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Fudgin [204]
3 years ago
8

How many universes out there?​

Physics
2 answers:
bonufazy [111]3 years ago
8 0
Only one but there are theories that there are more
Montano1993 [528]3 years ago
7 0

Answer:

onlu 1 universe

Explanation:

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What is the equivalent resistance for a series circuit with three resistors : 5.0 ohms, 2.0 ohms, and 12.0 ohms
vesna_86 [32]

Answer:19ohms

Explanation:

equivalent resistance=5+2+12

equivalent resistance=19ohms

8 0
3 years ago
Two blocks are connected by a massless cable which goes through the center of a rotating turntable. The blocks have masses M1 =
Airida [17]

Answer:

If the final question is; at what velocity will the first block start to move outward in m/s?

v = 3.5596 \frac{m}{s}

Explanation:

The motion have the velocity that will make the block move using:

F_{1}*F_{r}+ F_{2}= Ec \\Ec= \frac{M*v^{2} }{r}

m_{1} = 1.2 Kg

m_{2} = 2.8 Kg

r = 0,45 m

μs= 0,54

Resolving:

\frac{m_{1}*v^{2}  }{r} = us*m_{1}* g + m_{2} * g

v^{2} = \frac{((us *m_{1} + m_{2})*g )* r}{m_{1} }

v^{2} = \frac{((0.54 *1.2 + 2.8)*9.8 )* 0.45}{1.2}

v^{2} = 12.6714 \frac{m^{2} }{s^{2} }

v = 3.559 \frac{m}{s}

4 0
3 years ago
A bird is about 6.26.2 in.​ long, with a​ thin, dark bill and a​ wide, white wing stripe. If the bird can fly 9292 mi with the w
Trava [24]

Answer:

209 mph

Explanation:

V = Speed of bird in still air

v = Speed of wind = 44 mph

Consider the motion of the bird with the wind

D_{1} = distance traveled with the wind = 9292 mi

t_{1} = time taken to travel the distance with wind

Time taken to travel the distance with wind is given as

t_{1} = \frac{D_{1}}{V + v}

t_{1} = \frac{9292}{V + 44}                              eq-1

Consider the motion of the bird with the wind

D_{2} = distance traveled against the wind = 6060 mi

t_{2} = time taken to travel the distance against wind

Time taken to travel the distance against wind is given as

t_{2} = \frac{D_{2}}{V + v}

t_{2} = \frac{6060}{V - 44}                              eq-2

As per the question,

Time taken with the wind = Time taken against the wind

t_{1} = t_{2}

\frac{9292}{V + 44} = \frac{6060}{V - 44}

(9292) (V - 44) = (6060) (V + 44)

9292V - 408848 = 6060V + 266640

3232V = 675488

V = 209 mph

5 0
3 years ago
The energy of an electromagnetic wave is related to its amplitude. <br> a. True<br> b. False
barxatty [35]
The answer to your question is True
6 0
3 years ago
Read 2 more answers
A shot-putter released the shot at an angle of 41.5 degrees and a height of 1.9 m with an initial velocity of 13.3 m/s. How far
liq [111]

Answer:

x = 17.88[m]

Explanation:

We can find the components of the initial velocity:

(v_{x})_{o}  = 13.3*cos(41.5)=9.96[m/s]\\(v_{y})_{o}  = 13.3*sin(41.5)=8.81[m/s]

We have to remember that the acceleration of gravity will be worked with negative sign, since it acts in the opposite direction to the movement in direction and the projectile upwards.

g = - 9.81[m/s^2]

Now we must find the time it takes for the projectile to hit the ground, as the problem mentions that it does not impact on the board.

y=y_{o} +(v_{y} )_{o} *t-0.5*g*(t)^{2} \\0=1.9+(8.81*t)-(4.905*t^{2})\\-1.9=8.81*t*(1-0.5567*t)\\t=0\\t=1.796[s]

With this time we can calculate the horizontal distance:

x=(v_{x})_{o} *t\\x=9.96*1.796\\x=17.88[m]

3 0
3 years ago
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