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polet [3.4K]
3 years ago
15

A positive test charge of 5.00 E-5 C is places in an electric field. The force on it is 0.751 N. The magnitude of the electric f

ield at the location of the test charge is
1.50 E4 N/C
1.52 E5 N/C
3.75 E4 N/C
3.75 E5 N/C
6.75 E4 N/C
Physics
2 answers:
zalisa [80]3 years ago
8 0

Explanation:

Charge=5.00 E-5

Force=0.751N

F=qE

0.751=5.00 E-5*E

E=1.502*10⁴

option a

Pavel [41]3 years ago
8 0

Answer:

\boxed{\mathrm{1.50 \: E^4  \: N/C}}

Explanation:

\displaystyle \mathrm{E=\frac{F_e}{q} }

\displaystyle \mathrm{Electric \: field \: strength \: (N/C)=\frac{Electric \: force \: (N)}{Charge \: (C)} }

\displaystyle \mathrm{E=\frac{0.751}{5.00\:  E^{-5}} }

\displaystyle \mathrm{E=\frac{0.751}{0.00005} }

\displaystyle \mathrm{E=15020}

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Charra [1.4K]

Answer:

b) q large and m small

Explanation:

q is large and m is small

We'll express it as :

q > m

As we know the formula:

F = Eq

And we also know that :

F = Bqv

F = \frac{mv^{2} }{r}

Bqv = \frac{mv^{2} }{r}

or Eq = \frac{mv^{2} }{r}

Assume that you want a velocity selector that will allow particles of velocity v⃗  to pass straight through without deflection while also providing the best possible velocity resolution. You set the electric and magnetic fields to select the velocity v⃗ . To obtain the best possible velocity resolution (the narrowest distribution of velocities of the transmitted particles) you would want to use particles with q large and m small.

6 0
3 years ago
A model engine accelerated forward from rest along a straight track at 10.0 m/s2 for 3.0 seconds. It then accelerates coward at
Mrac [35]

Answer:

17.2 seconds

Explanation:

Given:

v₀ = 0 m/s

a₁ = 10.0 m/s²

t₁ = 3.0 s

a₂ = 16 m/s²

t₂ = 5.0 s

a₃ = -12 m/s²

v₃ = 0 m/s

Find: t

First, find v₁:

v₁ = a₁t₁ + v₀

v₁ = (10.0 m/s²) (3.0 s) + (0 m/s)

v₁ = 30 m/s

Next, find v₂:

v₂ = a₂t₂ + v₁

v₂ = (16 m/s²) (5.0 s) + (30 m/s)

v₂ = 110 m/s

Finally, find t₃:

v₃ = a₃t₃ + v₂

(0 m/s) = (-12 m/s²) t₃ + (110 m/s)

t₃ = 9.2 s

The total time is:

t = t₁ + t₂ + t₃

t = 3.0 s + 5.0 s + 9.2 s

t = 17.2 s

Round as needed.

4 0
3 years ago
Which one should be my pfp?
Zepler [3.9K]

Explanation:

I think the third one coz it's so good

7 0
2 years ago
Read 2 more answers
Electromagnetic waves, which include light, consist of vibrations of electric and magnetic fields, and they all travel at the sp
Alekssandra [29.7K]

Answer:

2.92682 m

1.5\times 10^{18}\ Hz

250000000000 Hz

2.88462\times 10^{-8}\ m

0.21126 m

0.12244 m

Explanation:

c = Speed of light = 3\times 10^8\ m/s

\lambda = Wavelength

f = Frequency

Wavelength is given by

\lambda=\dfrac{c}{f}\\\Rightarrow \lambda=\dfrac{3\times 10^8}{102.5\times 10^6}\\\Rightarrow \lambda=2.92682\ m

The wavelength is 2.92682 m

Frequency is given by

f=\dfrac{c}{\lambda}\\\Rightarrow f=\dfrac{3\times 10^8}{0.2\times 10^{-9}}\\\Rightarrow f=1.5\times 10^{18}\ Hz

The frequency is 1.5\times 10^{18}\ Hz

f=\dfrac{c}{\lambda}\\\Rightarrow f=\dfrac{3\times 10^8}{1.2\times 10^{-3}}\\\Rightarrow f=250000000000\ Hz

The frequency is 250000000000 Hz

\lambda=\dfrac{c}{f}\\\Rightarrow \lambda=\dfrac{3\times 10^8}{1.04\times 10^{16}}\\\Rightarrow \lambda=2.88462\times 10^{-8}\ m

The wavelength is 2.88462\times 10^{-8}\ m

\lambda=\dfrac{c}{f}\\\Rightarrow \lambda=\dfrac{3\times 10^8}{1.42\times 10^{9}}\\\Rightarrow \lambda=0.21126\ m

The wavelength is 0.21126 m

\lambda=\dfrac{c}{f}\\\Rightarrow \lambda=\dfrac{3\times 10^8}{2.45\times 10^9}\\\Rightarrow \lambda=0.12244\ m

The wavelength is 0.12244 m

8 0
3 years ago
Situation 6.1 A 13.5-kg box slides over a rough patch 1.75 m long on a horizontal floor. Just before entering the rough patch, t
GenaCL600 [577]
B) 14.0 N    

The way to solve this problem is to determine the kinetic energy the box had before and after the rough patch of floor. The equation for kinetic energy is: 

 E = 0.5 M V^2 

 where 

 E = Energy 

 M = Mass 

 V = velocity   

 Substituting the known values, let's calculate the before and after energy. 

 Before: 

 E = 0.5 M V^2 

 E = 0.5 13.5kg (2.25 m/s)^2 

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 E = 34.17188 kg*m^2/s^2 = 34.17188 joules   

 After: 

 E = 0.5 M V^2 

 E = 0.5 13.5kg (1.2 m/s)^2 

 E = 6.75 kg 1.44 m^2/s^2 

 E = 9.72 kg*m^2/s^2 = 9.72 Joules   

 So the box lost 34.17188 J - 9.72 J = 24.451875 J of energy over a distance of 1.75 meters. Let's calculate the loss per meter by dividing the loss by the distance.   

 24.451875 J / 1.75 m = 13.9725 J/m = 13.9725 N   

 Rounding to 1 decimal place gives 14.0 N which matches option "B".
8 0
2 years ago
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