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2 years ago

A friend frequently misses the morning

Physics

1 answer:

poizon [28]2 years ago

7 0

Answer: Heyaa! ~

State of the problem is that your friend routinely misses the school bus.

Explanation:

What are the causes for his missing the school bus:

Not waking early enough

Too long breakfast

Too far to the bus stop.

Deliberately tries to miss the bus.

Hopefully this helps you!

-Matthew ^^

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a particle moves along the x axis with an acceleration of a=18t, where a has units if m/s2. if the particle at time t=0 is at th

ludmilkaskok [199]

Answer:

Position at t= 4 seconds is 144 m

Explanation:

It is given that acceleration, a = 18 t, where t is the time.

We know that Velocity, $v = \int { a} \, dt$

Substituting value of a,

Velocity, $v = \int {18t} \, dt=\frac{18t^2}{2} +c=9t^2+c$

We know that at t = 0, v = -12 m/s

So, $9*0^2+c=-12\\ \\ c=-12m/s$

So velocity, $v = (9t^2-12)m/s$

We also know that displacement, $x = \int { v} \, dt$

Substituting value of v,

Displacement, $x=\int {(9t^2-12)} \, dt=\frac{9t^3}{3} -12t+c=3t^3-12t+c$

We know that at t = 0, particle is at origin, x =0.

So, $0=3*0^3-12*0+c\\ \\ c=0$

Displacement, $x = 3t^3-12t$

At t = 4 seconds

$x = 3*4^3-12*4=192--48=144m$

Position at t= 4 seconds is 144 m

4 0

3 years ago

Forest protect the soil from erosion ? True or False

Ivenika [448]

Answer:

Forests helps in binding the soil together. Thus, prevent soil from erosion.

4 0

2 years ago

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A bike first accelerates from 0.0m/s to 4.5m/s in 4.5 s, the continues at this constant speed for another 6.0 s. What is the tot

Alex_Xolod [135]

Answer:

37.125 m

Explanation:

Using the equation of motion

s=ut+0.5at^{2} where s is distance, u is initial velocity, t is time and a is acceleration

Distance during acceleration

Acceleration, a=\frac {V_{final}-V_{initial}}{t} where V_{final} is final velocity and V_{initial} is initial velocity.

Substituting 0.0 m/s for initial velocity and 4.5 m/s for final velocity, acceleration will be

a=\frac {4.5 m/s-0 m/s}{4.5 s}=1 m/s^{2}

Then substituting u for 0 m/s, t for 4.5 s and a for 1 m/s^{2} into the equation of motion

s=0*4.5+ 0.5*1*4.5^{2}=0+10.125 =10.125 m

Distance at a constant speed

At a constant speed, there's no acceleration and since speed=distance/time then distance is speed*time

Distance=4.5 m/s*6 s=27 m

Total distance

Total=27+10.125=37.125 m

3 0

3 years ago

Find the fundamental frequency and the next three frequencies that could cause standing-wave patterns on a string that is 30.0 m

maksim [4K]

Answer:

0.786 Hz, 1.572 Hz, 2.358 Hz, 3.144 Hz

Explanation:

The fundamental frequency of a standing wave on a string is given by

$f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}$

where

L is the length of the string

T is the tension in the string

$\mu$ is the mass per unit length

For the string in the problem,

L = 30.0 m

$\mu=9.00\cdot 10^{-3} kg/m$

T = 20.0 N

Substituting into the equation, we find the fundamental frequency:

$f=\frac{1}{2(30.0)}\sqrt{\frac{20.0}{(9.00\cdot 10^{-3}}}=0.786 Hz$

The next frequencies (harmonics) are given by

$f_n = nf$

with n being an integer number and f being the fundamental frequency.

So we get:

$f_2 = 2 (0.786 Hz)=1.572 Hz$

$f_3 = 3 (0.786 Hz)=2.358 Hz$

$f_4 = 4 (0.786 Hz)=3.144 Hz$

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4 years ago

When an object is at a distance of twice the focal length from a concave lens, the image produced is virtual and smaller than th

Elenna [48]

The right answer for the question that is being asked and shown above is that: "The image produced is virtual and of the same size as the object." the image if the object is shifted closer to the lens to a point one focal length away from it is that The image produced is virtual and of the same size as the object.

3 0

4 years ago

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