How many days are in 345000 minutes? (Set up by dimensional analysis)

ITS TIMED PLEASE HELP !!!!!

dusya [7]

Answer: direct linear

Explanation:

3 0

3 years ago

Please......... I need desperate help

PIT_PIT [208]

' A ' is one crest of the wave. After every wavelength, there's another one.

' B ' . . . the vertical arrow under B shows the amplitude of the wave

' C ' is one trough of the wave. After every wavelength, there's another one.

' D ' . . . the horizontal arrow over D shows the wavelength.

3 0

3 years ago

A 125kg bumper car going 12m/s hits a 235kg bumper car going -13m/s.if the first car bounces back at -12.5m/s what is the veloci

vovikov84 [41]

According to the law of conservation of momentum:

$m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}'$

m1 = mass of first object
m2 = mass of second object
v1 = Velocity of the first object before the collision
v2 = Velocity of the second object before the collision
v'1 = Velocity of the first object after the collision
v'2 = Velocity of the second object after the collision

Now how do you solve for the velocity of the second car after the collision? First thing you do is get your given and fill in what you know in the equation and solve for what you do not know.

m1 = 125 kg v1 = 12m/s v'1 = -12.5m/s
m2 = 235kg v2 = -13m/s v'2 = ?

$m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}'$
$(125kg)(12m/s)+(235kg)(-13m/s)=(125kg)(-12.5m/s)+(235kg)(v_{2}'$
$1,500kg.m/s+(-3055kg.m/s)=(-1562.5kg.m/s)+(235kg)(v_{2}')$
$-1,555kg.m/s=(-1562.5kg.m/s)+(235kg)(v_{2}')$

Transpose everything on the side of the unknown to isolate the unknown. Do not forget to do the opposite operation.

$-1,555kg.m/s + 1562.5kg.m/s=(235kg)(v_{2}')$
$7.5kg.m/s=(235kg)(v_{2}')$
$(7.5kg.m/s)/(235kg)=(v_{2}')$
$0.03m/s=(v_{2}')$

The velocity of the 2nd car after the collision is 0.03m/s.

5 0

3 years ago

A 53.0-kg skater is traveling due east at a speed of 2.90 m/s. A 72.0-kg skater is moving due south at a speed of 6.20 m/s. They

soldier1979 [14.2K]

Answer:

a. $\thta=71^{\circ}$

b. $v_f=3.78 m/s$

Explanation:

We are given that

$m_1=53 kg$

$v_1=2.9 m/s$

$m_2= 72 kg$

$v_2=6.2 m/s$

a.We have to find the angle

$\theta=tan^{-1}{\frac{m_2v_2}{m_1v_1}=\frac{72(6.2)}{53(2.9)}=71^{\circ}$

$\theta=71^{\circ}$

b. We have to find the speed $v_f$

According to law of conservation of momentum

$m_1v_1=(m_1+m_2)v_fcos\theta$

$53(2.9)=(53+72)v_fcos 71^{\circ}=40.7 v_f$

$v_f=\frac{153.7}{40.7}=3.78 m/s$

3 0

3 years ago

How does inertia relate to coin dropping????

vekshin1

Well Inertia means something wants to stay in place, and in reality that coin wants to stay in one place, If you placed it on an index card on a cup, and SLOWLY pulled it, it wouldn't be fast enough to overcome that force, if you pulled it quickly that coin would stay in place and drop into the cup.

3 0

3 years ago

How many days are in 345000 minutes? (Set up by dimensional analysis)​

How many days are in 345000 minutes? (Set up by dimensional analysis)