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morpeh [17]
3 years ago
13

what logging method greatest impact on forest ecosystems

Physics
1 answer:
lora16 [44]3 years ago
3 0
Are there options to choose e
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What is the maximum speed with which a 1200-kg car can round a turn of radius 88.0 m on a flat road if the coefficient of static
boyakko [2]

Answer:

18.6 m/s

Explanation:

The frictional force acting on the car is given by:

F_f = \mu m g=(0.40 )(1200 kg)(9.81 m/s^2)=4709 N

But the frictional force also corresponds to the centripetal force that keeps the car in circular motion in the turn:

F_f = F_c = m \frac{v^2}{r}

Where v is the maximum speed the car can achieve remaining in the turn. Substituting r=88.0 m and re-arranging the formula, we can find the value of v:

v=\sqrt{\frac{Fr}{m}}=\sqrt{\frac{(4709 N)(88.0 m)}{1200 kg}}=18.6 m/s

6 0
3 years ago
I’m still lost on this please help me on this I know it’s not D I got it wrong
RSB [31]

Before you even look at the questions, look over the graph, so you know what kind of information is there.

The x-axis is "time".  OK.  You know that as the graph moves from left to right, it shows what's happening as time goes on.

The y-axis is "speed" of something.  OK.  When the graph is high, the thing is moving fast.  When the graph is low, the thing is moving slow.  When the graph slopes up, the thing is gaining speed.  When the graph slopes down, the thing is slowing down.  When the graph is flat, the speed isn't changing, so the thing is moving at a constant speed.

NOW you can look at the questions.

OMG !  It's only ONE question:  What's happening from 'c' to 'd' ?  Well I don't know.  Perhaps we can figure it out if we LOOK AT THE GRAPH !

-- Between c and d, the graph is flat.  The speed is not changing.  It's the same speed at d as it was back at c .

What speed is it ?

-- Look back at the y-axis.  The speed at the height of c and d is 'zero' .

-- The 2nd and 4th choices are both correct.  From c to d, <em>the speed is constant</em>.  The constant speed is zero.  <em>The car is not moving</em>.

5 0
3 years ago
By experiment, determine what makes a force attractive or repulsive. Describe your experiments and observations with some exampl
drek231 [11]

The charge present determines a force to be attractive or repulsive.

The charges acquired by two bodies determines the Force as Attractive Or Repulsive.

Electric Force applied due to Electrical charges is same in magnitude but opposite in direction. This corresponds this phenomenon equivalent to the Newton's Third Law.

Examples of the experiments and observations:

  • On combing hair through a comb and then keeping it close to small pieces of paper shows attraction of paper pieces towards the comb.

This occurs due to the Electric charges present in the comb that induces charge in paper pieces leading to their attraction.

  • In both Gravitational Force and Coulomb force, the force remains inversely proportional to the square of the distance following the Inverse Square Law being the Central Force system. This only differs by the fact that in Gravitational Force, masses are used and in Coulomb force, charges are used.

The more the distance between the charges, the less is the Electric Force.

The lesser the distance between the charges, the more is the Electric Force.

If both the objects are charged the same i.e. either positive or negative then the Force is Repulsive and if the charges are Oppositely charged then the force is attractive.

Hence, the charge present determines a force to be attractive or repulsive.

Learn more about Coulomb Force here, brainly.com/question/15451944

#SPJ4

6 0
2 years ago
Consider two displacements, one of magnitude 3 m and another of magnitude 4 m. Show how the displacement vectors may be combined
Rufina [12.5K]

Answer:

Explanation:

A = 3m

B = 4 m

let the angle between the two vectors is θ.

the resultant of two vectors is given by

R=\sqrt{A^{2}+B^{2}+2ABCos\theta }

(a) R = 7 m

So, 7=\sqrt{3^{2}+4^{2}+2\times 3\times 4 Cos\theta }

49 = 9 + 16 + 24 Cosθ

Cosθ = 1

θ = 0°

Thus, the two vectors are inclined at 0°.

(b) R = 1 m

So, 1=\sqrt{3^{2}+4^{2}+2\times 3\times 4 Cos\theta }

1 = 9 + 16 + 24 Cosθ

Cosθ = - 1

θ = 180°

Thus, the two vectors are inclined at 180°.

(c) R = 5 m

So, 5=\sqrt{3^{2}+4^{2}+2\times 3\times 4 Cos\theta }

25 = 9 + 16 + 24 Cosθ

Cosθ = 0

θ = 90°

Thus, the two vectors are inclined at 90°.

3 0
3 years ago
PLEASE HELP!!! 50 points!!!!!! SEE ATTACHMENT!!! In which direction is there a net force of 200 N? left, right, up, down
Sladkaya [172]

The answer is right. The point and the 200 N sign maybe on the left side but the direction can be going right. I just took the quiz today, I wouldn't give the wrong answer cause i'm not a liar.  -_-

5 0
3 years ago
Read 2 more answers
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