An electric field of 710,000 N/C points due west at a certain spot. What is the magnitude of the force that acts on a charge of -6.00 C at this spot? (14C - 10 6C) Give your answer in Si unit rounded to two decimal places

Answers

Answer 1
Answer:

Answer:

4.26*10^6N

Explanation:

A charge within an electric field E experiences a force proportional to the field whose module is F = qE, whose direction is the same, if the charge is negative, it experiences a force in the opposite direction to the field and if the charge is positive, experience a force in the same direction of the field.

In our case we are interested in the magnitude of the force, therefore the sign of the charge has no relevance

\left | F \right |=\left |q  \right |  \left |E\right |\n\left | F \right |=6.00C*710000(N)/(C)=4.26*10^6N


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A model airplane with a mass of 0.741kg is tethered by a wire so that it flies in a circle 30.9 m in radius. The airplane engine provides anet thrust of 0.795 N perpendicular tothe tethering wire.(a) Find the torque the net thrust producesabout the center of the circle.
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(b) Find the angular acceleration of the airplane when it is inlevel flight.
rad/s2

(c) Find the linear acceleration of the airplane tangent to itsflight path.
m/s2

Answers

(a) 24.6 Nm

The torque produced by the net thrust about the center of the circle is given by:

\tau = F r

where

F is the magnitude of the thrust

r is the radius of the wire

Here we have

F = 0.795 N

r = 30.9 m

Therefore, the torque produced is

\tau = (0.795 N)(30.9 m)=24.6 N m

(b) 0.035 rad/s^2

The equivalent of Newton's second law for a rotational motion is

\tau = I \alpha

where

\tau is the torque

I is the moment of inertia

\alpha is the angular acceleration

If we consider the airplane as a point mass with mass m = 0.741 kg, then its moment of inertia is

I=mr^2 = (0.741 kg)(30.9 m)^2=707.5 kg m^2

And so we can solve the previous equation to find the angular acceleration:

\alpha = (\tau)/(I)=(24.6 Nm)/(707.5 kg m^2)=0.035 rad/s^2

(c) 1.08 m/s^2

The linear acceleration (tangential acceleration) in a rotational motion is given by

a=\alpha r

where in this problem we have

\alpha = 0.035 rad/s^2 is the angular acceleration

r = 30.9 m is the radius

Substituting the values, we find

a=(0.035 rad/s^2)(30.9 m)=1.08 m/s^2

WILL MARK BRAINLIEST PLS HELPPP -- Which of Newton’s Laws explains why the satellite would collide with the moon if gravity is “turned off?”picture attached

Answers

Answer:

b

Explanation:

B is the answer sorry for the late response

A uniformly dense solid disk with a mass of 4 kg and a radius of 4 m is free to rotate around an axis that passes through the center of the disk and perpendicular to the plane of the disk. The rotational kinetic energy of the disk is increasing at 21 J/s. If the disk starts from rest through what angular displacement (in rad) will it have rotated after 3.3 s?

Answers

Answer:

3.44 rad

Explanation:

The rotational kinetic energy change of the disk is given by ΔK = 1/2I(ω² - ω₀²) where I = rotational inertia of solid sphere = MR²/2 where m = mass of solid disk = 4 kg and R = radius of solid disk = 4 m, ω₀ = initial angular speed of disk = 0 rad/s (since it starts from rest) and ω = final angular speed of disk

Since the kinetic energy is increasing at a rate of 21 J/s, the increase in kinetic energy in 3.3 s is  ΔK = 21 J/s × 3.3 s = 69.3 J

So, ΔK = 1/2I(ω² - ω₀²)

Since ω₀ = 0 rad/s

ΔK = 1/2I(ω² - 0)

ΔK = 1/2Iω²

ΔK = 1/2(MR²/2)ω²

ΔK = MR²ω²/4

ω² = (4ΔK/MR²)

ω = √(4ΔK/MR²)

ω = 2√(ΔK/MR²)

Substituting the values of the variables into the equation, we have

ω = 2√(ΔK/MR²)

