If two opposing forces are equal, then the net force is 0 N.<br><br> true or false?

PLS HELP!!!- The moon rotates around the Earth every 28 days. What is the frequency of the moon’s revolution?

Naily [24]

Answer: the frequency is every 27.322 days

6 0

3 years ago

A truck using a rope to tow a 2230-kg car accelerates from rest to 13.0 m/s in a time of 15.0s. How strong must the rope be? μk

Leokris [45]

Answer:

The rope must have a force of 10084,21 N

Explanation

Acceleration calculation

The car acceleration is equal to the acceleration of the truck

ac: car acceleration $\frac{m}{s^{2} }$

at: truck acceleration $\frac{m}{s^{2} }$ )

$ac = at= \frac{vf-vi}{t-ti}$ equation(1)

Known information:

vi = Initial speed = 0, ti = initial time = 0

vf = Final speed = 13 $\frac{m}{s}$ , t = final time =5 s

We replaced the known information in the equation(1):

$ac = at = \frac{13-0}{15-0}$

$ac=ac=\frac{13}{15} \frac{m}{s}$

Dynamic analysis

The forces acting on the car are the following:

Wc: Car weight

N: normal force, road force on the car

Ff: Friction force

T: Force of tension

Car weight calculation:

$Wc=mc*g$

mc = Car mass = 2230kg

g = Gravity acceleration=9.8 $\frac{m}{s^{2} }$

$Wc= 2230*9.8$

$Wc=21854 N$

Normal force calculation:

Newton's first law

$sum Fy= 0$

$N-W=0$

$N=W$

$N=21854 N$

Friction force calculation (Ff):

We have the formula to calculate the friction force:

Ff = μk * N Equation (3)

μk kinetic coefficient of friction

We know that μk = 0.373and N= 21854N ,then:

$Ff=0.373*21854$

$Ff=8151.54 N$

Calculation of the tension force in the rope (T):

Newton's Second law

$sum Fx= mc*ac$

$T-Ff=mc*ac$

$T=2230(\frac{13}{15}) + 8151.54$

T=10084,21 N

Answer: The rope must have a force of 10084,21 N

8 0

3 years ago

Choose the correct words from the box to complete the definition of resistance.

zzz [600]

Answer:

Opposition of passing a electric circuit

6 0

3 years ago

If a marble is released from a height of 10 meters how long would it take to hit the ground?

zlopas [31]

S = ut + 0.5at^2

<span>10 = 0 + 0.5(9.81)t^2 {and if g = 10 then t^2 = 2 so t ~1.414} </span>

<span>t^2 ~ 2.04 </span>

<span>t ~ 1.43 seconds</span>

3 0

3 years ago

Assume that the body's muscle mechanism can be approximated by a spring with a uniform continuous mass distribution that follows

tatiyna

Based on Hooke's law, the spring constant of the the body's muscle mechanism is the ratio of force to extension, the effective mass is m/3 and the potential energy that can be stored is ke^2 / 2.

<h3>What is the spring constant?</h3>

The spring constant or stiffness constant of an elastic spring is constant which describes the extent a bit forceapplied to an elastic spring will extend it.

Spring constant, K = force/extension

Assuming, a body's muscle mechanism is a spring obeying Hooke's law, the effective mass of the spring with mass m is 1/3 of the mass of the spring = m/3

The potential energy that can be stored = ke^2 / 2

where K is spring constant and e is the extension produced.

Therefore, the spring constant of the the body's muscle mechanism is the ratio of force to extension, the effective mass is m/3 and the potential energy that can be stored is ke^2 / 2.

Learn more about Hooke's law at: brainly.com/question/12253978

5 0

2 years ago

If two opposing forces are equal, then the net force is 0 N. true or false?