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3 years ago

The amount of steering wheel movement needed to turn will ____________ the faster you go.

Physics

2 answers:

Naddika [18.5K]3 years ago

4 0

Answer:

The answer to your question is Decrease

Svetach [21]3 years ago

3 0

Answer:

<h2>The amount of steering wheel movement needed to turn will <u><em>decrease</em></u> the faster you go.</h2>

Explanation:

A steering wheel, a driving wheel or a hand wheel allow us to control the vehicle, it's part of the steering. So the faster the vehicle goes, less movement of the wheel is needed, because the movement of the vehicle makes easier to handle the wheel. Also, when the vehicle is almost not moving or moving slowly, we have to put more effort to move the wheel.

In addition, this want of the reasons why when we are driving too fast, we must put attention on our wheel, because a simple and short movement can get us out of the road.

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If you cut a apple and leave it in the open air and it turns brown what kind of chemical change is that

barxatty [35]

Answer:

Oxidation

Explanation:

The browning of the apple after you cut it undergoes a natural chemical change called oxidation, wherein the apple's enzymes react with the oxygen in the environment.

3 0

3 years ago

A 1400 kg car starts from rest on a horizontal road and gains a speed of 61 km/h in 19 s. (a) what is its kinetic energy at the

lana [24]

(a) Let's convert the final speed of the car in m/s:
$v_f = 61 km/h = 16.9 m/s$
The kinetic energy of the car at t=19 s is
$K= \frac{1}{2}mv_f^2= \frac{1}{2}(1400 kg)(16.9 m/s)^2=2.00 \cdot 10^5 J$

(b) The average power delivered by the engine of the car during the 19 s is equal to the work done by the engine divided by the time interval:
$P= \frac{W}{\Delta t}$
But the work done is equal to the increase in kinetic energy of the car, and since its initial kinetic energy is zero (because the car starts from rest), this translates into
$P= \frac{K}{\Delta t}= \frac{2.00 \cdot 10^5 J}{19 s}=1.05 \cdot 10^4 W$

(c) The instantaneous power is given by
$P_i = Fv_f$
where F is the force exerted by the engine, equal to F=ma.

So we need to find the acceleration first:
$a= \frac{v_f-v_i}{\Delta t}= \frac{16.9 m/s}{19 s}=0.89 m/s^2$
And the problem says this acceleration is constant during the motion, so now we can calculate the instantaneous power at t=19 s:
$P_i = Fv=(ma)v=(1400 kg)(0.89 m/s^2)(16.9 m/s)=2.11 \cdot 10^4 W$

5 0

3 years ago

What is the relationship between the temperature of a star and its luminosity?

Sladkaya [172]

The Luminosity of a star is proportional to its Effective Temperature to the 4th power and its Radius squared.

3 0

3 years ago

. Each valve A, B, and C, when open, releases water into a tank at its own constant rate. With all three valves open, the tank f

olasank [31]

Answer:

Time taken by A and B is 1.2 hr.

Explanation:

Given that

Time taken by tank when all(A+B+C) are open = 1 hr

Time taken by tank when A+C are open = 1.5 hr

Time taken by tank when B+C are open = 2 hr

If we treat as filling of tank is a work then

Work = time x rate

Lets take work is 1 unit

1 = 1(1/a+1/b+1/c) ---------1

1 = 1.5(1/a+1/c) ----------2

1 = 2(1/b+1/c) --------3

From equation 1 and 3

1=1(1/a+1/2)

a=2

Form equation 2

1 = 1.5(1/2+1/c)

c=6

From equation 3

1 = 2(1/b+1/6)

b=3

So time taken by

A is alone to fill tank is 2 hr

B is alone to fill tank is 3 hr

C is alone to fill tank is 6 hr

So $time\ taken\ by\ A\ and\ B =\dfrac{2\times 3}{2+3} hr$

Time taken by A and B is 1.2 hr.

7 0

3 years ago

In Thomson experiment, why was the glowing beam repelled by a negatively charged plate

algol13

<span>In Thomson experiment, why was the glowing beam repelled by a negatively charged plate, because the glowing beam was negatively charged. The glowing beam particles were attracted to the positive plate.

</span><span>J.JThomson proved that the cathode rays produced a stream of negatively charged particles called electrons. </span>

6 0

3 years ago

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