Two forces have magnitudes in the ratio

In a tug of war, two teams exerts a force of 30 N each. What is the net force acting on the rope?

Arte-miy333 [17]

The force exerted by each team is 30 N. The forces are in the direction opposite to each other.

Force exerted by team 1, $F_{1}=30 N$

Force exerted by team 2, $F_{2}=-30 N$

Net force, $F=F_{1} +F_{2}$

$=30+(-30)=0$

Therefore, net force on the rope is 0 N.

4 0

3 years ago

According to the graph, during which time interval are the particles in the air slowing down?

kenny6666 [7]

Answer b like the last time should be it

4 0

2 years ago

Read 2 more answers

The tires of a car make 60 revolutions as the car reduces its speed uniformly from 95.0 km/h to 60.0 km/h. The tires have a diam

yawa3891 [41]

Answer:

-2.869 rad/s2

Explanation:

Data given:

speed, vi at 95.0 km/h = 95 X (1 hour /3600 seconds) X (1000m / 1km)

Note that, for every 1 hour, there will be 60sec X 60sec = 3600 seconds

And for every 1km, there will be 1000m.

So, speed of 95.0 km/h = 26.389 m/s

speed, vi = r ω (radius X angular velocity)

angular velocity, ωi = v/r

ωi = 26.389 m/s ÷ half of 0.88 m diameter

= 59.975 rad/s

decelerating to speed, vf at 60.0 km/h = 60 X X (1 hour /3600 seconds) X (1000m / 1km)

= 16.667m/s

The angular velocity for this speed = 16.667m/s ÷ half of 0.88 m diameter

= 37.879rad/s

How far the car goes is equivalent to the angular acceleration which equals to (ωf^2 - ωi^2) ÷ 2θ

= (37.879rad/s)^2 - (59.975 rad/s)^2 ÷ 2 (60 rev X 2π rad/rev)

= -2.869 rad/s2

4 0

3 years ago

QUESTION 10

Elena L [17]

The maximum value of θ of such the ropes (with a maximum tension of 5,479 N) will be able to support the beam without snapping is:

$\theta =37.01^{\circ}$

We can apply the first Newton's law in x and y-direction.

If we do a free body diagram of the system we will have:

x-direction

All the forces acting in this direction are:

$T_{1}sin(\theta)-T_{2}sin(\theta)=0$ (1)

Where:

T(1) is the tension due to the rope 1
T(2) is the tension due to the rope 2

Here we just conclude that T(1) = T(2)

y-direction

The forces in this direction are:

$T_{1}cos(\theta)+T_{2}cos(\theta)-W=0$ (2)

Here W is the weight of the steel beam.

We equal it to zero because we need to find the maximum angle at which the ropes will be able to support the beam without snapping.

Knowing that T(1) = T(2) and W = mg, we have:

$T_{1}cos(\theta)+T_{1}cos(\theta)-m_{steel}g=0$

$2T_{1}cos(\theta)-m_{steel}g=0$

$2T_{1}cos(\theta)=m_{steel}g$

T(1) must be equal to 5479 N, so we have:

$cos(\theta)=\frac{m_{steel}g}{2T_{1}}$

$cos(\theta)=\frac{892*9.81}{2*5479}$

$cos(\theta)=0.80$

Therefore, the maximum angle allowed is θ = 37.01°.

You can learn more about tension here:

brainly.com/question/12797227

I hope it helps you!

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The primary organ of the digestive system that produces digestive enzymes is the:Small intestine Liver Gallbladder Pancreas

Juliette [100K]

Your pancreas is the organ that produces digestive enzymes

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