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3 years ago

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The behavior of which objects are better explained by Einstein’s general theory of relativity than Newton’s universal law of gra

vitation?
A. objects that accelerate
B. objects that are several meters apart
C. objects with small masses
D. objects moving at near the speed of light

2 answers:

liq [111]3 years ago

7 0

If I had to go with any of those answers, It would be A maybe D, But im not too sure on how to decide between them. Because Einstein mentioned the sun in his theory which has a very large mass <span> 1.989 x 10 with a exponent of 30 to be exact. Hope this helped though.</span>

Westkost [7]3 years ago

7 0

objects moving at near the speed of light

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On your first day at work as an electrical technician, you are asked to determine the resistance per meter of a long piece of wi

Lostsunrise [7]

Answer:

$0.06\Omega/m$

Explanation:

Firstly, when you measure the voltage across the battery, you get the emf,

E = 13.0 V

In order to proceed we have to assume that the voltmeter offers no loading effect, which is a valid assumption since it has a very high resistance.

Secondly, the wires must be uniform. So the resistance per unit length is constant (say z). Now, even though the ammeter has very little resistance it cannot be ignored as it must be of comparable value/magnitude when compared to the wires. This is can seen in the two cases when currents were measured. Following Ohm's law and the resistance of a length of wire being proportional to it's length, we should have gotten half the current when measuring with the 40 m wire with respect to the 20 m wire ( $I=\frac{V}{R}$ ). But this is not the case.

Let the resistance of the ammeter be r

Hence, using Ohm's law we get the following 2 equations:

$\frac{13}{20z+r} =7.6$ .......(1)

$\frac{13}{40z+r} =4.5$ ......(2)

Substituting the value of r from (2) in (1), we have,

$13=152z+7.6\times\frac{13-180z}{4.5}$

which simplifying gives us, $z=0.0589\Omega/m\approx0.06\Omega/m$ (which is our required solution)

putting the value of z in either (1) or (2) gives us, r = 0.5325 $\Omega$

3 0

3 years ago

A person's body is producing energy internally due to metabolic processes. If the body loses more energy than metabolic processe

Allushta [10]

To solve this problem it is necessary to apply the concepts related to the Stefan-Boltzman law that is responsible for calculating radioactive energy.

Mathematically this expression can be given as

$P = \sigma Ae\Delta T^4$

Where

A = Surface area of the Object

$\sigma =$ Stefan-Boltzmann constant

e = Emissivity

T = Temperature (Kelvin)

Our values are given as

$A = 1.36m^2$

$\Delta T^4 = T_2^4 -T_1^4 = 307^4-T_1^4$

$\sigma = 5.67*10^{-8} J/(s m^2 K^4)$

$P = 122J/s$

$e = 0.7$

Replacing at our equation and solving to find the temperature 1 we have,

$P = \sigma Ae\Delta T^4$

$P = \sigma Ae (T_2^4 -T_1^4)$

$122 = (5.67*10^{-8})(1.36)(0.7)(307^4-T_1^4)$

$T_1 = 285.272K = 12.122\°C$

Therefore the the temperature of the coldest room in which this person could stand and not experience a drop in body temperature is 12°C

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3 years ago

A compact disc (CD) is played by a cd player, which uses a laser to read the tracks on the disc. The disc spins initially at app

uranmaximum [27]

Answer:

a. The laser tracking mechanism experiences a changing tangential velocity

c. The laser tracking mechanism experiences a non-zero angular acceleration

d. The laser tracking mechanism experiences a non-zero tangential acceleration

Explanation:

a. The laser tracking mechanism experiences a changing tangential velocity

This is because the tangential velocity v = rω where r = radius of disc and ω = angular speed of discs. Since r is constant, v ∝ ω.

Since the angular speed changes from 200 rpm to 500 rpm, thus, the tangential velocity would also change.

So, the laser tracking mechanism experiences a changing tangential velocity

c. The laser tracking mechanism experiences a non-zero angular acceleration

Since angular acceleration, α = Δω/Δt where Δω = change in angular speed and Δt = change in time.

Since there is a change in angular speed from 200 rpm to 500 rpm in time Δt, there is thus a non-zero angular acceleration.

So, The laser tracking mechanism experiences a non-zero angular acceleration

d. The laser tracking mechanism experiences a non-zero tangential acceleration

Since tangential acceleration, a = rα where r = radius of disc and α = angular acceleration.

Since there is an angular acceleration of the disc, there is thus going to be a tangential acceleration given by a = rα.

So, the laser tracking mechanism experiences a non-zero tangential acceleration

Statement b is false because, the disc experiences a changing angular speed from 200 rpm to 500 rpm.

4 0

2 years ago

suppose a child walks from the outer edge of a rotating Merry-Go-Round to the inside does angular velocity of the Merry-Go-Round

Akimi4 [234]

Okay here initially child walks on the outer edge of the disc and then started to move inside

So here as the child and Merry go round is an isolated system so there is no external Torque on this system from outside

As here we can see there is external force acting on this system by the hinge of the Merry go round as well as due to gravity so we can not use momentum conservation to solve such type of questions.

But as we can say that there is no external torque on this system about the hinge point so we will use conservation of angular momentum for this system

Here as we know that

$\tau = \frac{dL}{dt}$

where L = angular momentum

since here torque is ZERO

$0 = \frac{dL}{dt}$

L = constant

so here we can write initial angular momentum of the system as

$L = (I_1 + I_2)*\omega$

here we know that

$I_1$ = moment of inertia of merry go round

$I_2$ = moment of inertia of child

so here we can say

$(I_1 + I_2)* \omega_1 = (I_1 + I_2')\omega_2$

so here as the child moves from edge to inside the disc it moment of inertia will decrease because as we know that moment of inertia of child is given as

$I_2 = mr^2$

here m = mass of child

r = distance of child from center

Since child is moving inside so his distance from center is decreasing

so here moment of inertia of child is decreasing as he starts moving inside

so final angular speed of merry go round will increase as child go inside

$\omega_2 = \frac{(I_1 + I_2)*\omega}{(I_1 + I_2')}$

so here as

$I_2' < I_2$

final angular speed will be more than initial speed as child moves inside

3 0

3 years ago

A 200 kg wood crate sits in the back of a truck. The coefficients of friction between the crate and the truck are ????s = 0.9 an

Xelga [282]

Answer:

Maximum acceleration of the truck is 5.25 m/s^2

Explanation:

To find the maximum acceleration the truck first we need to calculate friction between truck and wood. Because of wood is not moving, coefficient of c_{s} need to be used for it. Then the formula will be:

$F_{s}=c_{s}*200*9.8*cos(20)\\F_{s}=0.9*200*9.8*0.94\\F_{s}=1657.62$

1657.62 Newton is the friction.

So force against friction need to be at most 1657.62 N. Then equation will be:

$F_{s}=F_{a}+F_{m}\\1657.62=200*9.8*sin(20)+200*a*cos(20)\\1657.62=670.36+187.94*a\\987.26=187.94*a\\a=5.25$

Maximum acceleration of the truck need to be 5.25 m/s^2

4 0

3 years ago