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Advocard [28]
3 years ago
10

Newton’s Second Law establishes the relationship between mass, net applied force, and acceleration given by F=ma. Consider a 4 k

g box of holiday candy on a horizontal surface such as a table. There is a 10N applied force to the right and a 2N force to the left. The box accelerates with a magnitude of _________?
Physics
1 answer:
netineya [11]3 years ago
7 0

Answer:

a= 2 m/s^2

Explanation:

take to the right as positive

let Ftot be the total forces acting on the box , m be the mass of the box and a be the acceleration of the box.

Ftot = 10 - 2 = 8 N

and,

Ftot = ma

   a = Ftot/m

      = 8/4

      = 2 m/s^2

therefore, the acceleration of the box is of magnitude of 2 m/s^2.

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You perform a double‑slit experiment in order to measure the wavelength of the new laser that you received for your birthday. Yo
BigorU [14]

Answer:

\lambda = 6.25\times10^{-9}= 625 nm

Explanation:

We now that for

for maximum intensity(bright fringe) d sinθ=nλ n=0,1,2,....

d= distance between the slits, λ= wavelength of incident ray

for small θ, sinθ≈tanθ= y/D where y is the distance on screen and D is the distance b/w screen and slits.

Given

d=1.19 mm, y=4.97 cm,  and,   n=10,   D=9.47 m

applying formula

λ= (d*y)/(D*n)

putting values we get

\lambda = \frac{1.19\times10^{-3}\times4.97\times10^{-2}}{9.47\times10}

on solving we get

\lambda = 6.25\times10^{-9}= 625 nm

8 0
3 years ago
What is the equation for electric potential
Rama09 [41]

Answer:

The unit of charge is the Coulomb (C), and the unit of electric potential is the Volt (V), which is equal to a Joule per Coulomb (J/C).

Explanation:

4 0
3 years ago
In a television set, electrons are accelerated from rest through a potential difference of 20 kV. The electrons then pass throug
Svetradugi [14.3K]

Answer: Fmax = 5.54*10^-12 N

Explanation: From the question, we have the potential difference (V) =20kv = 20,000v and strength of magnetic field (B) =0.41 T.

The maximum force experienced by a charge of magnitude (q) is given as

Fmax = qvB

Where v = velocity of electron.

The velocity of the electron can be gotten by using the work energy theorem.

The kinetic energy of the electron (mv²/2) equals the work done needed to accelerate it.

mv²/2 = qV.

Where m = mass of an electronic charge = 9.11×10^-31 kg, q = magnitude of an electronic charge = 1.609×10^-19 c, v = velocity of electron, V = potential difference = 20,000v.

By substituting the parameters, we have that

(9.11×10^-31 × v²)/2 = 1.609×10^-19 × 20000

(9.11×10^-31 × v²) = 1.609×10^-19 × 20000 ×2

v² = (1.609×10^-19 × 20000 ×2)/9.11×10^-31

v² = 64.36*10^(-16)/9.11×10^-31

v² = 7.0647×10^15

v = √7.0647×10^15

v = 8.40×10^7 m/s

Fmax = 1.609×10^-19 × 8.40×10^7 × 0.41

Fmax = 5.54*10^-12 N

7 0
3 years ago
A 76-W incandescent light bulb operates at 120 V. How many electrons and coulombs flow through the bulb in one day?
Tema [17]

Answer:

43200c

Explanation:

7 0
3 years ago
Which sphere is NOT a part of the cycling of oxygen through Earths systems ?
stepan [7]
The answer would be Exosphere because, there are 3 main regions that circulate oxygen through the Earths system, which are the Biosphere, Atmosphere, and the Lithosphere. 
5 0
3 years ago
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