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Anton [14]
4 years ago
8

In a certain region of space, a uniform electric field has a magnitude of 4.30 x 104 n/c and points in the positive x direction.

find the magnitude and direction of the force this field exerts on a charge of +6.80 μc.
Physics
1 answer:
denis23 [38]4 years ago
5 0
The magnetic force exerted by a field E to a charge q is given by F=Eq. In this case, F=4.30*10^4*(6.80mu C). 1mu C=10^-6C, so F=4.30*6.80=10^-2=0.29N. The direction is in the x direction, the direction that the field is applied because the charge is positive.
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MatroZZZ [7]
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3-

An electroscope is made up of a metal detector knob on top which is connected to a pair of metal leaves hanging from the bottom of the connecting rod. When no charge is present the metals leaves hang loosely downward. But, when an object with a charge is brought near an electroscope, one of the two things can happen.
7 0
3 years ago
Sliding friction is affected by the weight of the object.<br><br> True or false
Dmitry [639]
Yes that is correct or in other form, True

7 0
3 years ago
Read 2 more answers
A sinusoidal wave of angular frequency 1,203 rad/s and amplitude 3.1 mm is sent along a cord with linear density 3.9 g/m and ten
kobusy [5.1K]

Answer:

18.7842493212 W

Explanation:

T = Tension = 1871 N

\mu = Linear density = 3.9 g/m

y = Amplitude = 3.1 mm

\omega = Angular frequency = 1203 rad/s

Average rate of energy transfer is given by

P=\dfrac{1}{2}\sqrt{T\mu}\omega^2y^2\\\Rightarrow P=\dfrac{1}{2}\sqrt{1871\times 3.9\times 10^{-3}}\times 1203^2\times (3.1\times 10^{-3})^2\\\Rightarrow P=18.7842493212\ W

The average rate at which energy is transported by the wave to the opposite end of the cord is 18.7842493212 W

7 0
3 years ago
Two blocks of masses M 1 and M 2 are connected by a massless string that passes over a massless pulley as shown in the figure. M
stealth61 [152]

The mass M1 is 7.8 kg

Explanation:

Block M1 is hanging on the string while block M2 is on the frictionless ramp.

We have to write the equations of motion for the two blocks.

- For M1, the only two forces acting on it are the force of gravity M_1 g (downward) and the tension in the string T (upward). So we can write

M_1 g - T = M_1 a

where

M_1 is the mass of the block

g=9.8 m/s^2 is the acceleration of gravity

a is the acceleration of the system

- For M2, the only two forces acting on it are the tension in the string T (acting up along the ramp) and the component of the gravity acting down along the ramp, M_2 g sin \theta. So the equation of motion is

T-M_2 g sin \theta = M_2 a

where

M_2 = 13.5 kg is the mass of the 2nd block

\theta=35.5^{\circ} is the angle of the ramp

In order for the two blocks to be in equilibrium, the acceleration must be zero:

a=0

So the two equations become:

M_1 g - T=0\\T-M_2 g sin \theta = 0

Isolating T from the 1st equation,

T=M_1 g

And substituting into the 2nd equation, we can find the value of the mass M_1:

M_1 g - M_2 g sin \theta = 0\\M_1 = M_2 sin \theta = (13.5)(sin 35.5^{\circ})=7.8 kg

Learn more about acceleration and forces:

brainly.com/question/11411375

brainly.com/question/1971321

brainly.com/question/2286502

brainly.com/question/2562700

#LearnwithBrainly

7 0
3 years ago
A linear network has a current input 7.5 cos(10t + 30°) A and a voltage output 120 cos(10t + 75°) V. Determine the associated im
Leona [35]

Answer:

16∠45° Ω

Explanation:

Applying,

Z = V/I................... Equation 1

Where Z = Impedance, V = Voltage output, I = current input.

Given: V = 120cos(10t+75°), = 120∠75°,  I = 7.5cos(10t+30) = 7.5∠30°

Substitute these values into equation 1

Z = 120cos(10t+75°)/7.5cos(10t+30)

Z = 120∠75°/ 7.5∠30°

Z = 16∠(75°-30)

Z = 16∠45° Ω

Hence the impedance of the linear network is 16∠45° Ω

8 0
3 years ago
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