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marishachu [46]

3 years ago

11

Can someone explain how exactly to calculate the net force? Do you subtract the smaller number from the larger number? Any help

would be appreciated!

1 answer:

aleksley [76]3 years ago

3 0

Explanation:

Let's say right is positive and left is negative.

F₁ = -150 N

F₂ = 220 N

Fnet = F₁ + F₂

Fnet = -150 N + 220 N

Fnet = 70 N

The magnitude of Fnet is 70 N, and since it's positive, the direction is to the right.

And since Fnet isn't 0, the force is unbalanced and the motion is changing.

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Two charges, each of 2.9 microC are placed at two corners of a square 50cm on a side, If the charges are on one side of the squa

anyanavicka [17]

Answer:

The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

Explanation:

Given that,

First charge $q_{1}= 2.9\mu C$

Second charge $q_{2}= 2.9\mu C$

Distance between two corners r= 50 cm

We need to calculate the electric field due to other charges at one corner

For E₁

Using formula of electric field

$E_{1}=\dfrac{kq}{r'^2}$

Put the value into the formula

$E_{1}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\sqrt{2}\times10^{-2})^2}$

$E_{1}=52200=52.2\times10^{3}\ N/C$

For E₂,

Using formula of electric field

$E_{1}=\dfrac{kq}{r^2}$

Put the value into the formula

$E_{2}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\times10^{-2})^2}$

$E_{2}=104400=104.4\times10^{3}\ N/C$

We need to calculate the horizontal electric field

$E_{x}=E_{1}\cos\theta$

$E_{x}=52.2\times10^{3}\times\cos45$

$E_{x}=36910.97=36.9\times10^{3}\ N/C$

We need to calculate the vertical electric field

$E_{y}=E_{2}+E_{1}\sin\theta$

$E_{y}=104.4\times10^{3}+52.2\times10^{3}\sin45$

$E_{y}=141310.97=141.3\times10^{3}\ N/C$

We need to calculate the net electric field

$E_{net}=\sqrt{E_{x}^2+E_{y}^2}$

Put the value into the formula

$E_{net}=\sqrt{(36.9\times10^{3})^2+(141.3\times10^{3})^2}$

$E_{net}=146038.69\ N/C$

$E_{net}=146.03\times10^{3}\ N/C$

We need to calculate the direction of electric field

Using formula of direction

$\tan\theta=\dfrac{141.3\times10^{3}}{36.9\times10^{3}}$

$\theta=\tan^{-1}(\dfrac{141.3\times10^{3}}{36.9\times10^{3}})$

$\theta=75.36^{\circ}$

Hence, The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

4 0

3 years ago

An LC circuit is built with a 20 mH inductor and an 8.0 PF capacitor. The capacitor voltage has its maximum value of 25 V at t =

Margaret [11]

Answer:

a) the required time is 0.6283 μs

b) the inductor current is 0.5 mA

Explanation:

Given the data in the question;

The capacitor voltage has its maximum value of 25 V at t = 0

i.e V $_m$ = V₀ = 25 V

we determine the angular velocity;

ω = 1 / √( LC )

ω = 1 / √( ( 20 × 10⁻³ H ) × ( 8.0 × 10⁻¹² F) )

ω = 1 / √( 1.6 × 10⁻¹³ )

ω = 1 / 0.0000004

ω = 2.5 × 10⁶ s⁻¹

a) How much time does it take until the capacitor is fully discharged for the first time?

V $_m$ = V₀sin( ωt )

we substitute

25V = 25V × sin( 2.5 × 10⁶ s⁻¹ × t )

25V = 25V × sin( 2.5 × 10⁶ s⁻¹ × t )

divide both sides by 25 V

sin( 2.5 × 10⁶ × t ) = 1

( 2.5 × 10⁶ × t ) = π/2

t = 1.570796 / (2.5 × 10⁶)

t = 0.6283 × 10⁻⁶ s

t = 0.6283 μs

Therefore, the required time is 0.6283 μs

b) What is the inductor current at that time?

$I$ (t) = V₀√(C/L) sin(ωt)

{ sin(ωt) = 1 )

$I$ (t) = V₀√(C/L)

we substitute

$I$ (t) = 25V × √( ( 8.0 × 10⁻¹² F ) / ( 20 × 10⁻³ H ) )

$I$ (t) = 25 × 0.00002

$I$ (t) = 0.0005 A

$I$ (t) = 0.5 mA

Therefore, the inductor current is 0.5 mA

8 0

3 years ago

Microwave ovens emit microwave energy with a wavelength of 12.2 cm. What is the energy of exactly one photon of this microwave r

Iteru [2.4K]

Answer:

$1.63\cdot 10^{-24} J$

Explanation:

The energy of a photon is given by:

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

where

h is the Planck constant

c is the speed of light

$\lambda$ is the wavelength of the photon

In this problem, we have a microwave photon with wavelength

$\lambda=12.2 cm=0.122 m$

Substituting into the equation, we find its energy:

$E=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{0.122 m}=1.63\cdot 10^{-24} J$

5 0

3 years ago

Hi..Please answer questions???

Nadya [2.5K]

Answer:

A

Explanation:

Becuse its complete number

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3 years ago

When the rubber sheet is pulled down, the volume of the bell jar is ____________ and the pressure in the bell jar is ___________

Makovka662 [10]

Answer:

increases, decreases, respiration

8 0

2 years ago