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Inessa [10]
2 years ago
6

A force is represented by a ____ the choice of a ____ is necessary​

Physics
1 answer:
Vlad [161]2 years ago
3 0

Answer:

a force is represented by a<u> vector </u>the choice of a <u>reference frame</u> is necessary​

Explanation:

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You are standing 1 meter from a squawking parrot. If you move to a distance three meters away, the sound
spayn [35]
Use the inverse square law, thus if you move a distance of 3m away, the sound intensity decrease by 1/3^2= 1/9
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3 years ago
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A uniform rod of length 50cm and mass 0.2kg is placed on a fulcrum at a distance of 40cm from the left end of the rod. At what d
oksano4ka [1.4K]

Answer:

x = 45 cm

Explanation:

Given that,

The length of a rod, L = 50 cm

Mass, m₁ = 0.2 kg

It is at 40cm from the left end of the rod.

We need to find the distance from the left end of the rod should a 0.6kg mass be hung to balance the rod.

The centre of mass of the rod is at 25 cm.

Taking moments of both masses such that,

15\times 0.2=x\times 0.6\\\\x=\drac{3}{0.6}\\\\x=5\ cm

The distance from the left end is 40+5 = 45 cm.

Hence, at a distance of 45 cm from the left end it will balance the rod.

5 0
2 years ago
Refer to Concept Simulation 4.4 for background relating to this problem. The drawing shows a large cube (mass = 28.9 kg) being a
Usimov [2.4K]

Answer:

smallest magnitud is P=33.3 N

Explanation:

We are analyze the situation as an external force is applied and there is a friction force. We have a problem with Newton's second law.

          F = ma

As the two blocks go together they must have the same acceleration, so we can calculate this for the entire system

        P = (m1 + m2) a

        a = P / (m1 + m2)

In this case there is no friction force because the small block does not touch the ground.

In order to calculate the friction force, we must analyze each system component separately.

The large block on the X axis has an applied P force and as it moves feels a force from the small block.  In the Y axis has the weight (W1) and the reaction to normal (N1)

For the small block on the X axis, the force it feels is the thrust of the large block, note that this is an action and reaction force between the two blocks, it is the same definition we have of the normal one, so we can call this force (N)

Y axis it has the weight (W2) down, the force of friction (fr) that opposes the movement, so it is directed upwards. we write these equations

       N = m2 a

       fr -W2 = 0    

       fr = W2

       

The definition of friction force is

       fr = μ N

       

Let's replace and calculate

       μ (m2 a) = m2 g

       μ (P / (m1 + m2)) = g

       P = g /μ  (m1 + m2)

Let's calculate the value of this force

       P = 9.8 / 0.710 (28.9 +4.4)

       P = 13.80 (33.3)

       P = 33.3 N

This is the minimum friction force that prevents the block from sliding down

6 0
2 years ago
A baseball rolls off of a .7 m high desk and strikes the floor .25 m always how fast was the ball rolling
labwork [276]

Answer:

the ball's velocity was approximately 0.66 m/s

Explanation:

Recall that we can study the motion of the baseball rolling off the table in vertical component and horizontal component separately.

Since the velocity at which the ball was rolling is entirely in the horizontal direction, it doesn't affect the vertical motion that can therefore be studied as a free fall, where only the constant acceleration of gravity is affecting the vertical movement.

Then, considering that the ball, as it falls covers a vertical distance of 0.7 meters to the ground, we can set the equation of motion for this, and estimate the time the ball was in the air:

0.7 = (1/2) g t^2

solve for t:

t^2 = 1.4 / g

t = 0.3779  sec

which we can round to about 0.38 seconds

No we use this time in the horizontal motion, which is only determined by the ball's initial velocity (vi) as it takes off:

horizontal distance covered = vi * t

0.25 = vi * (0.38)

solve for vi:

vi = 0.25/0.38  m/s

vi = 0.65798  m/s

Then the ball's velocity was approximately 0.66 m/s

4 0
2 years ago
During a plane showcase, a pilot makes circular "looping" with a speed equal to the sound speed (340 m/s). However, the pilot ca
Dmitriy789 [7]

Answer:

1472.98 m

Explanation:

Data provided:

Speed of circular looping, v = 340 m/s

Acceleration, a = 8g

here,

g is the acceleration due to the gravity = 9.81 m/s²

Now,

the centripetal acceleration is given as,

a=\frac{v^2}{r}

r is the radius of the loop

on substituting the respective values, we get

8\times9.81=\frac{340^2}{r}

or

r = 1472.98 m

5 0
3 years ago
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