Ask question

Ask question

All categories

2 years ago

6

A car moving with a velocity of 20 meters/second has

1 answer:

kenny6666 [7]2 years ago

5 0

Kinetic energy is the energy of an object in motion. It is expressed as the product of its mass and the square of its velocity. The opposite of this energy is called the potential energy, the energy of an object at rest. Calculations are as follows:
<span>
KE = mv^2
</span><span>1.8x10^5 J = m(20 m/s)^2
</span><span>m = 450 kg</span>

You might be interested in

A unit for measuring frequency is the

jonny [76]

A.) Hertz is the measuring unit for frequency!!
[Your explanation is correct]

Hope this helps!

4 0

2 years ago

A racecar drives at a constant speed down a straight track. The car is in _?_

vagabundo [1.1K]

Answer:

the answer is

the car is in motion

7 0

2 years ago

The following table lists the work functions of a few common metals, measured in electron volts. Metal Φ(eV) Cesium 1.9 Potassiu

Citrus2011 [14]

A. Lithium

The equation for the photoelectric effect is:

$E=\phi + K$

where

$E=\frac{hc}{\lambda}$ is the energy of the incident light, with h being the Planck constant, c being the speed of light, and $\lambda$ being the wavelength

$\phi$ is the work function of the metal (the minimum energy needed to extract one photoelectron from the surface of the metal)

K is the maximum kinetic energy of the photoelectron

In this problem, we have

$\lambda=190 nm=1.9\cdot 10^{-7}m$ , so the energy of the incident light is

$E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{1.9\cdot 10^{-7} m}=1.05\cdot 10^{-18}J$

Converting in electronvolts,

$E=\frac{1.05\cdot 10^{-18}J}{1.6\cdot 10^{-19} J/eV}=6.5 eV$

Since the electrons are emitted from the surface with a maximum kinetic energy of

K = 4.0 eV

The work function of this metal is

$\phi = E-K=6.5 eV-4.0 eV=2.5 eV$

So, the metal is Lithium.

B. cesium, potassium, sodium

The wavelength of green light is

$\lambda=510 nm=5.1\cdot 10^{-7} m$

So its energy is

$E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5.1\cdot 10^{-7} m}=3.9\cdot 10^{-19}J$

Converting in electronvolts,

$E=\frac{3.9\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=2.4 eV$

So, all the metals that have work function smaller than this value will be able to emit photoelectrons, so:

Cesium

Potassium

Sodium

C. 4.9 eV

In this case, we have

- Copper work function: $\phi = 4.5 eV$

- Maximum kinetic energy of the emitted electrons: K = 2.7 eV

So, the energy of the incident light is

$E=\phi+K=4.5 eV+2.7 eV=7.2 eV$

Then the copper is replaced with sodium, which has work function of

$\phi = 2.3 eV$

So, if the same light shine on sodium, then the maximum kinetic energy of the emitted electrons will be

$K=E-\phi = 7.2 eV-2.3 eV=4.9 eV$

7 0

2 years ago

Which factor would reduce the size of a forest population

zhuklara [117]

C. Lack of mates. If they cannot reproduce enough, their size will reduce.

7 0

2 years ago

Read 2 more answers

Two balls move away from each other, both traveling at 7 m/s. One has a mass of 2 kg and the other has a mass of 3 kg

zavuch27 [327]

A 300-kg bear grasping a vertical tree slides down at constant velocity. The friction force between the
tree and the bear is

5 0

2 years ago