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3 years ago

A girl on a skateboard going around a corner at a speed of 3 m/sec.

1 answer:

5 0

When going round a corner your direction changes which means your velocity changes which means there is an acceleration.

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A world class sprinter is travelling with speed 12.0 m/s at the end of a 100 meter race. Suppose he decelerates at the rate of 2

Solnce55 [7]

Answer:

after 6 second it will stop

he travel 36 m to stop

Explanation:

given data

speed = 12 m/s

distance = 100 m

decelerates rate = 2.00 m/s²

so acceleration a = - 2.00 m/s²

to find out

how long does it take to stop and how far does he travel

solution

we will apply here first equation of motion that is

v = u + at ......1

here u is speed 12 and v is 0 because we stop finally

put here all value in equation 1

0 = 12 + (-2) t

t = 6 s

so after 6 second it will stop

and

for distance we apply equation of motion

v²-u² = 2×a×s ..........2

here v is 0 u is 12 and a is -2 and find distance s

put all value in equation 2

0-12² = 2×(-2)×s

s = 36 m

so he travel 36 m to stop

3 0

2 years ago

Two objects, with different sizes, masses, and temperatures, are placed in thermal contact. in which direction does the energy t

s344n2d4d5 [400]

<span>(c) energy travels from the object at higher temperature
to the object at lower temperature.

Size and mass have no effect.</span>

3 0

3 years ago

Two coconuts fall freely from rest at the same time, one from a tree twice as high as the other. If the coconut from the shorter

valentinak56 [21]

Wee can use here kinematics

as we know that

$y = v*t + \frac{1}{2} at^2$

for shorter tree we know that

$y = 0 + \frac{1}{2}*9.8 * 2^2$

$y = 19.6 meter$

now since we know that other tree is twice high

So height of other tree is y = 39.2 m

now again by above equation

$y = v*t + \frac{1}{2} at^2$

$39.2 = 0 + \frac{1}{2}*9.8 * t^2$

$t = 2.83 s$

so the time taken is 2.83 s

4 0

3 years ago

PLEASE HELP ME

posledela

Low levels of fecal coliform

5 0

3 years ago

Read 2 more answers

At what speed does a 1,248 kg compact car have the same kinetic energy as a 18,777 kg truck going 26 km/h?

notka56 [123]

The speed of car is 100.8km/h

$KE = \frac{1}{2} m(v \: truck ) {}^{2}$

$= 0.5 \times 18777 \times \frac{26}{3.6} \times \frac{26}{3.6}$

$= 489708.79j$

$= \frac{1}{2} \times 1248 \times( v \: car) {}^{2}$

$= 624v {}^{2}$

$so \: v {}^{2} = \frac{489708.7}{624}$

$= 784$

$v = \sqrt{784}$

$v = 28m/s$

v car= 28×3.6

=100.8km/h

Hence, the speed of the car is 100.8km/h

learn more about speed from here:

brainly.com/question/28326855

#SPJ4

3 0

1 year ago