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3 years ago

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You must give a signal either by hand and arm or by a signal device: Only if other traffic is affected by your movement Only at

night Anytime you change lanes Only if you are driving a truck or car

1 answer:

Vesnalui [34]3 years ago

5 0

Hello there!

Anytime you change lanes.

Hope this helps :)

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Certain rifles can fire a bullet with a speed of 950 m/s just as it leaves the muzzle (this speed is called the muzzle velocity)

TiliK225 [7]

Answer:

a) By $v^2 = u^2 + 2as$ => a= 70291.70.

(b)By $v = u + at$ => t= 1.58 ms.

(c)By $v^2 = u^2 - 2gh$ => H = 46045.92 m.

Explanation:

a) By $v^2 = u^2 + 2as$

$(950)^2 = 0 + 2 \times a \times 0.75\\a = 601666.67 m/s^2\\a/g = 688858.70/9.8 = 70291.70$

(b)By $v = u + at$

$950 = 0 + 601666.67 \times t\\t = 1.58 x 10^-3 sec = 1.58 ms$

(c)By $v^2 = u^2 - 2gh$

$0 = (950)^2 - 2 \times9.8 \times H\\H = 46045.92 m$

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3 years ago

Blood is tissue .how?

Levart [38]

Yes, blood is a tissue. It is a tissue because it is a group of similar cells that have functions.

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2 years ago

Read 2 more answers

Star #1 is approaching the Earth with speed v. Star #2 is receding from the Earth with the same speed v. Measurements of the sam

leva [86]

Answer:

b. Both stars will have the same shift.

Explanation:

It's a very simple problem to solve. Star 1 is approaching toward Earth with a speed v, so let's assume that the change in Doppler Shift is +F and Star 2 is moving away so the change in Doppler shift is -F. But it's time to notice the speed of both stars and that is same but only directions are different. speed is the main factor here. The magnitude of both shifts is F as we can see and + and - are showing there direction of motion. So, because of same amount of speed, both stars will have same shift magnitude. (Just the directions are different)

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3 years ago

An electric field is produced by the very long, uniformly charged rod drawn above. If the strength of the electric field is E1 a

Debora [2.8K]

Answer: hello the complete question is attached below

answer :

r2 = 4r1

Explanation:

Electric field strength = F / q

we will assume the rod has an infinite length

For an infinitely charged rod

E ∝ 1/ r

considering two electric fields E1 and E2 at two different locations as described in the question

E1/E2 = r1/r2 ----- ( 2 )

<u>Calculate for r2 when E2 = E1/4 </u>

back to equation 2

E1 / (E1/4) = r1 / r2

∴ r2 = 4r1

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3 years ago

A girl throws a pebble into a deep well at 4.0 m/s (downward). It hits the water in 2.0 s. How far below the ground is the water

alexandr402 [8]

Answer:

27.6 m
13.8 m/s

Explanation:

(b) The initial velocity is added to that due to acceleration by gravity. The velocity is increased linearly by gravity at the rate of 9.8 m/s². The average velocity of the pebble will be its velocity halfway through the 2-second time period.* That is, it will be ...

4 m/s + (9.8 m/s²)(2 s)/2 = 13.8 m/s . . . . average velocity

__

(a) The distance covered in 2 seconds at an average velocity of 13.8 m/s is ...

d = vt

d = (13.8 m/s)(2 s) = 27.6 m

The water is about 27.6 m below ground.

_____

* We have chosen to make use of the fact that the velocity curve is linear, so the average velocity is half the sum of initial and final velocities:

vAvg = (vInit + vFinal)/2 = (vInit + (vInit +at))/2 = vInit +at/2

__

If you work this in a straightforward way, you would find distance as the integral of velocity, then find average velocity from the distance and time.

$\displaystyle d=\int_0^t{(v_0+at)}\,dt=v_0t+\dfrac{1}{2}at^2=t\left(v_0+a\dfrac{t}{2}\right)\\\\v_{avg}=\dfrac{d}{t}=v_0+a\dfrac{t}{2}\qquad\text{the formula we started with}$

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3 years ago