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3 years ago

Find acceleration. Will give brainliest!

Physics

2 answers:

Art [367]3 years ago

8 0

Answer:

$\boxed{\mathrm{4.5 \: kmph/s \: or \: 1.25 \: m/s^2 }}$

Explanation:

$\displaystyle \mathrm{acceleration = \frac{change \: in \: velocity}{time \: taken}}$

$\displaystyle \mathrm{a = \frac{v - u}{t}}$

$\displaystyle \mathrm{v=final \: velocity}\\\displaystyle \mathrm{u=initial \: velocity}$

$\displaystyle \mathrm{a = \frac{45- 0}{10}}$

$\displaystyle \mathrm{a = \frac{45}{10}}$

$\displaystyle \mathrm{a = 4.5}$

$\mathrm{4.5 \: kmph/s = 1.25 \: m/s^2 }$

zalisa [80]3 years ago

7 0

Answer:

16200 km/s

270 km/min

4.5 km/h

Explanation:

Acceleration Formula: Average Acceleration = Δv/Δt (change in velocity over change in time)

Simply plug in our known variables and solve:

a = (45.0 - 0)/10

a = 45.0/10

a = 4.5 km/h

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A hiker determines the length of a lake by listening for the echo of her shout reflected by a cliff at the far end of the lake.

Nataliya [291]

To solve the exercise it is necessary to take into account the definition of speed as a function of distance and time, and the speed of air in the sound, as well

$v=\frac{d}{t}$

Where,

V= Velocity

d= distance

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Re-arrange the equation to find the distance we have,

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Replacing with our values

$d= (343)(3.7)$

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$L_{lake} = \frac{d}{2}$

$L_{lake} = \frac{1269.1}{2}$

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A child of mass 27 kg swings at the end of an elastic cord. At the bottom of the swing, the child's velocity is horizontal, and

snow_tiger [21]

Answer:

The magnitude of the rate of change of the child's momentum is 794.11 N.

Explanation:

Given that,

Mass of child = 27 kg

Speed of child in horizontal = 10 m/s

Length = 3.40 m

There is a rate of change of the perpendicular component of momentum.

Centripetal force acts always towards the center.

We need to calculate the magnitude of the rate of change of the child's momentum

Using formula of momentum

$\dfrac{dp}{dt}=F$

$\dfrac{dP}{dt}=\dfrac{mv^2}{r}$

Put the value into the formula

$\dfrac{dP}{dt}=\dfrac{27\times10^2}{3.40}$

$\dfrac{dP}{dt}=794.11\ N$

Hence, The magnitude of the rate of change of the child's momentum is 794.11 N.

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3 years ago

Frim the castle wall 20 m high shot an arrow. The initial speed of the bow is 45 m/s directed 40 degrees above horizontal. Find

True [87]

Answer:

Range of arrow = 225.09 meter

Final horizontal velocity = 34.47 m/s

Explanation:

We have equation of motion $s=ut+\frac{1}{2} at^2$ , where u is the initial velocity, t is the time taken, a is the acceleration and s is the displacement.