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3 years ago

13

Technician A says that the pushrods should be rotating while the engine is running if the camshaft and lifters are okay. Technic

ian B says that the camshaft rotates at one-quarter the crankshaft speed. Who is right

1 answer:

Karo-lina-s [1.5K]3 years ago

6 0

Answer:

Technician A

Explanation:

camshafts rotate at one-half the crankshaft speed

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Which process generates energy in the Sun?1) nuclear fusion2) nuclear fission3) chain reaction4) transmutation

Ber [7]

1) nuclear fusion

During nuclear fusion, the high pressure and temperature in the sun's core cause nuclei to separate from their electrons. During this process, radiant energy is released.

5 0

1 year ago

A block of 200 g is attached to a light spring with a force constant of 5 N / m and freely in a horizontal plane vibrates. The m

Finger [1]

Answer:

m  200 g , T  0.250 s,E 2.00 J

;

2 2 25.1 rad s

T 0.250

 

   

(a)

 

2 2

k m    0.200 kg 25.1 rad s 126 N m

(b)

 

2

2 2 2.00 0.178 mm  200 g , T  0.250 s,E 2.00 J

;

2 2 25.1 rad s

T 0.250

 

   

(a)

 

2 2

k m    0.200 kg 25.1 rad s 126 N m

(b)

 

2

2 2 2.00 0.178 m

Explanation:

That is a reason

8 0

3 years ago

A 20.0 cm tall object is placed 50.0 cm in front of a convex mirror with a radius of curvature of 34.0 cm. Where will the image

Neko [114]

Answer:

1.Theimage will be located at -0.13m or -13 cm

2.The height of the image will be 0.052m or 5.2cm

Explanation:

Given that;

Height of object, h=20 cm = 0.2m

Object distance in front of convex mirror, o,= 50 cm =0.5m

Radius of curvature, r, =34 cm =0.34m

Let;

Image distance, i,=?

Image height, h'=?

You know that focal length,f, is half the radius of curvature,hence

f=r/2 = 0.34/2 = 0.17m ( this length is inside the mirror, in a virtual side, thus its is negative)

f= -0.17m

Apply the relationship that involves the focal length;

$=\frac{1}{o} +\frac{1}{i} =\frac{1}{f}$

$=\frac{1}{0.5} +\frac{1}{i} =-\frac{1}{0.17}$

Re-arrange to get i

$\frac{1}{i} =-2-5.88\\\\\\\frac{1}{i} =-7.88\\\\i=-0.13m$

This is a virtual image formed at a negative distance produced through extension of drawing rays behind the mirror if you use rays to locate the image behind the mirror

Apply the magnification formula

magnification, m=height of image÷height of object

$m=\frac{h'}{h} =-\frac{i}{o}$

substitute the values to get the height of image h'

$\frac{h'}{0.20} =-\frac{-0.13}{0.5} \\\\\\h'=\frac{0.13*0.20}{0.5} \\\\\\h'=\frac{0.025}{0.5} =0.052m\\\\\\h'=5.2cm$

5 0

3 years ago

g A 2.3 kg block is attached to the spring, and it is released from rest 0.7 m from the spring's equilibrium position. Neglectin

DedPeter [7]

Answer:

3.71 m/s

Explanation:

From the law of conservation of linear momentum, since we are neglecting minor energy losses due to friction then we can express it as $mgh=0.5mv^{2}$ since all the potential energy is transformed to kinetic energy

Making v the subject of the formula then $v=\sqrt {2gh}$ and here m is the mass of the block, g is acceleration due to gravity, h is the height. Substituting 0.7 m for h and 9.81 for g then we obtain that $v=\sqrt {2\times 9.81\times 0.7}=3.705941176 m/s\approx 3.71 m/s$

6 0

3 years ago

If the half-life of tritium (hydrogen-3) is 12.3 years, how much of

tatyana61 [14]

Answer:

0.0000076 grams

Explanation:

We're given the half life of Tritium to be 12.3 years. In order to find out the amount of substabce remaining:

Let's first find how many 'half lives' are in 250 years.

$n = \frac{250}{12.3} = 20.325$

Now what is half life? It means the time taken for a given quantity of an element to lose half it's mass.

So in 12.3 years we can find that The amount of 250 g of Tritium will be 250/2 = 125 g. In 24.6 years we'll have 125/2 = 62.5 g

So now we can devise a formula:

$m = \frac{original \: amount}{ {2}^{n} }$

Where m is the remaining amount and n is th number of half lives in the time given.

Using this formula we can calculate:

$m = \frac{10}{ {2}^{20.325} }$

Doing this calculation we get:

$m = 0.0000076 \: g$

As we can see a very small value remains.

4 0

3 years ago