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Oksana_A [137]
3 years ago
8

The braking effect of the engine is greatest when the engine is _____ the governed rpm and the transmission is in the ______ gea

rs.
Physics
1 answer:
Arisa [49]3 years ago
3 0

When the engine is near the governed RPMand the transmission is in the lower gears, the braking effect of the engine is at its greatest .


It is specified that the legal requirements for the braking system that:

A reasonable distance should be practiced in stopping your vehicle.


Adjournment from the jar or shudder should be smooth and free.


A good braking system’s other necessities are:

Unaffected by heat, water, road grit, dust etc.

Constant continued application while moving downwards from a hill should not decrease its effectiveness and it must have good anti fade characteristics


Adequate durability with economical maintenance and adjustment Its durability must be enough and must have proper adjustment and maintenance. 

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The Electron Cloud Model shows shaded regions of probabilty where the electons most likely are at a given point in time. True Fa
melisa1 [442]

Answer:

True

Explanation:

The statement is true because According to Schrödinger it's not possible to get the exact position of where an electron is located. However, he showed from an equation that we can at determine the most likely area where an electron will be and he said they are most likely to be in the most dense area of the clouds.

7 0
3 years ago
Read 2 more answers
Choose the law each sentence describes. This law relates a planet's orbital period and its average distance to the Sun. The orbi
hram777 [196]

These are the Kepler's laws of planetary motion.

This law relates a planet's orbital period and its average distance to the Sun. - Third law of Kepler.

The orbits of planets are ellipses with the Sun at one focus. - First law of Kepler.

The speed of a planet varies, such that a planet sweeps out an equal area in equal time frames. - Second law of Kepler.

7 0
3 years ago
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An athlete needs to lose weight and decides to do it by "pumping iron." (a) How many times must an 60.0 kg weight be lifted a di
Dennis_Churaev [7]

Answer:

a) 37171

b) 20650.5 sec

Explanation:

m = mass of the weight being lifted = 60.0 kg

d = distance by which the weight is lifted = 0.670 m

E = Energy available to burn = 1 lb = 3500 kcal = 3500 x 4184 J

g = acceleration due to gravity = 9.8 m/s²

n = number of times the weight is lifted

Energy available to burn is given as

E = n m g d

3500 x 4184 = n (60) (9.8) (0.670)

n = 37171

b)

T = time period for each lift up = 1.80 s

t = total time taken

Total time taken is given as

t = \frac{n}{T}

t = \frac{37171}{1.80}

t = 20650.5 sec

8 0
3 years ago
A beam of protons is moving toward a target in a particle accelerator. This beam constitutes a current whose value is. (a) How m
Gelneren [198K]

Answer:

a. 5 × 10¹⁹ protons b. 2.05 × 10⁷ °C

Explanation:

Here is the complete question

A beam of protons is moving toward a target in a particle accelerator. This beam constitutes a current whose value is 0.42 A. (a) How many protons strike the target in 19 seconds? (b) Each proton has a kinetic energy of 6.0 x 10-12 J. Suppose the target is a 17-gram block of metal whose specific heat capacity is 860 J/(kg Co), and all the kinetic energy of the protons goes into heating it up. What is the change in temperature of the block at the end of 19 s?

Solution

a.

i = Q/t = ne/t

n = it/e where i = current = 0.42 A, n = number of protons, e = proton charge = 1.602 × 10⁻¹⁹ C and t = time = 19 s

So n = 0.42 A × 19 s/1.602 × 10⁻¹⁹ C

       = 4.98 × 10¹⁹ protons

       ≅ 5 × 10¹⁹ protons

b

The total kinetic energy of the protons = heat change of target

total kinetic energy of the protons = n × kinetic energy per proton

                                                         = 5 × 10¹⁹ protons × 6.0 × 10⁻¹² J per proton

                                                         = 30 × 10⁷ J

heat change of target = Q = mcΔT ⇒ ΔT = Q/mc where m = mass of block = 17 g = 0.017 kg and c = specific heat capacity = 860 J/(kg °C)

ΔT = Q/mc = 30 × 10⁷ J/0.017 kg × 860 J/(kg °C)

     = 30 × 10⁷/14.62

     = 2.05 × 10⁷ °C

5 0
3 years ago
Describe Darwin’s<br> observations on the Galápagos islands<br> during his voyage on the HMS Beagle.
Helen [10]

Answer:

During the voyage Charles Darwin explored the Galapagos islands and noticed the same species have different adaptations in places. ... Charles noticed that each species has the same ancestor but they evolve to adapt over time so they can live longer.

Explanation:

4 0
4 years ago
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