inertial systems. There is no preferential system.

2. The speed of light in vacuum has the same value for all inertial systems.

Focusing on the first postulate, it can be affirmed that any measurement on a body is made with reference to the system in which it is being measured.

Now, taking into account that an inertial reference system is the one that complies with the principle of inertia:

"For a body to have acceleration, an external force must act on it."

The correct answer is

Within any frame of reference that is accelerating

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4 years ago

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Charge is uniformly distributed around a ring of radius R and the resulting electric field magnitude E is measured along the rin

Arte-miy333 [17]

Answer:

$x=\dfrac{r}{\sqrt2}$

Explanation:

Given that

Radius =r

Electric filed =E

Q=Charge on the ring

The electric filed at distance x given as

$E=K\dfrac{Q}{(r^2+x^2)^{3/2}}$

For maximum condition

$\dfrac{dE}{dx}=0$

$E=K{Q}{(r^2+x^2)^{-3/2}}$

$\dfrac{dE}{dx}=K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}$

For maximum condition

$\dfrac{dE}{dx}=0$

$K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}=0$

$r^2+x^2-3x^2=0$

$x=\dfrac{r}{\sqrt2}$

At $x=\dfrac{r}{\sqrt2}$ the electric field will be maximum.

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3 years ago

A sample of argon gas (molar mass 40 g) is at four times the absolute temperature of a sample of hydrogen gas (molar mass 2 g).

qaws [65]

To solve this problem, let us recall that the formula for gases assuming ideal behaviour is given as:

rms = sqrt (3 R T / M)

where

R = gas constant = 8.314 Pa m^3 / mol K

T = temperature

M = molar mass

Now we get the ratios of rms of Argon (1) to hydrogen (2):

rms1 / rms2 = sqrt (3 R T1 / M1) / sqrt (3 R T2 / M2)

rms1 / rms2 = sqrt ((T1 / M1) / (T2 / M2))

rms1 / rms2 = sqrt (T1 M2 / T2 M1)

Since T1 = 4 T2

rms1 / rms2 = sqrt (4 T2 M2 / T2 M1)

rms1 / rms2 = sqrt (4 M2 / M1)

and M2 = 2 while M1 = 40

rms1 / rms2 = sqrt (4 * 2 / 40)

rms1 / rms2 = 0.447

Therefore the ratio of rms is:

rms_Argon / rms_Hydrogen = 0.45

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