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2 years ago

What is the mystery behind black hole

2 answers:

lisabon 2012 [21]2 years ago

5 0

Well since the gravitational pull is so strong, nothing can escape from it which means we haven't been able to get close enough to really know everything about them. There's no way of knowing what truly goes on inside of them since there is no way out.

muminat2 years ago

3 0

Answer:

A black hole is a region of spacetime where gravity is so strong that nothing—no particles or even electromagnetic radiation such as light—can escape from it. The theory of general relativity predicts that a sufficiently compact mass can deform spacetime to form a black hole.

Explanation:

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A thick steel sheet of area 100 in.2 is exposed to air near the ocean. After a one-year period it was found to experience a weig

vladimir1956 [14]

Answer:

Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.

Explanation:

The corrosion rate is the rate of material remove.The formula for calculating CPR or corrosion penetration rate is

$CPR=\frac{KW}{DAT}$

K= constant depends on the system of units used.

W= weight =485 g

D= density =7.9 g/cm³

A = exposed specimen area =100 in² =6.452 cm²

K=534 to give CPR in mpy

K=87.6 to give CPR in mm/yr

mpy

$CPR=\frac{KW}{DAT}$

$=\frac{534\times( 485g)\times( 10^3mg/g)}{(7.9g/cm^3) \times (100in^2)\times (24h/day)\times (365day/yr)\times 1yr}$

=37.4mpy

mm/yr

$CPR=\frac{KW}{DAT}$

$=\frac{87.6\times (485g)\times (10^3 mg/g)}{(7.9g/cm^3)\times (100in^2)\times(2.54cm/in)^2\times (24h/day)\times (365day/yr)\times 1yr}$

=0.952 mm/yr

Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.

3 0

2 years ago

Two long, straight wires are separated by a distance of 9.15 cm . One wire carries a current of 2.79 A , the other carries a cur

Dafna1 [17]

Answer:

The force is the same

Explanation:

The force per meter exerted between two wires carrying a current is given by the formula

$\frac{F}{L}=\frac{\mu_0 I_1 I_2}{2\pi r}$

where

$\mu_0$ is the vacuum permeability

$I_1$ is the current in the 1st wire

$I_2$ is the current in the 2nd wire

r is the separation between the wires

In this problem

$I_1=2.79 A\\I_2=4.36 A\\r = 9.15 cm = 0.0915 m$

Substituting, we find the force per unit length on the two wires:

$\frac{F}{L}=\frac{(4\pi \cdot 10^{-7})(2.79)(4.36)}{2\pi (0.0915)}=2.66\cdot 10^{-5}N$

However, the formula is the same for the two wires: this means that the force per meter exerted on the two wires is the same.

The same conclusion comes out from Newton's third law of motion, which states that when an object A exerts a force on an object B, then object B exerts an equal and opposite force on object A (action-reaction). If we apply the law to this situation, we see that the force exerted by wire 1 on wire 2 is the same as the force exerted by wire 2 on wire 1 (however the direction is opposite).

3 0

2 years ago

Which of the physical variables listed below will change when you change the area of the capacitor plates (while keeping the bat

RSB [31]

Answer:

a. Capacitance

b. Charge on the plates

e. Energy stored in the capacitor

Explanation:

Let A be the area of the capacitor plate

The capacitance of a capacitor is given as;

$C = \frac{Q}{V} = \frac{\epsilon _0 A}{d} \\\\$

where;

V is the potential difference between the plates

The charge on the plates is given as;

$Q = \frac{V\epsilon _0 A}{d}$

The energy stored in the capacitor is given as;

$E = \frac{1}{2} CV^2\\\\E = \frac{1}{2} (\frac{\epsilon _0 A}{d} )V^2$

Thus, the physical variables listed that will change include;

a. Capacitance

b. Charge on the plates

e. Energy stored in the capacitor

3 0

2 years ago

A car travels up a hill at a constant speed of 38 km/h and returns down the hill at a constant speed of 66 km/h. Calculate the a

mojhsa [17]

Answer:

Average speed will be 48.23 km/h

Explanation:

Let the distance up to hill is = d km

Speed when car goes to hill = 38 km/h

So time required $t=\frac{distance}{speed}=\frac{d}{38}hour$

Speed when car return from hill = 66 km/h

So time required to return fro hill $t=\frac{d}{66}h$

Total time $t_{total}=\frac{t}{38}+\frac{t}{66}$

Total distance = d+d =2d

So average speed $=\frac{total\ distance}{total\ time}=\frac{2d}{\frac{d}{38}+\frac{d}{66}}=48.23km/h$

8 0

2 years ago

Small rockets are used to make small adjustments in the speed of satellites. One such rocket has a thrust of 42 N. If it is fire

Over [174]

To solve this problem we will apply the concepts related to Newton's second law that relates force as the product between acceleration and mass. From there, we will get the acceleration. Finally, through the cinematic equations of motion we will find the time required by the object.

If the Force (F) is 42N on an object of mass (m) of 83000kg we have that the acceleration would be by Newton's second law.

$F = ma \rightarrow a = \frac{F}{m}$