ω = 2√(69.3 J/( 4 kg × (4 m)²))

ω = 2√(69.3 J/[ 4 kg × 16 m²])

ω = 2√(69.3 J/64 kgm²)

ω = 2√(1.083 J/kgm²)

ω = 2 × 1.041 rad/s

ω = 2.082 rad/s

The angular displacement θ is gotten from

θ = ω₀t + 1/2αt² where ω₀ = initial angular speed = 0 rad/s (since it starts from rest), t = time of rotation = 3.3 s and α = angular acceleration = (ω - ω₀)/t = (2.082 rad/s - 0 rad/s)/3.3 s = 2.082 rad/s ÷ 3.3 s = 0.631 rad/s²

Substituting the values of the variables into the equation, we have

θ = ω₀t + 1/2αt²

θ = 0 rad/s × 3.3 s + 1/2 × 0.631 rad/s² (3.3 s)²

θ = 0 rad + 1/2 × 0.631 rad/s² × 10.89 s²

θ = 1/2 × 6.87159 rad

θ = 3.436 rad

θ ≅ 3.44 rad

Which one of the following statements concerning the Stefan-Boltzmann equation is correct? The equation can be used to calculate the power absorbed by any surface. The equation applies only to perfect radiators. The equation applies only to perfect absorbers. The equation is valid with any temperature units. The equation describes the transport of thermal energy by conduction.

Answers

"The equation can be used to calculate the power absorbed by any surface" statement concerning the Stefan-Boltzmann equation is correct.

Answer: Option A

Explanation:

According to Stefan Boltzmann equation, the power radiated by black body radiation source is directly proportionate to the fourth power of temperature of the source. So the radiation transferred is absorbed by another surface and that absorbed power will also be equal to the fourth power of the temperature. So the equation describes the relation of net radiation loss with the change in temperature from hotter temperature to cooler temperature surface.  

                            P=e \sigma A\left(T^(4)-T_(c)^(4)\right)

So this law is application for calculating power absorbed by any surface.

15.Restore the battery setting to 10 V. Now change the number of loops from 4 to 3. Explain what happens to the magnitude and direction of the magnetic field. Now change to 2 loops, then to 1 loop. What do you observe the relationship to be between the magnitude of the magnetic field and the number of loops for the same current

Answers

Answer:

we see it is a linear relationship.

Explanation:

The magnetic flux is u solenoid is

      B = μ₀ N/L   I

where N is the number of loops, L the length and I the current

By applying this expression to our case we have that the current is the same in all cases and we can assume the constant length. Consequently we see that the magnitude of the magnetic field decreases with the number of loops

      B = (μ₀ I / L)  N

the amount between paracentesis constant, in the case of 4 loop the field is worth

      B = cte 4

N       B

4       4 cte

3       3 cte

2       2 cte

1        1 cte

as we see it is a linear relationship.

In addition, this effect for such a small number of turns the direction of the field that is parallel to the normal of the lines will oscillate,

A rocket with a mass of 62,000 kg (including fuel) is burning fuel at the rate of 150 kg/s and the speed of the exhaust gases is 6,000 m/s. If the rocket is fired vertically upward from the surface of the Earth, determine its height after 744 kg of its total fuel load has been consumed. Since the mass of fuel consumed is small compared to the total mass of the rocket, you can consider the mass of the rocket to be constant for the time interval of interest.

Answers

Answer:

h≅ 58 m

Explanation:

GIVEN:

mass of rocket M= 62,000 kg

fuel consumption rate =  150 kg/s

velocity of exhaust gases v= 6000 m/s

Now thrust = rate of fuel consumption×velocity of exhaust gases

=6000 × 150 = 900000 N

now to need calculate time t = amount of fuel consumed÷ rate

= 744/150= 4.96 sec

applying newton's law

M×a= thrust - Mg

62000 a=900000- 62000×9.8

acceleration a= 4.71 m/s^2

its height after 744 kg of its total fuel load has been consumed

h= (1)/(2)at^2

h= (1)/(2)4.71*4.96^2

h= 58.012 m

h≅ 58 m