If a 100 N net force act on a 50-kg car, what will the acceleration be

an airplane is flying through a thundercloud at a height of 2000m (this is very dangerous thing to do because of updrafts, turbu

Vedmedyk [2.9K]

Answer

In this question we have given,

Height of plane, h1=2000m

Height at which charge concentration is 40C, h2=3000m

Height at which charge concentration is -40C, h3=1000m

charge concentaration, q1=40C

charge concentaration, q2=-40C

let the charge concentrations at height h2 and h3 as point charges

Now we will first find the electric feild on plane due to positive charge q1=40

$E1= k*q1/(h1-h2)..............(1)$

Here k=8.98755*10^9N.m^2/C^2

q1=40C

put values of k, q1 , h1 and h2 in equation 1

$[tex]E1=(8.98755*10^9)*(40)/(2000-3000)^2$ \\

$E1=[tex]E= 359502+359502\\E=719004 V/m$ V/m[/tex]

similarly electric feild due to negative charge q2=-40

$[tex]E2=(8.98755*10^9)*(-40)/(2000-1000)^2$ \\

$E2=359502V/m$

Total electric feild E at the aircraft is given as

$E= E1+ E2\\$ ...............(2)

Put values of E1 and E2 in equation2

$\\E=359502+359502\\E= 719004V/m\\$

therefore s Total electric feild E at the aircraft is E= 719004V/m

3 0

2 years ago

What is the energy in joules of a mole of photons associated with visible light of wavelength 486 nm?

ivann1987 [24]

Answer:

$2.46\cdot 10^5 J$

Explanation:

The energy of a single photon is given by:

$E=\frac{hc}{\lambda}$

where

h is the Planck constant

c is the speed of light

$\lambda$ is the wavelength

For the photon in this problem,

$\lambda=486 nm=4.86\cdot 10^{-7}m$

So, its energy is

$E_1=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{4.86\cdot 10^{-7}m}=4.09\cdot 10^{-19} J$

One mole of photons contains a number of photons equal to Avogadro number:

$N_A = 6.022\cdot 10^{23}$

So, the total energy of one mole of photons is

$E=N_A E_1 = (6.022\cdot 10^{23})(4.09\cdot 10^{-19} J)=2.46\cdot 10^5 J$

7 0

3 years ago

Mercury is characterized by

cupoosta [38]

It's close to the sun without much atmosphere, so it's characterized by very extreme temperatures.

Happy studying ^_^

8 0

2 years ago

A certain piece of metal (density 9.45 grams per cubic centimeter) has the shape of a hockey puck with a diameter of 13 cm and a

atroni [7]

Answer:

0.0195 m

Explanation:

$\rho _{p}$ = density of hockey puck = 9.45 gcm⁻³ = 9450 kgm³

$d_{p}$ = diameter of hockey puck = 13 cm = 0.13 m

$h_{p}$ = height of hockey puck = 2.8 cm = 0.028 m

$\rho _{m}$ = density of mercury = 13.6 gcm⁻³ = 13600 kgm³

$d$ = depth of puck below surface of mercury

According to Archimedes principle, the weight of puck is balanced by the weight of mercury displaced by puck

Weight of mercury displaced = Weight of puck

$\rho _{m} (0.25)(\pi d_{p}^{2} d ) g = \rho _{p} (0.25)(\pi d_{p}^{2} h_{p} ) g\\\rho _{m} ( d ) = \rho _{p} ( h_{p} )\\(13600) d = (9450) (0.028)\\d = 0.0195 m$

7 0

2 years ago

Two carts, one of mass 2m and one of mass m, approach each other with the same speed, v. When the carts collide, they hook toget

Nikitich [7]

Answer:

Second option

Explanation:

Linear Momentum

The linear momentum of an object of mass m and speed v is

P=mv

If two or more objects are interacting in the same axis, the total momentum is

$P_t=m_1v_1+m_2v_2+...$

Where the speeds must be signed according to a fixed reference

The images show a cart of mass 2m moves to the left with speed v since our reference is positive to the right

$P_1=-2mv$

The second cart of mass m goes to the right at a speed v

$P_2=mv$

The total momentum before the impact is

$P_t=-2mv+mv=-mv$

The total momentum after the collision is negative, both carts will join and go to the left side

The first option shows both carts with the same momentum before the collision and therefore, zero momentum after. It's not correct as we have already proven

The third option shows the 2m cart has a positive greater momentum than the other one. We have proven the 2m car has negative momentum. This option is not correct either

The fourth option shows the two carts keep separated after the collision, which contradicts the condition of the question regarding "they hook together".

The second option is the correct one because the mass $m_2$ has a negative momentum and then the sum of both masses keeps being negative

3 0

3 years ago

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