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4 years ago

What is dark energy and what evidence do astronomers have that it is an important component of the universe?

Physics

1 answer:

PSYCHO15rus [73]4 years ago

8 0

Answer:

The existance of dark energy was suggested to help explain measurements, using Type Ia supernovae as distance indicators, that the expansion of the universe is speeding up and its acceleration requires a source of energy. Scientists suggested that It may be a new form of energy for which there is not yet a theoretical explanation. quantum mechanics predict that it may be the vacuum energy associated with "empty" space itself.

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How much energy is required to move 2 electrons through a potential difference of 1.0 x 10^ 2 volts?

Ksenya-84 [330]

Charge of electron = 1.6×10−¹⁹

(1.6×10−¹⁹)(1×10²) (2e)

= 3.2×10−¹⁷ J

3 0

3 years ago

What are tectonic plates, and which of Earht's layers are they composed of?

motikmotik

The tectonic plates are made up of Earth's crust and the upper part of the mantle layer underneath. Together the crust and upper mantle are called the lithosphere. hope this helps :)

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3 years ago

A car travels 82 meters do North and 14 seconds the car turns around and travels 44 m due south in four seconds what is the magn

Yakvenalex [24]

Answer:

2.11 m/s

Explanation:

Take north to be positive and south to be negative.

Average velocity = displacement / time

v = (82 m + -44 m) / (14 s + 4 s)

v = 2.11 m/s

The velocity is positive, so it is 2.11 m/s north. The magnitude of the velocity is 2.11 m/s.

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3 years ago

A tennis ball travels the length of the court in 24 m in .5 seconds find its average speed

maxonik [38]

Speed= Distance/Time
Distance=24m
Time= 5seconds
Speed= 24/5= 4.8m/s

3 0

3 years ago

Read 2 more answers

A hanging weight, with a mass of m1 = 0.365 kg, is attached by a string to a block with mass m2 = 0.825 kg as shown in the figur

morpeh [17]

The speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

<h3>Angular Speed of the pulley </h3>

The angular speed of the pulley after the block m1 fall through a distance, d, is obatined from conservation of energy and it is given as;

K.E = P.E

$\frac{1}{2} mv^2 + \frac{1}{2} I\omega^2 = mgh\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2(m_1R^2_2 + m_2R_2^2) + \frac{1}{2} \omega^2( \frac{1}{2} MR_1^2 + \frac{1}{2} MR_2^2) = m_1gd- \mu_km_2gd\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2[R_2^2(m_1 + m_2)+ \frac{1}{2} M(R_1^2 + R_2^2)] = gd(m_1 - \mu_k m_2)\\\\$

$\frac{1}{2} m_2v_0 + \frac{1}{4} \omega^2[2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = gd(m_1 - \mu_k m_2)\\\\2m_2v_0 + \omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2)\\\\\omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2\\\\\omega^2 = \frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)} \\\\\omega = \sqrt{\frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)}} \\\\$

Substitute the given parameters and solve for the angular speed;

$\omega = \sqrt{\frac{ 4(9.8)(0.7)(0.365 \ - \ 0.25\times 0.825) - 2(0.825)(0.82)^2}{2(0.03)^2(0.365 \ + \ 0.825)\ \ +\ \ 0.35(0.02^2\ + \ 0.03^2)}} \\\\\omega = \sqrt{\frac{3.25}{0.00214\ + \ 0.000455 } } \\\\\omega = 35.39 \ rad/s$

<h3>Linear speed of the block</h3>

The linear speed of the block after travelling 0.7 m;

v = ωR₂

v = 35.39 x 0.03

v = 1.1 m/s

Thus, the speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

Learn more about conservation of energy here: brainly.com/question/24772394

5 0

2 years ago