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1 year ago

An earthmover slows from 42.0 km/h to 4.00 km/h in 4.75 s. What is its rate of acceleration in m/s2

Physics

1 answer:

Simora [160]1 year ago

6 0

acceleration (a) = change in velocity / change in time

Δv = final velocity - initial velocity = 4 km/h - 42 km/h = -38 km/h

Δt = 4.75 s

convert km/h to m/s

-38/ 3.6 = -10.56 m/s

a = -10.56 m/s / 4.75 s = -2.23 m/s^2

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Two sound waves are traveling through the same medium. They have the same amplitude, wavelength, and direction of travel. If the

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Answer:

Constructive interference

Explanation:

Here, the medium is same, same wavelength, same frequency, same amplitude and same direction of propagation.

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3 years ago

A parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 67 m/s*2. If he reaches the gr

KIM [24]

Answer:

3.49 seconds

3.75 seconds

-43200 ft/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$s=ut+\frac{1}{2}at^2\\\Rightarrow 50=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{50\times 2}{9.81}}\\\Rightarrow t=3.19\ s$

Time the parachutist falls without friction is 3.19 seconds

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 50+0^2}\\\Rightarrow v=31.32\ m/s$

Speed of the parachutist when he opens the parachute 31.32 m/s. Now, this will be considered as the initial velocity

$v=u+at\\\Rightarrow 11=31.32+9.81t\\\Rightarrow t=\frac{11-31.32}{-67}=0.3\ s$

So, time the parachutist stayed in the air was 3.19+0.3 = 3.49 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=0t+\frac{1}{2}\times a\times t^2\\\Rightarrow \frac{s}{2}=\frac{1}{2}at^2$

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=u1.1+\frac{1}{2}\times a\times 1.1^2$

Now the initial velocity of the last half height will be the final velocity of the first half height.

$v=u+at\\\Rightarrow v=at$

Since the height are equal

$\frac{1}{2}at^2=u1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow \frac{1}{2}at^2=at1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow 0.5t^2-1.1t-0.605=0\\\Rightarrow 500t^2-1100t-605=0$

$t=\frac{11\left(1+\sqrt{2}\right)}{10},\:t=\frac{11\left(1-\sqrt{2}\right)}{10}\\\Rightarrow t=2.65, -0.45$

Time taken to fall the first half is 2.65 seconds

Total time taken to fall is 2.65+1.1 = 3.75 seconds.

When an object is thrown with a velocity upwards then the velocity of the object at the point to where it was thrown becomes equal to the initial velocity.

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-240^2}{2\times \frac{8}{12}}\\\Rightarrow a=-43200\ ft/s^2$

Magnitude of acceleration is -43200 ft/s²

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3 years ago

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Natali [406]

Near Greenland in the northern hemisphere <span />

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3 years ago

S A block of mass M is connected to a spring of mass m and oscillates in simple harmonic motion on a frictionless, horizontal tr

Elina [12.6K]

The period of oscillation is T = 2 * pi * sqrt ( ( m2/3 + m1) / k )

<h3>What is period of oscillation?</h3>

This is the time in seconds it takes to complete one oscillation. where an oscillation is a repetitive to and fro motion. period if the inverse of frequency and both are basic when calculation motion in simple harmonic motion.

The period of oscillation is given as T

T = 2 * pi * sqrt ( m / k )

where

m = mass on this case mass of the spring will be inclusive to the mass of the block such that we have:

m1 = mass of the block

m2 = mass pf the spring

k = force constant of the spring

including the two masses to the period gives

T = 2 * pi * sqrt ( ( m2/3 + m1) / k )

Read more on period of oscillation here: brainly.com/question/22499336

#SPJ4

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2 years ago