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3 years ago

Two objects are interacting but stay stationary. Which best describes what is happening to he action and reaction forces

Physics

2 answers:

weeeeeb [17]3 years ago

4 0

Answer: B

Explanation: The forces are equal and opposite each other.

Elena L [17]3 years ago

3 0

Answer:

The forces are equal and opposite each other

Explanation:

If two objects are interacting then a force is being applied to them, but because the objects are staying stationary the objects must be exerting the same amount of force but in opposite directions.

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A 5.80kΩ5.80kΩ resistor in a circuit has two mesh currents flowing through it. The first current measures 3.10mA3.10mA. The seco

77julia77 [94]

Answer:

power = 23.2 mW

Explanation:

given data

resister r = 5.80 kΩ

current measures I1 = 3.10 m A

measure current I2 = 1.10mA

solution

we get here I total

I = current 1 - current 2

I = 3.10 - 1.10

I = 2 mA

so here power will be

power = I²×R .............1

power = 2² × 5.80

power = 23.2 mW

3 0

3 years ago

What is the spring constant of a rubber band that exerts a force of 35 N when it is

bezimeni [28]

K = F/x = 35/0.1 = 350 N/m

8 0

4 years ago

Read 2 more answers

When Marie sees the red traffic light, Marie brakes to a halt from a speed of 70 m/s in just 2 seconds. What is her deceleration

SashulF [63]

Answer:

-35 $m/s^2$

Explanation:

Deceleration is the complete opposite of acceleration and is calculated by adding a negative sign to the formula for finding acceleration.

Since acceleration is calculated by diving the change in velocity with time, deceleration will become:

Deceleration = -Δv/t, where Δv = change in velocity and t = time

In this case, Δv = 70 m/s and t = 2 s, hence;

Deceleration = -70/2 = -35 $m/s^2$

The deceleration is 35 $m/s^2$ .

7 0

3 years ago

A charge +q is located at the origin, while an identical charge is located on the x axis at x = +0.57 m. A third charge of +2q i

Jlenok [28]

Answer:

The third charge placed is 0.80 m.

Explanation:

Given that,

Distance = 0.57 m

First charge = q

Third charge = 2q

We need to calculate the electrostatic force on charge q₁ due to q₂

Using formula of electrostatic force

$F_{21}=\dfrac{kq_{1}q_{2}}{r_{1}^2}$

When placed another charge q₃ at certain distance from origin, then the net force on charge q₁ due to both charges is

$F_{net}=F_{21}+F_{31}$

The net electrostatic force on the charge at the origin doubles.

$2F_{21}=F_{21}+F_{31}$

$F_{31}=F_{21}$

$\dfrac{kq_{3}q_{1}}{r_{2}^2}=\dfrac{kq_{2}q_{1}}{r_{1}^2}$

$\dfrac{q_{3}}{r_{2}^2}=\dfrac{q_{2}}{r_{1}^2}$

$r_{2}^2=\dfrac{q_{3}}{q_{2}\timesr_{1}^2}$

$r_{2}=\sqrt{\dfrac{q_{3}}{q_{2}}}r_{1}$

Put the value into the formula

$r_{2}=\sqrt{\dfrac{2q}{q}}\times0.57$

$r_{2}=\sqrt{2}\times0.57$

$r_{2}=0.80\ m$

Hence, The third charge placed is 0.80 m from origin in x-axis.

4 0

3 years ago

A glider of mass 5.0 kg hits the end of a horizontal rail and bounces off with the same speed, in the opposite direction. The co

NeTakaya

To solve the exercise it is necessary to apply the concepts given in Newton's second law and the equations of motion description.

Let's start by defining acceleration based on speed and time, that is

$a = \frac{v}{t}$

On the other hand according to Newton's second law we have to

F=ma

where

m= Mass

a = Acceleration

Replacing the value of acceleration in this equation we have

$F=m(\frac{v}{t})$

Substituting with our values we have

$100N=(5Kg)\frac{v}{0.2}$

Re-arrange to find v

$v=\frac{100*0.2}{5}$

$v = 2m/s$

Therefore the speed of the glider is 2m/s

5 0

3 years ago