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ioda
3 years ago
14

If your brakes give out, why can't you just pull the keys out of the ignition?

Physics
1 answer:
Pachacha [2.7K]3 years ago
5 0

There should be a small amount of play in the wheel when the steering is locked. Gently pull the key from the ignition while you slowly jiggle the steering wheel back and forth. If this is the cause of the problem, the key should come out after a little effort.

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А
bagirrra123 [75]

Explanation:

The reading on the scale is

W = m(g + a)

= (77 kg)(9.8 m/s^2 + 2 m/s^2)

= 908.6 N

7 0
3 years ago
If a fish experiences an a of 9.8m/s2 over a period of 1.84s, what was the 4v?
777dan777 [17]

Answer:

It’s 18.0 m/s

Explanation:

Use acceleration formula then plug in 9.8 and 1.84s

6 0
3 years ago
The melting of polar ice is one effect of
Stels [109]
The melting of polar ice is one effect of the greenhouse effect, or also global warming.

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5 0
3 years ago
What is the measurement 111.009 mm rounded off to four significant digits?
dedylja [7]
111.0 because 111.009 rounds off to 111.01, thus rounding again off to 111.0
4 0
3 years ago
An amusement park ride consists of a rotating circular platform 11.1 m in diameter from which 10 kg seats are suspended at the e
frozen [14]

To solve this problem we will use the relationship given between the centripetal Force and the Force caused by the weight, with respect to the horizontal and vertical components of the total tension given.

The tension in the vertical plane will be equivalent to the centripetal force therefore

Tsin\theta= \frac{mv^2}{r}

Here,

m = mass

v = Velocity

r = Radius

The tension in the horizontal plane will be subject to the action of the weight, therefore

Tcos\theta = mg

Matching both expressions with respect to the tension we will have to

T = \frac{\frac{mv^2}{r}}{sin\theta}

T = \frac{mg}{cos\theta}

Then we have that,

\frac{\frac{mv^2}{r}}{sin\theta} =  \frac{mg}{cos\theta}

\frac{mv^2}{r} = mg tan\theta

Rearranging to find the velocity we have that

v = \sqrt{grTan\theta}

The value of the angle is 14.5°, the acceleration  (g) is 9.8m/s^2 and the radius is

r = \frac{\text{diameter of rotational circular platform}}{2} + \text{length of chains}

r = \frac{11.1}{2}+2.41

r = 7.96m

Replacing we have that

v = \sqrt{(9.8)(7.96)tan(14.5\°)}

v = 4.492m/s

Therefore the speed of each seat is 4.492m/s

6 0
3 years ago
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