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3 years ago

TRUE OR FALSE The nearby galaxy in the spectrum below is moving toward the close star.

Physics

1 answer:

katen-ka-za [31]3 years ago

5 0

Answer:

i think its false tell me if im wrong

Explanation:

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an ice sheet 5m thick covers a lake that is 20m deep. at what is the temperature of the water at the bottom of the lake?

muminat

Answer:

4°C

Explanation:

Water is densest at 4°C. Since dense water sinks, the bottom of the lake will be 4°C.

6 0

4 years ago

A straight vertical wire carries a current of 1.0 A into the page in a region where the magnetic field strength is 0.60 T. What

Advocard [28]

Answer:

force on the wire of section cm will be $6\times 10^{-3}N$

Direction of force on the wire will be in south direction

Explanation:

We have given current in the wire i = 1 A

Magnetic field strength B = 0.6 T

We have to find the force on 1 cm section of the wire so l = 1 cm = 0.01 m

Force on the wire containing current is equal to

$F=iBL$

$F=1\times 0.6\times 0.01=6\times 10^{-3}N$

So force on the wire of section cm will be $6\times 10^{-3}N$

Direction of force on the wire will be in south direction

5 0

3 years ago

Calculate the potential V(r) for r>rb. (Hint: The net potential is the sum of the potentials due to the individual spheres.)

Marat540 [252]

Answer:

The potential for r > rb is equal to zero.

Explanation:

For r > rb, the potential is:

$V=\frac{Kq}{r}$

Then, the net potential is:

$V_{(r)} =\frac{K(+\epsilon )}{r} +\frac{K(-\epsilon )}{r}$

$K=\frac{1}{4\pi \epsilon _{o} }$

$V_{(r)} =\frac{K(+\epsilon )}{r} -\frac{K(\epsilon )}{r}\\V_{(r)}=0$

8 0

4 years ago

A fisherman is fishing from a bridge and is using a "50.0-N test line." In other words, the line will sustain a maximum force of

umka21 [38]

Answer:(a) 50 N

(b)38.34 N

Explanation:

Given

Maximum tension(T) in line 50 N

(a)If line is moving up with constant velocity i.e. there is no acceleration

This will happen when Tension is equal to weight of Fish

T-mg=0

T=mg

Maximum weight in this case will be 50 N

(b)acceleration of magnitude $2.98 m/s^2$

T-mg=ma

$50=m\left ( g+a\right )$

$50=m\left ( 12.78\right )$

m=3.91

Therefore weight is $3.91\times 9.8=38.34 N$

7 0

3 years ago

The total pressure inside a 1.43 L cylinder at 28C is 1.03 atm. If the temperatures drops to -35C, and the volume remains the sa

Art [367]

Answer:

0.81452 atm

Explanation:

$P_1$ = Initial pressure = 1.43 L

$V_1=V_2$ = Volume

$T_1$ = Initial temperature = (28+273.15) K

$T_2$ = Final temperature = (-35+273.15) K

$P_2$ = Final pressure

From the ideal gas law we have

$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow P_2=\dfrac{P_1V_1T_2}{T_2V_2}\\\Rightarrow P_2=\dfrac{1.03\times 1.43\times (273.15-35)}{(273.15+28)1.43}\\\Rightarrow P_2=0.81452\ atm$

The pressure in the cylinder is 0.81452 atm

8 0

3 years ago