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Norma-Jean [14]
3 years ago
5

Consider the circuit below, which is powered by a 8-v battery. switch s is opened at t = 0 after having been closed for a long t

ime.
Physics
1 answer:
GrogVix [38]3 years ago
4 0
The battery will be full still a 8v bc of no time comparison
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How does the change in the temperature of the universe provide evidence for universe expansion that supports the Big Bang Theory
xeze [42]

Question:

1) The universe is cooling which, according to the Big Bang Theory, is expected to happen as the cosmos accumulates.

2) The universe is warming which, according to the Big Bang Theory, is expected to happen as the cosmos disperses.

3) The universe is cooling which, according to the Big Bang Theory, is expected to happen as the cosmos disperses.

4) The universe is warming which, according to the Big Bang Theory, is expected to happen as the cosmos accumulates.​

Answer:

The correct option is;

3) The Universe is cooling which, according to the Big Bang Theory, is expected to happen as the cosmos disperses

Explanation:

With the temperature measurement carried out using the CSIRO radio telescope, Astronomers have been able to determine a temperature difference in the universe from 5.08 Kelvin 7.2 billion light years away to 2.73 Kelvin in the Universe today, which is in support of the Big Bang theory that as the Universe expanded from a state of extreme temperature that cools down as the Universe expands or the cosmos disperses.

8 0
2 years ago
A projectile is launched from ground level at an angle of 30 degrees above the horizontal. Neglect air resistance and consider t
Oduvanchick [21]

Answer:

just before landing the ground

Explanation:

Let the velocity of projection is u and the angle of projection is 30°.

Let T is the time of flight and R is the horizontal distance traveled. As there is no force acting in horizontal direction, so the horizontal velocity remains constant. Let the particle hits the ground with velocity v.

initial horizontal component of velocity, ux = u Cos 30

initial vertical component of velocity, uy = u Sin 30

Time of flight is given by

T = \frac{2u Sin\theta }{g}

Final horizontal component of velocity, vx = ux = u Cos 30

Let vy is teh final vertical component of velocity.

Use first equation of motion

vy = uy - gT

v_{y}=u_{y}- g \times \frac{2u Sin\theta }{g}

v_{y}=u Sin 30 - 2u Sin 30

vy = - u Sin 30

The magnitude of final velocity is given by

v = \sqrt{v_{x}^{2}+v_{y}^{2}}

v = \sqrt{\left (uCos 30  \right )^{2}+\left (uSin 30  \right )^{2}}

v = u

Thus, the velocity is same as it just reaches the ground.

6 0
3 years ago
On his fishing trip Justin takes the boat 25 km south. The fish aren’t biting so he goes 10 km west. He follows a school of fish
S_A_V [24]

Distance is 50 km

Displacement is 10 km

<u>Explanation:</u>

Given:

Distance toward south, x = 25 km

Distance towards west, y = 10 km

Distance towards north, z = 15 km

(a) Total distance, D = ?

Total distance, D = x + y + z

                       D = 25 + 10 + 15

                       D = 50km

(b) Displacement, d = ?

Displacement = final position - initial position

                      = 10 - 0 km

                      = 10km

8 0
2 years ago
Im confused on number one
mr Goodwill [35]

Explanation:

v=?, u=0, a=?, S=22m.

Using the formula, S=ut+½at²

22={0×5}+(½.a.5²)

22=½.a.5²

a=44/25 = 1.76m/s².

Therefore, net force = work done = ma = 48×1.76 = 84.48N.

therefore, power = work done/time = 84.48/5 = 16.896W.

hope this helps you.

5 0
2 years ago
Coherent light with wavelength 610 nm passes through two very narrow slits, and theinterference pattern is observed on a screen
andriy [413]

Answer:

= 1220 nm

= 1.22 μm

Explanation:

given data:

wavelength \lambda = 610 nm = 610\times  10 ^{-9} m

distance of screen from slits D = 3 m

1st order bright fringe is 4.84 mm

condition for 1 st bright is

d sin \theta =\lambda     ---( 1)

andtan \theta = \frac{y}{ D}

\theta = tan^{-1}\frac{(y }{D})

= 0.0924 degrees

plug theta value in equation 1 we get

d sin ( 0.0924) = 610 \times 10 ^{-9}

d = 3.78\times 10^{-4} m

condition for 1 st dark fringe

d sin \theta =\frac{λ'}{2}

\lambda '= 2 d sin\theta

= 2λ    since from eq (1)

= 1220 nm

= 1.22 μm

7 0
3 years ago
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