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faltersainse [42]

3 years ago

14

Which statement best defines a motor unit? A. a motor nerve and the muscle cells it stimulates B. a nerve supplying a muscle ori

ginating from the spinal cord C. branches located in each individual muscle cell D. single nerves found in each muscle

1 answer:

Gennadij [26K]3 years ago

4 0

It is A.
There's no doubt

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In the parallelogram shown, AE = t + 2, CE = 3t − 14, and DE = 2t + 8.

solniwko [45]

The answer would be 48

7 0

3 years ago

Read 2 more answers

Stars of spectral type A and F are considered ________.

LekaFEV [45]

Answer:

<u>B. the stars of spectral type A and F are considered reasonably to have habitable planets but much less likely to have planets with complex plant - or animal - like life.</u>

Explanation:

The appropriate spectral range for habitable stars is considered to be "late F" or "G", to "mid-K" or even late "A". <em>This corresponds to temperatures of a little more than 7,000 K down to a little less than 4,000 K</em> (6,700 °C to 3,700 °C); the Sun, a G2 star at 5,777 K, is well within these bounds. "Middle-class" stars (late A, late F, G , mid K )of this sort have a number of characteristics considered important to planetary habitability:

• They live at least a few billion years, allowing life a chance to evolve. <em>More luminous main-sequence stars of the "O", "B", and "A" classes usually live less than a billion years and in exceptional cases less than 10 million.</em>

• They emit enough high-frequency ultraviolet radiation to trigger important atmospheric dynamics such as ozone formation, but not so much that ionisation destroys incipient life.

• They emit sufficient radiation at wavelengths conducive to photosynthesis.

• Liquid water may exist on the surface of planets orbiting them at a distance that does not induce tidal locking.

<u><em>Thus , the stars of spectral type A and F are considered reasonably to have habitable planets but much less likely to have planets with complex plant - or animak - like life.</em></u>

4 0

3 years ago

Which one of the following energy conversion devices has the lowest energy efficiency?

Anna35 [415]

Answer:

A) incandescent ligth bulb, its efficiency is about 10%

Explanation:

The incandescent bulb, that is, the well-known focus with its warm light, was one of the most useful inventions of the 19th century although its use is currently considered very inefficient. These lamps waste between 80 and 90 percent of the total electricity they consume by turning it into heat. The metal filament thus heated and which is the central part of the bulb, only converts the remaining energy into light. Its service life ranges from 750 to 1,000 hours.

This is why they are used in ovens for food preparation, because of the large amount of heat they generate.

The steam boiler in a power plant depends on the fuel that it is using, but a coal-fired power plant with modern technology its efficiency is about 40%

Electric motor are around 85-92%

In order to better understand the concept of efficiency it is as if we pay 100 dollars of gasoline for our weekly use, but of that 100 dollars the car only uses 10 dollars to do that activity the rest of the money the 90 dollars were lost because of the inefficiencies of the vehicle.

3 0

3 years ago

A car of mass 1000 kg is moving at 25 m/s. It collides with a car of mass 1200 kg moving at 30 m/s. When the cars collide, they

Alinara [238K]

Answer:

The total momentum of the cars before the collision is 61,000 kg.m/s

The total momentum of the cars after the collision is 61,000 kg.m/s

The velocity of the cars after the collision is 27.727 m/s

Explanation:

Given;

mass of the first car, m₁ = 1000 kg

initial velocity of the car, u₁ = 25 m/s

mass of the second car, m₂ = 1200 kg

initial velocity of the second car, u₂ = 30 m/s

The common velocity of the cars after collision = v

The total momentum of the cars before collision is calculated as;

P₁ = m₁u₁ + m₂u₂

P₁ = (1000 x 25) + (1200 x 30)

P₁ = 61,000 kg.m/s

The total momentum of the cars after collision is calculated as;

P₂ = m₁v + m₂v

where;

v is the common velocities of the cars after collision since they stick together.

P₂ = v(m₁ + m₂)

To determine "v" apply the principle of conservation of linear momentum for inelastic collision.

m₁u₁ + m₂u₂ = v(m₁ + m₂)

(1000 x 25) + (1200 x 30) = v(1000 + 1200)

61,000 = 2,200v

v = 61,000/2,200

v = 27.727 m/s

The total momentum after collsion = v(m₁ + m₂)

= 27.727(1000 + 1200)

= 61,000 kg.m/s

Thus, momentum before and after collsion are equal.

8 0

3 years ago

Define vector quantity giving examples

Mama L [17]

Answer:

Quantity with direction and magnitude

Displacement

Explanation:

6 0

2 years ago