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pav-90 [236]
3 years ago
5

How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and

Physics
1 answer:
aev [14]3 years ago
4 0

Answer:

h= 46.66 m

Explanation:

Given that

Initial speed of the car ,u = 110 km/h

We know that

1 km/h= 0.277 m/s

u= 30.55 m/s

lets height gain by car is h.

The final speed of the car will be zero at height h.

v²=u²- 2 g h

v= 0 m/s

0²=30.55²- 2 x 10 x h           ( g = 10 m/s²)

h= 46.66 m

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What is meant by the term illegal alien
anyanavicka [17]

Answer:

The term illegal alien refers to a foreign person who is living in a country to which they are not an official legal citizen.

5 0
3 years ago
Read 2 more answers
A person walks 60 meters in 1 minute. Then walks 30 meters in 2 minutes. What is thier average speed over the 3 minute walk?
Gennadij [26K]
Your answer would be 75 meters
8 0
3 years ago
Read 2 more answers
A sound has a sound level of 30 dB. Its intensity is what multiple of the standard reference level for intensities?
arsen [322]
<h2>Answer:</h2>

1000th multiple of the standard reference level for intensities.

<h2>Explanation:</h2>

The sound intensity level (β), measured in decibels, of a sound with an intensity of I is defined as follows;

β = 10 log (I / I₀)       --------------------(i)

Where;

I₀ = reference intensity

Given from the question;

β = sound level = 30dB

Substitute this value into equation (i) as follows;

30 = 10 log (I / I₀)

Divide both sides by 3;

3 = log (I / I₀)

Take antilog of both sides;

10^(3) = (I / I₀)

1000 = I / I₀

Solve for I;

I = 1000I₀

Therefore the intensity of the sound is 1000 times the standard reference level for intensities (I₀)

7 0
3 years ago
On the way home from school, Taylor's car runs out of gas. He has to walk 25m north and 10m west in order to reach the nearest g
spin [16.1K]

Answer:

<em>The distance is 35 m and the magnitude of the displacement is 26.93 m</em>

Explanation:

<u>Displacement  and Distance</u>

These are two related concepts. A moving object constantly travels for some distance at defined periods of time. The total distance is the sum of each individual distance the object traveled. It can be written as:

dtotal=d1+d2+d3+...+dn

This sum is calculated independently of the direction the object moves.

The displacement only takes into consideration the initial and final positions of the object. The displacement, unlike distance, is a vectorial magnitude and can even have magnitude zero if the object starts and ends the movement at the same point.

Taylor walks 25 m north and 10 m west. The total distance is the sum of both numbers:

d = 25 m + 10 m = 35 m

To calculate the displacement, we need to know the final position with respect to the initial position. If we set the coordinates of Taylor's car as the origin (0,0), then his final position is (-10,25), assuming the west direction is negative and the north direction is positive.

The magnitude of the displacement is the distance from (0,0) to (-10,25):

D=\sqrt{(25-0)^2+(-10-0)^2}

D=\sqrt{625+100}=\sqrt{725}

D = 26.93 m

The distance is 35 m and the magnitude of the displacement is 26.93 m

8 0
3 years ago
I need to know the right answer to that question
ad-work [718]
The answer is C 8.87*10^4 m/s (it shouldn't be m/s^2 though as velocity is in m/s)

Since you know the acceleration is 12 m/s^2, the initial velocity is 2.39*10^4 m/s and the time (you have to convert to seconds) is 5400 seconds, then you can use the equation

v = vo + at

When you plug in the values you get

v = 2.39*10^4 + 5400*12 . so v = 8.87*10^4 m/s. C is your answer.
8 0
4 years ago
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