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Nezavi [6.7K]
3 years ago
11

An automobile is traveling on a long, straight highway at a steady 80.0 mi/h when the driver sees a wreck 190 m ahead. At that i

nstant, she applies the brakes (ignore reaction time). Between her and the wreck are two different surfaces. First there is 100 m of ice, where the deceleration is only 1.20 m/s2 . From then on, it is dry concrete, where the deceleration is a more normal 6.80 m/s2 .
a) What was the car’s speed just after leaving the icy portion of the road?
b)What is the total distance her car travels before it comes to a stop?
c)What is the total time it took the car to stop?
Physics
1 answer:
kozerog [31]3 years ago
6 0

Answer:

a) The car’s speed just after leaving the icy portion of the road is the first part

Explanation:

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Light passing through the center of a lens will carry on undeviated.
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I think the answer is B true

4 0
3 years ago
An object moving in circular motion has a mass of 15 kg and a centripetal acceleration of 10 m/s2. What is the centripetal force
sweet-ann [11.9K]

Answer:

1) A

2) C

3) B

4) A

5) Incomplete information(picture missing)

6) Incomplete information(picture missing)

7) Incomplete information(picture missing)

8) A

9) C

10) C

Explanation:

1) m = 15kg, a = 10ms^{-2}, F = ma = 15*10 = 150N

2) m = 3kg, v = 4ms^{-1}, r = 4m, F = \frac{mv^{2} }{r}

\frac{3*4^{2} }{4} = 12N

3) a = 10ms^{-2}, r = 10m, v=?

F = \frac{mv^{2} }{r} and F = ma

equating the two equations and cancelling a, we have:

\frac{v^{2} }{r} = a

making v the subject of formula, we have:

v = \sqrt{ar}

= \sqrt{100}

= 10ms^{-1}

4) r = 10m, v = 5ms^{-1}, a = ?

F = \frac{mv^{2} }{r}

F = ma

equating the above equations and making a subject of formula, we get:

a = \frac{v^{2} }{r}

a = 25/10 = 2.5ms^{-2}

5) I can't find the picture associated with this question

6) I can't find the picture associated with this question

7) I can't find the picture associated with this question

8) F = \frac{mv^{2} }{r}

assuming m and r is unity, that is the values are 1 respectively, the formula simplifies to:

F = v^{2}

Now, if v is tripled

F = (3v)^{2}

F = 9v^{2}

We can see that the force will be 9X greater than it was.

9) F = \frac{mv^{2} }{r}

assuming m and r is unity, that is the values are 1 respectively, the formula simplifies to:

F = v^{2}

Now, if v is doubled

F = (2v)^{2}

F = 4v^{2}

We can see that the force will be 4X greater than it was.

10) F = \frac{mv^{2} }{r}

assuming m and v is unity, that is the values are 1 respectively

F = 1/r

if r is doubled,

F = 1/2 * 1/r

We can see that the force is 1/2 as big as it was

7 0
3 years ago
What is the density of an object with a mass of 60g and a volume of 2cm3?
zaharov [31]

m= 60g = 60/1000 Kg = 0.06 Kg v = 2cm3 = 2 * (0.01^3) m3 = 2 *10^-6 m3 Density= m/v = 6 * 10^-2 / 2 *10^-6  =  3 *10^4 Kg/m3

8 0
3 years ago
ਭਾਰਤ ਵਿੱਚ ਸਭ ਤੋਂ ਵੱਧ ਵਰਖਾ ਕਿੱਥੇ<br>ਹੁੰਦੀ ਹੈ।​
astra-53 [7]

Answer:

I don’t understand:(

3 0
3 years ago
A jet plane leaves the stratosphere and returns to the ground. How does the temperature outside the jet change during its descen
finlep [7]

Answer : A. It decreases and then increases.

Explanation :  Troposphere is the lowermost layer of atmosphere.

Stratosphere is next layer up to the troposphere. As the jet descends from stratosphere towards the troposphere, the temperature initially decreases and then at troposphere is roughly constant and then steadily increases.

So, option (A) is correct.

8 0
2 years ago
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