All categories

2 years ago

Help on this question ‼️

Physics

2 answers:

Yuri [45]2 years ago

4 0

Answer:

I think the option b

Is the answer

ipn [44]2 years ago

4 0

Answer:

Explanation:

after a CCW rotation it would become a waning gibbous, so after a Clockwise rotation it would be a waxing gibbous (opposite)

You might be interested in

Explain why theories may be changed or replaced overnight time.

vlabodo [156]

They may be changed because they may find evidence of some thing that will change their perspective on things.

5 0

3 years ago

A metal rod has a moves with a constant velocity of 40 cm/s along two parallel metal rails through a magnetic field of 0.575 T.

love history [14]

Answer:

2.12/R mW

Explanation:

The electrical power, P generated by the rod is

P = B²L²v²/R where B = magnetic field = 0.575 T, L = length of metal rod = separation of metal rails = 20 cm = 0.2 m, v = velocity of metal rod = 40 cm/s = 0.4 m/s and R = resistance of rod = ?

So, the induced emf on the conductor is

E = BLv

= 0.575 T × 0.2 m × 0.4 m/s

= 0.046 V

= 46 mV

The electrical power, P generated by the rod is

P = B²L²v²/R

= B²L²v²/R

So, P = (0.575 T)² × (0.2 m)² × (0.4 m/s)²

= 0.002116/R W

= 2.12/R mW

3 0

3 years ago

Objects with masses of 165 kg and a 465 kg are separated by 0.340 m. (a) Find the net gravitational force exerted by these objec

Dmitrij [34]

Answer:

(a) $F_{net}$ = 4.19 x $10^{-5}$ N, and its direction is towards $m_{2}$ .

(b) It must be placed inside a hollow shell.

Explanation:

Let, $m_{1}$ = 165 kg, $m_{2}$ = 465 kg, $m_{3}$ = 60 kg, and the distance between $m_{1}$ and $m_{2}$ is 0.340 m.

(a) Since $m_{3}$ is placed midway between $m_{1}$ and $m_{2}$ , then its distance to both masses is 0.170 m.

From the Newton's law of universal gravitation,

F = $\frac{Gm_{1}m_{2} }{r^{2} }$

Where all variables have their usual meaning.

Then,

a. $F_{net}$ = $F_{23}$ - $F_{13}$

$F_{13}$ = $\frac{6.67*10^{-11}*165*60 }{(0.17)^{2} }$

= 2.25 x $10^{-5}$ N

$F_{23}$ = $\frac{6.67*10^{-11}*465*60 }{(0.17)^{2} }$

= 6.44 x $10^{-5}$ N

∴ $F_{net}$ = = 6.44 x $10^{-5}$ - 2.25 x $10^{-5}$

= 4.19 x $10^{-5}$ N

The net force exerted by the two masses on the 60 kg object is 4.19 x $10^{-5}$ N.

(ii) / $F_{net}$ / = / $F_{23}$ / - / $F_{13}$ /

= 6.44 x $10^{-5}$ - 2.25 x $10^{-5}$

= 4.19 x $10^{-5}$ N

(iii) The direction of the net force is to the right i.e towards $m_{2}$ .

(b) For the net force experienced by the 60 kg object to be zero, it must be placed inside a hollow shell.

7 0

3 years ago

A spring vibrates 120 times in 2 mins find its frequency and time period

german

2 minutes is 120 seconds, so if you were finding vibrations per minute, it would be 60 times a minute.

3 0

2 years ago

Three basic classes of collisions include all but which of the following?

pickupchik [31]

Is mantle plate versus lithospheric plate ,one of the choices.?

3 0

3 years ago