Which has a larger resistance a 60 w lightbulb or a 100 w lightbulb?

The motion of an object looks different to observers in different

Positions. Happy to help! Please mark as Brainliest!

7 0

2 years ago

a 4357 kg roller coaster car starts from rest at the top of a 36.5 m high track. determine the speed of the car at the top of a

andrey2020 [161]

The correct answer is 17.24 m/s. You get the answer by subtracting the two heights of the tracks which are 36.5 and 10.8 m, and the answer is 25.7. Since you already know the height at which the kinetic energy will be coming from, you then divide the amount of weight the roller coaster has to the distance it needs to travel in order for you to determine the speed of the car. So that is, 4,357 kg and 25.7 m and the answer is 169 kg/m. Dividing it to the earth's gravity of 9.8 m/s you'll get 17.24 m/s.

4 0

2 years ago

A particle in the first excited state of a one-dimensional infinite potential energy well (with U = 0 inside the well) has an en

nataly862011 [7]

Answer:

The energy of this particle in the ground state is E₁=1.5 eV.

Explanation:

The energy $E_{n}$ of a particle of mass m in the nth energy state of an infinite square well potential with width L is:

$E_{n}=\frac{n^{2}h^{2}}{8mL^{2}}$

In the ground state (n=1). In the first excited state (n=2) we are told the energy is E₂= 6.0 eV. If we replace in the above equation we get that:

$E_{1}=\frac{h^{2}}{8mL^{2}}$

$E_{2}=\frac{h^{2}}{2mL^{2}}$

So we can rewrite the energy in the ground state as:

$E_{1}=\frac{1}{4}(\frac{h^{2}}{2mL^{2}})$

$E_{1}=\frac{1}{4} E_{2}$

$E_{1}=\frac{1}{4} ( 6.0\ eV)$

Finally

$E_{1}=1.5\ eV$

6 0

2 years ago

You have negotiated with the Omicronians for a base on the planet Omicron Persei 7. The architects working with you to plan the

steposvetlana [31]

Answer:

5.724 meters / second^2

Explanation:

We are given two pieces of information, 5.24 flurg = 1 meter, 1 grom = 0.493 second. If that is so, we can say that there are two possible conversion units, 5.25 flurg / meter, and 0.493 second / grom.

_____

We want to convert 7.29 flurg / grom^2 ( I believe? ) to the units meters / second^2. But, let's break this down into bits. It would be convenient to first convert 7.29 flurg / grom^2 to the units meters / grom^2, by dividing the conversion factors as to cancel out the appropriate things, which we will go into detail on a bit later ( using the first conversion factor ). Respectively we can convert meters / grom^2 to meters / grom * s, canceling out the flurg ( through the second conversion factor ). And now we would need to get rid of the grom, dividing similarly.

_____

( 1 ) ( flurg / grom^2 ) / ( flurg / meters ) - first conversion unit

= flurg / grom^2 * meters /flurg

= ( meters * flurg ) / ( grom^2 * flurg )

= meters /grom^2,

7.29 flurg / grom^2 / 5.24 flurg / meter = ( About ) 1.39 meter / grom^2

( 2 ) ( meter / grom^2 ) / ( second / grom ) - second conversion unit

= meter / grom^2 * grom / second

= ( meter * grom ) / ( grom^2 * second )

= meter / ( grom * second ),

( 1.39 meter / grom^2 ) / 0.493 second / grom = ( About ) 2.82195 meter / grom * second

( 3 ) ( 2.82195 meter / ( grom * second ) ) / 0.493 second / grom = 5.724 meter / second^2

( And thus, the value of gOP7 in the units the architects will use should be about 5.724 meters / second^2 )

8 0

2 years ago

You are called as an expert witness to analyze the following auto accident: Car B, of mass 2000 kg, was stopped at a red light w

OLga [1]

Answer:

a). va=17.23 $\frac{m}{s}$ or 38.54 mph

b). v=38.54 mph and limit is 35 mph

c). Completely inelastic

d). Eka=192.967 kJ

Ekt=76.071 kJ

Explanation:

$m_{a}=1300kg\\m_{b}=2000kg\\x_{f}=7.25m\\u_{k}=0.65$

The motion is an inelastic collision so

$m_{a}*v_{a}+m_{b}*v_{b}=(m_{a}+m_{b})*v_{f}$

The force of the motion is contrarest by the force of friction so

$F-F_{uk} =0\\F=F_{uk}\\F_{uk}=u_{k}*m*g\\F=m*a\\a=\frac{F}{m}\\ a=\frac{F_{uk}}{m}\\a=\frac{u_{k}*m*g}{m}\\a=u_{k}*g\\a=0.65*9.8\frac{m}{s^{2}} \\a=6.39\frac{m}{s^{2}}$

Now with the acceleration can find the time and the velocity final that make the distance 7.25m being united

$x_{f}=x_{o}+v_{o}*t+2*a*t^{2}\\x_{o}=0\\v_{o}=0\\x_{f}=2*a*t^{2}\\t^{2}=\frac{x_{f}}{2*a}\\t=\sqrt{\frac{7.25m}{6.37\frac{m}{s^{2} } } } \\t=1.06s$

So the velocity final can be find using this time

$v_{f}=v_{o}+a*t\\v_{o}=0\\v_{f}=6.37\frac{m}{s^{2} } *1.06s\\v_{f}=6.79 \frac{m}{s}$

a).

Replacing in the first equation the final velocity can find the initial velocity

$m_{a}*v_{a}+m_{b}*v_{b}=(m_{a}+m_{b})*v_{f}$

$v_{b}=0$

$v_{a}= \frac{(m_{a}+m_{b)*v_{f}}}{m_{a}}\\v_{a}= \frac{(1300+2000)*6.37}{1300}\\v_{a}=17.23 \frac{m}{s}$

b).

$35mph*\frac{1m}{0.000621371mi} *\frac{1h}{3600s}=15.646\frac{m}{s}$

Velocity limit in m/s is 15.646 m/s and the initial velocity is 17.23 m/s

so is exceeding the speed limit in about 1.58 m/s

or in miles per hour

3.5 mph

c).

The collision is complete inelastic because any mass can be returned to the original mass, so even they are no the same mass however in the moment they move the distance 7.25m as a same mass the motion is considered completely inelastic

d).

$Ek=\frac{1}{2}*m*(v)^{2}\\ Eka=\frac{1}{2}*1300kg*(17.23\frac{m}{s})^{2}\\Eka=192.967 kJ\\Ekt=\frac{1}{2}*m*(v)^{2}\\Ekt=\frac{1}{2}*3300kg*(6.79\frac{m}{s})^{2}\\Ekt=76.071 kJ$

8 0

2 years ago