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NikAS [45]
3 years ago
8

A car of mass 1200Kilograms moving at 50 m/s the driver applies the brakes for 0.08 seconds and the castles down to 10 meter per

second calculate the The change of moment
Physics
2 answers:
Inessa05 [86]3 years ago
8 0

Initial velocity=u=15m/s

Final velocity=v=10m/s

Time=0.08s

\\ \sf\longmapsto Acceleration=\dfrac{v-u}{t}

\\ \sf\longmapsto Acceleration=\dfrac{10-15}{0.08}

\\ \sf\longmapsto Acceleration=\dfrac{-5}{0.08}

\\ \sf\longmapsto Acceleration=-62.5m/s^2

never [62]3 years ago
8 0

Answer:

accleration =  \frac{v - u}{t}  \\  =  \frac{10 - 15}{0.08}  \\  =  \frac{ - 5}{0.08}  =  - 62.5m \: per \: sec {}^{2} .

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A brick is dropped from a height of 31.9 meters
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Explanation:

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What event, which occurs twice a year when the sun crosses Earth's equator, recently
andrey2020 [161]

Answer:

<u>the autumnal equinox (when day and night are of equal length)</u>

<u>Explanation:</u>

The first event occurs around the 22nd or 23rd in the month of September. What usually happens is that regions around the Northern Hemisphere (which includes all of North America and Europe) experience an equal length of day and night.

While the second event occurs around March 20th or 21st mainly observable in countries found in the Southern Hemisphere.

5 0
3 years ago
A cart is given an initial velocity of 5.0 m/s and
larisa86 [58]

Answer:

cart displacement is 66 m

Explanation:

given data

velocity = 5 m/s

acceleration = 2 m/s²

time = 6 s

to find out

What is the

magnitude of cart displacement

solution

we will apply here equation of motion to find displacement that is

s = ut + 0.5×at²    .............1

here s id displacement and u is velocity and a is acceleration and time is t here

put all value in equation 1

s = ut + 0.5×at²

s = 5(6) + 0.5×(2)×6²

s = 66

so cart displacement is 66 m

5 0
4 years ago
The surface of the dock is 6 feet above the water. If you pull the rope in at a rate of 2 ft/sec, how quickly is the boat approa
Oduvanchick [21]

Answer:

The boat is approaching the dock at a rate of <u>2.5 ft/s</u>.

Explanation:

Let the rope length be 'l' at any time 't', the distance of boat from dock be 'b' at any time 't'.

Given:

The height of dock above water (h) = 6 feet

Rate of pull of rope or rate of change of rope is, \frac{dl}{dt}=2\ ft/s

As clear from the question, the height is fixed and only the length 'l' and distance 'b' varies with time 't'.

Now, the above situation represents a right angled triangle as shown below.

Using Pythagoras Theorem, we have:

l^2=h^2+b^2\\\\l^2=6^2+b^2\\\\l^2=36+b^2----------(1)

Now, differentiating the above equation with time 't', we get:

2l\frac{dl}{dt}=0+2b\frac{db}{dt}\\\\l\frac{dl}{dt}=b\frac{db}{dt}\\\\\frac{db}{dt}=\frac{l}{b}\frac{dl}{dt}------(2)

Now, the distance 'b' can be calculated using 'l=10 ft' in equation (1). This gives,

b^2=10^2-36\\\\b=\sqrt{64}=8\ ft

Now, substituting all the given values in equation (2) and solve for \frac{db}{dt}. This gives,

\frac{db}{dt}=\frac{10}{8}\times 2\\\\\frac{db}{dt}=2.5\ ft/s

Therefore, the boat is approaching the dock at a rate of 2.5 ft/s.

7 0
3 years ago
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