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3 years ago

Do wormholes exist ?? is it possible to traverse to the other side of the universe ?

Physics

1 answer:

Nastasia [14]3 years ago

8 0

Black holes is exists
Black holes is highly gravitional wave field in space

You might be interested in

Jan ran 4 miles north in 28 minutes. What was Jan's average velocity?

fenix001 [56]

Answer:

3.83 m/s

Explanation:

Given that,

Distance covered by Jan, d = 4 miles

1 mile = 1609.34 m

4 miles = 6437.38 m

Time, t = 28 minutes = 1680 s

Jan's average speed,

v = d/t

$v=\dfrac{6437.38\ \text{m}}{1680\ \text{s}}\\\\v=3.83\ \text{m/s}$

Hence, the average velocity of Jan is 3.83 m/s.

6 0

3 years ago

What mass of a material with density Ï is required to make a hollow spherical shell having inner radius r1 and outer radius r2?

inysia [295]

<span>Yes, there are! r1 and r2 are numbers. The volume of the hollow shell is 4 π 3 ( r 3 1 − r 3 2 ) 4π3(r13−r23). Now multiply by ρ to get the mass.</span>

3 0

2 years ago

Consider an opaque horizontal plate that is well insulated on its back side. The irradiation on the plate is 2500 W/m2, of which

GuDViN [60]

Answer:

i. 0.34

ii. 0.4

iii. 1700 w/m²

iv. 2211.36 w/m²

Explanation:

Given that

Irradiation of the plate, G = 2500 w/m²

Reflected rays, p = 500 w/m²

Emissive power, E = 1200 w/m²

See attachment for calculations

8 0

3 years ago

Can you make the work output of a machine greater than the work input?

drek231 [11]

Nearly equal the output work is greater than the input work because of friction.All machines use some amount of input work to overcome friction.The only way to increase the work output is to increase the work you put into the machine.You cannot get more work out of a machine than you put into it.

4 0

3 years ago

A wire lying along a y axis form y=0 to y=0.25 m carries a current of 2.0 mA in the negative direction of the axis. The wire ful

balu736 [363]

Answer:

FB = 0.187 N

Explanation:

To find the magnetic force FB in the wire you use the following formula:

$|\vec{F_B}|=ILBsin\theta\\\\L=0.25m\\\\|\vec{B}|=\sqrt{(0.3y)^2+(0.4y)^2}=0.5y \ T$

the angle between B and L is given by:

$\vec{L}\cdot\vec{B}=LBcos\theta\\\\\theta=cos^{-1}(\frac{\vec{L}\cdot\vec{B}}{LB})=cos^{-1}(\frac{0*0.3y+0.25*0.4y}{0.25*0.5y})=36.86\°$

Due to B depends on "y" you take into account the contribution of each element dy of the wire to the magnitude of the magnetic force. Thus, you have to integrate the following expression:

$|\vec{F_B}|=Isin\theta\int_0^{0.25}B(y)dy=Isin\theta\int_0^{0.25}(0.5y)dy\\\\|\vec{F_B}|=(2.0*10^{-3}A)(sin36.86\°)(0.5T)[\frac{0.25^2}{2}m]=0.187\ N$

hence, the magnitude of the magnetic force is 0.187N

4 0

2 years ago