All categories

3 years ago

You are riding on a school bus and settling get thrown forward what did the bus just do

Physics

2 answers:

Ksivusya [100]3 years ago

5 0

hit its brakes aka as going straight but abruptly stops

hope this helps :)

Arturiano [62]3 years ago

5 0

It accelerated. Lol I just had something like this it is so stupid

You might be interested in

Balance the equation- Al+Mn02 ———-> Mn + Al2O3

joja [24]

Answer:

Explanation:

3 0

3 years ago

The equation of a transverse wave traveling along a string is 1 1 y (2.00 mm)sin[(20 m )x (600 s )t] − − = − Find the (a) amplit

Lelu [443]

Answer:

a)Amplitude ,A = 2 mm

b)f=95.49 Hz

c)V= 30 m/s ( + x direction )

d) λ = 0.31 m

e)Umax= 1.2 m/s

Explanation:

Given that

$y=2\ mm\ sin[(20m^{-1})x-(600s^{-1})t]$

As we know that standard form of wave equation given as

$y=A sin(\phi -\omega t)$

A= Amplitude

ω=Frequency (rad /s)

t=Time

Φ = Phase difference

$y=2\ mm\ sin[(20m^{-1})x-(600s^{-1})t]$

So from above equation we can say that

Amplitude ,A = 2 mm

Frequency ,ω= 600 rad/s (2πf=ω)

ω= 2πf

f= ω /2π

f= 300/π = 95.49 Hz

K= 20 rad/m

So velocity,V

V= ω /K

V= 600 /20 = 30 m/s ( + x direction )

V = f λ

30 = 95.49 x λ

λ = 0.31 m

We know that speed is the rate of displacement

$U=\dfrac{dy}{dt}$

$U=2\ mm\ sin[(20m^{-1})x-(600s^{-1})t]$

$U=1200\ cos[(20m^{-1})x-(600s^{-1})t]\ mm/s$

The maximum velocity

Umax = 1200 mm/s

Umax= 1.2 m/s

8 0

3 years ago

It says find the slope for each line I'm stuck on number one can you help me

Allushta [10]

$\text{slope}=\frac{y_2-y_1}{x_2-x_1}$

(-2, -1)(3, 1)

Therefore,

$\text{slope}=\frac{1-(-1)}{3-(-2)}=\frac{1+1}{3+2}=\frac{2}{5}$

7 0

1 year ago

565,900 seconds into days

alukav5142 [94]

Answer:

Explanation:

4 0

3 years ago

The radii of the sprocket assemblies and the wheel of the bicycle in the figure are:

Furkat [3]

To solve this task we have to make a proportion, but firstly we have to set up all the main points : so, the distance is s=r(B), that has its r=radius,B=angle in rad velocity v=ds/dt= w(r)
Do not forget about w = angular speed in rad/s and $w1 = 1 revolution/sec = 2Pi (rad/s)$
Now we can go to proportion
$v1=v2$
$w1*r1 = w2r2$ $w2 = w1 * r1/r2 = 2w1 = 4Pi (rad/s)$
$w2 = w3 (which is the angular velocity of the rear wheel) 
$
SOLVING FOR A : $v3 = w3 * r3 = 4pi * 14 (inch/s) = 14.66 ft/sec$
$v3 = 14.66 ft/sec(1 mile/5280 ft)( 3600 sec/h)$ $= 9.99$ or something about 10 mph --- SOLVING FOR B.
I'm sure it helps!

7 0

3 years ago