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worty [1.4K]
4 years ago
12

Some people wish that we lived in a recollapsing universe that would eventually stop expanding and start contracting. For this t

o be the case, which of the following would have to be true (based on current understanding)?)? (A) Dark energy is the dominant form of energy in the cosmos. (B) Dark energy does not exist and there is much more dark matter than we are aware of to date. (C) Neither dark energy nor dark matter really exist. (D) Dark energy exists but dark matter does not.
Physics
1 answer:
Monica [59]4 years ago
4 0

Answer:

(B) Dark energy does not exist and there is much more matter than current evidence suggests.

Explanation:

The repulsive force which is accelerating expansion of the universe is called as dark energy. Most of matter present in the universe is the dark matter of about eighty five percent.

So, a collapsing universe would not have the dark energy and there is more matter which is not the dark matter. This theory is rejected because expansion of the universe is observable.

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vA child is in danger of drowning in the Merimac river. The Merimac river has a current of 3.1 km/hr to the east. The child is 0
ikadub [295]

Answer:

d = 2.26 km

Explanation:

Let the child is moving with speed same as the speed of water flow

So here the position of child with respect to flow must be zero

And if the boat start at an angle with the vertical

so its relative speed with flow of water is given as

v_x = 24.8 sin\theta

v_y = 24.8 cos\theta

now the time to reach the child is given as

\frac{0.6}{24.8 cos\theta} = \frac{2.5}{24.8 sin\theta }

so now we have

\theta = 76.5 degree

So the time to catch the child is given as

t = \frac{0.6}{24.8 cos78.2}

t = 0.104 h

So distance moved by it in 0.104 h

distance moved by the boat in upstream direction given as

x = (24.8 sin 74.8 - 3.1)(0.104)

x = 2.18 km

In y direction the displacement of boat is

y = 0.6 km

net displacement of the ball is given as

d = \sqrt{x^2 + y^2}

d = \sqrt{2.18^2 + 0.6^2}

d = 2.26 km

4 0
3 years ago
Consider dropping a ball from rest. This ball moves from astate of high gravitational potential energy to one of lowgravitationa
Leni [432]

Answer:

From the negative to the positive cable.

Explanation:

The electrons have negative charge, which means that the negative terminal of the battery will suply the electrons, thus they are present in excess on the negative cable and will jump from it to the positive cable. This current direction is called real current.

3 0
3 years ago
Because energy is conserved, the “lost” energy has actually been changed into other forms. Looking at the two sleds, what effect
Shkiper50 [21]

Answer:

The answer is the kinetic energy of sled B after it crash is 6439j.

3 0
2 years ago
You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=
velikii [3]

Answer:

Approximately 4.2\; {\rm s} (assuming that the projectile was launched at angle of 35^{\circ} above the horizon.)

Explanation:

Initial vertical component of velocity:

\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}.

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing y_{1} is the same as the altitude y_{0} at which this projectile was launched: y_{0} = y_{1}.

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is 20.6\; {\rm m\cdot s^{-1}} (upwards,) the vertical velocity right before landing would be (-20.6\; {\rm m\cdot s^{-1}}) (downwards.) The change in vertical velocity is:

\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}.

Since there is no drag on this projectile, the vertical acceleration of this projectile would be g. In other words, a = g = -9.81\; {\rm m\cdot s^{-2}}.

Hence, the time it takes to achieve a (vertical) velocity change of \Delta v_{y} would be:

\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}.

Hence, this projectile would be in the air for approximately 4.2\; {\rm s}.

8 0
1 year ago
Read 2 more answers
A change in position of an object relating to time
MaRussiya [10]
The answer is: Motion!

Have a great day
3 0
3 years ago
Read 2 more answers
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