The rate at which work is done is called

I will give brainiest to the first person that gives a somewhat sensible answer, but for the love of god, someone answer!

Olegator [25]

It’s c Because is the one the makes the most sense

7 0

3 years ago

Using F=ma, explain why a train is more difficult to start moving and more difficult to bring to a stop than an average sized ca

seropon [69]

Answer:

M_Train>> m_car a_train <a_car

Explanation:

To start the movement of the train or the car, the motorcycle applies a force on the wheels, which starts the acceleration in the case of the train, it has a much greater mass than that of the car, for which to obtain the same acceleration necessary a much greater force

a = F / m

as the mass of the train is greater than that of the car.

a_train <a_car

Something similar happens when the vehicles stop, the engine stops applying force forward and the brakes apply a force backward that creates a negative acceleration that slows down, again as the mass of the train is much greater than the of the car its negative acceleration is much less.

It is good to clarify that to compensate for this the trains have a braking system on all wheels

6 0

3 years ago

10. A triply ionized beryllium atom is in the ground state. It absorbs energy and makes a transition to the n = 5 excited state.

Xelga [282]

Answer:

$\lambda=1282\ nm$

Explanation:

$E_n=-2.179\times 10^{-18}\times \frac{1}{n^2}\ Joules$

For transitions:

$Energy\ Difference,\ \Delta E= E_f-E_i =-2.179\times 10^{-18}(\frac{1}{n_f^2}-\frac{1}{n_i^2})\ J=2.179\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J$

$\Delta E=2.179\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J$

Also, $\Delta E=\frac {h\times c}{\lambda}$

Where,

h is Plank's constant having value $6.626\times 10^{-34}\ Js$

c is the speed of light having value $3\times 10^8\ m/s$

So,

$\frac {h\times c}{\lambda}=2.179\times 10^{-18}(|\frac{1}{n_i^2} - \dfrac{1}{n_f^2}|)\ J$

$\lambda=\frac {6.626\times 10^{-34}\times 3\times 10^8}{{2.179\times 10^{-18}}\times (|\frac{1}{n_i^2} - \dfrac{1}{n_f^2}|)}\ m$

So,

$\lambda=\frac {6.626\times 10^{-34}\times 3\times 10^8}{{2.179\times 10^{-18}}\times (|\frac{1}{n_i^2} - \dfrac{1}{n_f^2}|)}\ m$

Given, $n_i=5\ and\ n_f=3$

$\lambda=\frac{6.626\times 10^{-34}\times 3\times 10^8}{{2.179\times 10^{-18}}\times (\frac{1}{5^2} - \dfrac{1}{3^2})}\ m$

$\lambda=\frac{10^{-26}\times \:19.878}{10^{-18}\times \:2.179\left(|\frac{1}{25}-\frac{1}{9}\right)|}\ m$

$\lambda=\frac{19.878}{10^8\times \:2.179\left(|-\frac{16}{225}\right|)}\ m$

$\lambda= 1.2828\times10^{-6}$

1 m = 10⁻⁹ nm

$\lambda=1282\ nm$

6 0

3 years ago

On a straight road (taken to be in the x direction) you drive for an hour at 60 km per hour, then quickly speed up to 120 km per

Luda [366]

Answer:

The average velocity is 180 km/hr

Explanation:

Given;

initial velocity, u = 60 km per hour

final velocity, v = 120 km per hour

initial time = 1 hour

final time = 2 hour

Initial position = 60 km/h x 1 hour = 60 km

final position = 120 km/h x 2 hour = 240 km

The average velocity is given by;

$V_{avg} = \frac{Final \ position\ - \ Initial \ position}{final \ time\ - \ initial \ time}\\\\V_{avg} = \frac{240km \ - \ 60km}{2hr\ - \ 1hr} \\\\V_{avg} = \frac{180 \ km}{1hr} \\\\V_{avg}= 180 \ km/hr$

Therefore, the average velocity is 180 km/hr

3 0

3 years ago

What is the weight of a 45 kg box? N

Aleks04 [339]

45 kg * 9.8 m/s^2 = <u>441 N.</u>

7 0

3 years ago

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