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andrew-mc [135]

2 years ago

14

Check out this app! It’s millions of students helping each other get through their schoolwork. https://brainly.app.link/qpzV02Ma

wvnVBx

2 answers:

Veronika [31]2 years ago

7 0

Why did you send a link to this app lol?

gayaneshka [121]2 years ago

3 0

Answer: kay.

Explanation:

You might be interested in

The energy that generates wind comes from what source?

mamaluj [8]

The energy that generates wind on an individual basis originates with
one's habitual diet. On a world-wide basis, it comes from the sun.

6 0

3 years ago

A Van de Graaff generator is one of the original particle accelerators and can be used to accelerate charged particles like prot

Rzqust [24]

Answer:

595391.482946 m/s

$3.21875\times 10^{6}$

Explanation:

E = Energy = 1.85 keV

I = Current = 5.15 mA

e = Charge of electron = $1.6\times 10^{-19}\ C$

t = Time taken = 1 second

m = Mass of proton = $1.67\times 10^{-27}\ kg$

Velocity of proton is given by

$v=\sqrt{\dfrac{2E}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 1.85\times 10^3\times 1.6\times 10^{-19}}{1.67\times 10^{-27}}}\\\Rightarrow v=595391.482946\ m/s$

The speed of the proton is 595391.482946 m/s

Current is given by

$I=\dfrac{\Delta Q}{t}\\\Rightarrow \Delta Q=It\\\Rightarrow \Delta Q=5.15\times 10^{-3}\times (1\ sec)\\\Rightarrow Q=5.15\times 10^{-3}\ C$

Number of protons is

$n=\dfrac{Q}{e}\\\Rightarrow n=\dfrac{5.15\times 10^{-3}}{1.6\times 10^{-19}}\\\Rightarrow n=3.21875\times 10^{6}\ protons$

The number of protons is $3.21875\times 10^{6}$

3 0

3 years ago

Does the same battery always deliver the same amount of flow to any circuit? Mention two observations of any circuits in this la

V125BC [204]

Answer:

Yes

Explanation:

Given that the battery is the same the PD ( potential difference ) in the circuit will also be the same likewise the flow of charge in the circuit,

Hence the same amount of charge flow is delivered to any circuit.

attached below are examples

6 0

3 years ago

In a two-slit experiment using coherent light, the distance between the slits and the screen is 1.10 m, and the distance between

Paul [167]

Answer:

D) 763 nm

Explanation:

Calculation for the wavelength of light

Using this formula

Wavelength of light=Delta Y*Distance / Length

Where,

Delta Y represent the 2nd order bright fringe

Length represent the distance between both the slits and the screen

Distance represent the Distance between the slits

Let note that cm to m = (4.2) x 10^-2 and mm to m= ( 0.0400x 10^-3)

Now Let plug in the formula

Wavelength of light=[(4.2 x 10^-2m)(0.0400 x 10^-3m) / 2(1.1m)]*10^-7 meters

Wavelength of light=[(0.042m) (0.0004m)/2.2m]*10^-7 meters

Wavelength of light =(0.0000168m/2.2m)*10^-7 meters

Wavelength of light =7.63 *10^-7 meters

Wavelength of light =763 nm

Therefore the Wavelength of light will be 763 nm

3 0

2 years ago

During an experiment of momentum, trolley, X, of mass (2.34 ± 0.01) kg is moving away from another trolley, Y, of mass (2.561 ±

Alla [95]

Answer:

P = 1 (14,045 ± 0.03 ) k gm/s

Explanation:

In this exercise we are asked about the uncertainty of the momentum of the two carriages

Δ (Pₓ / Py) =?

Let's start by finding the momentum of each vehicle

car X

Pₓ = m vₓ

Pₓ = 2.34 2.5

Pₓ = 5.85 kg m

car Y

Py = 2,561 3.2

Py = 8,195 kgm

How do we calculate the absolute uncertainty at the two moments?

ΔPₓ = m Δv + v Δm

ΔPₓ = 2.34 0.01 + 2.561 0.01

ΔPₓ = 0.05 kg m

Δ $P_{y}$ = m Δv + v Δm

ΔP_{y} = 2,561 0.01+ 3.2 0.001

ΔP_{y} = 0.03 kg m

now we have the uncertainty of each moment

P = Pₓ / $P_{y}$

ΔP = ΔPₓ/P_{y} + Pₓ ΔP_{y} / P_{y}²

ΔP = 8,195 0.05 + 5.85 0.03 / 8,195²

ΔP = 0.006 + 0.0026

ΔP = 0.009 kg m

The result is

P = 14,045 ± 0.039 = (14,045 ± 0.03 ) k gm/s

7 0

3 years ago