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andrew-mc [135]

1 year ago

15

Your vehicle engine cranks but does not start what is the first corrective action you should perform

1 answer:

Trava [24]1 year ago

4 0

Answer:

Gas !

Explanation:

Check to see if you have gas in the tank !

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a car collides with a wall. compare the forces exerted by the car on the wall and the wall on the car

NeTakaya

Mass multiplied by acceleration produces force.

The acceleration is (v - 0)/t in this situation, where t seems to be the time it takes automobile A to come to a stop. According to Newton's third law of motion, the automobile produces this turning force of the wall, however the wall, which really is static and indestructible, forces an equal force back on the car.

According to Newton's third law, each action has an equal and opposite response. On this basis, you may deduce that a car driving into a wall would exert force on the wall. However, since the wall did not move, the automobile receives an equivalent force, causing it to collapse.

<h3>Learn more:</h3>

brainly.com/question/13952508?referrer=searchResults

5 0

2 years ago

Read 2 more answers

Describe an object's velocity when an acceleration-time graph is zero?

svp [43]

Anything times zero is zero

7 0

3 years ago

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Parallel rays of monochromatic light with wavelength 571nm illuminate two identical slits and produce an interference pattern on

Natasha2012 [34]

Answer:

$8.8\times 10^{-6} W/m^2$

Explanation:

We are given that

Wavelength, $\lambda=571 nm=571\times 10^{-9} m$

$1 nm=10^{-9} m$

R=75 cm= $\frac{75}{100}=0.75 m$

1 m=100 cm

$d=0.640 mm=0.64\times 10^{-3} m$

$1 mm=10^{-3} m$

$a=0.434 mm=0.434\times 10^{-3} m$

$y=0.830 mm=0.830\times 10^{-3} m$

$I_0=5.00\times 10^{-4}W/m^2$

$tan\theta=\frac{y}{R}$

$\theta=\frac{0.830\times 10^{-3}}{0.75\times 10^{-2}}$

$\theta=1.1\times 10^{-3}rad$

$tan\theta\approx \theta$

Because $\theta$ is small.

$\phi=\frac{2\pi dsin\theta}{\lambda}$

$\sin\theta\approx \theta$ ,

Therefore

$\phi=\frac{2\times\pi\times 0.64\times 10^{-3}\times 1.1\times 10^{-3}}{571\times 10^{-9}}$

$\phi=7.74 rad$

$\beta=\frac{2\pi a\theta}{\lambda}$

$\beta=5.3 rad$

$I=I_0cos^2(\frac{\phi}{2})(\frac{sin\frac{\beta}{2}}{\frac{\beta}{2}})^2$

$I=5\times 10^{-4}cos^2(\frac{7.74}{2})(\frac{sin\frac{5.3}{2}}{\frac{5.3}{2}})^2$

$I=8.8\times 10^{-6} W/m^2$

6 0

3 years ago

The nucleus of an atom is composed of unchanged particles as well as of the following

Ivan

I suppose that you wanted write "uncharged". The particles without electrical charge present in the nucleus are called neutrons.

7 0

3 years ago

3 A 100 g steel ball falls from a height of 1.8 m on to a metal plate and rebounds to a height of 1.25 m.

BigorU [14]

Given values:

Mass of the steel ball, m = 100 g = 0.1 kg

Height of the steel ball, h1 = 1.8 m

Rebound height, h2 = 1.25 m

a. PE= mgh

0.1 x 9.8 x 1.8 =

1.764 Joules

b. KE = PE ->

1.764 Joules

c. KE= 1/2 mv square

so v = square root 2ke/m

square root 2 x 1.764/ 0.1

= 5.93 m/s

d. KE=PE=mgh square

0.1 x 9.8 x 1.21 =

1.186 joules

velocity of rebond is square root 2x 1.186/ 0.1 = 4.87 m/s

6 0

3 years ago

Read 2 more answers