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inysia [295]
2 years ago
5

Object A represents fixed negatively charged particle and Object B represents fixed. positively-charged particle. Object ( shows

a
moving positively charged particle. Identify which arrow would correctly show the force of attraction or repulsion on object C caused

by the other two objects?
Physics
1 answer:
abruzzese [7]2 years ago
8 0

Answer:

Moving electrons by cringing the object near another charged object ... This means that there is a different number of protons and electrons ... the attraction or repulsive electrostatic force between positive and negative charges ... the amount of charge on the particle and the distance between the particles ... fixed in position.ation:

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Four springs with the following spring constants, 113.0 N/m, 65.0 N/m, 102.0 N/m, and 101.0 N/m are connected in series. What is
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Answer:

K_e_q=22.75878093\frac{N}{m}

f=1.363684118Hz

Explanation:

In order to calculate the equivalent spring constant we need to use the next formula:

\frac{1}{K_e_q} =\frac{1}{K_1} +\frac{1}{K_2} +\frac{1}{K_3} +\frac{1}{K_4}

Replacing the data provided:

\frac{1}{K_e_q} =\frac{1}{113} +\frac{1}{65} +\frac{1}{102} +\frac{1}{101}

K_e_q=22.75878093\frac{N}{m}

Finally, to calculate the frequency of oscillation we use this:

f=\frac{1}{2(pi)} \sqrt{\frac{k}{m} }

Replacing m and k:

f=\frac{1}{2(pi)} \sqrt{\frac{22.75878093}{0.31} } =1.363684118Hz

4 0
3 years ago
In recent years, assistive technologies have been developed to give people with disabilities equal access to resources. Apex
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Answer: as the public need for equal access for people with disabilities became understood, laws were enacted to mandate assistive technologies inclusion in almost all public spaces

Explanation: I just took the test but my teachers a füçkįñg čūńt so don't know if the this will help

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3. Study 1 differed from Study 2 in which of the follow-
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I believe the correct answer is C
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2 years ago
A multimeter in an RL circuit records an rms current of 0.600 A and a 50.0-Hz rms generator voltage of 110 V. A wattmeter shows
jeka57 [31]

Answer:

(a) The impedance in the circuit is Z=183.33\Omega.

(b)The resistance is R=38.89\Omega.

(c) The inuctance is 0.57 H.

Explanation:

(a)

The expression for the impedance is as follows:

Z=\frac{V_rms}{I_rms}

Here, V_rms is the rms voltage and I_rms is the rms current.

PutV_rms=110 V and I_rms=0.600 A.

Z=\frac{110}{0.600}

Z=183.33\Omega

Therefore, the impedance in the circuit is Z=183.33\Omega.

(b)

The expression for the average power is as follows;

P_{a}=I_{rms}^{2}R

Here, P_{a} is the average power and R is the resistance.

Calculate the resistance by rearranging the above expression.

R=\frac{P_{a}}{I_{rms}^{2}}

Put P_{a}=14W and

R=\frac{14}{{0.600}^{2}}

R=38.89\Omega

Therefore, the resistance is R=38.89\Omega.

(c)

The expression for the impedance is as follows;

Z^{2}=R^{2}+X_{L}^{2}

Here,X_{L} is the inductive reactance.

Put Z=183.33\Omega and R=38.89\Omega.

(183.33)^{2}=(38.89)^{2}+X_{L}^{2}

X_{L}=179.16\Omega

The expression for the inductive reactance in terms of  frequency is as follows;

X_{L}=2\pi fL

Here, L is the inductance.

Calculate the inductance by rearranging the above expression.

L=\frac{X_{L}}{2\pi f}

Put X_{L}=179.16\Omega and f=50Hz.

L=\frac{179.16}{2\pi (50)}

L=0.57 H

Therefore, the inuctance is 0.57 H.

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garik1379 [7]

Answer:

A spiral galaxy

Explanation:

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