What is the magnitude of the net force on the block (the moment it is released?

A cat dozes on a stationary merry-go-round, at a radius of 4.4 m from the center of the ride. The operator turns on the ride and

monitta

Answer:

The coefficient of static friction is 0.29

Explanation:

Given that,

Radius of the merry-go-round, r = 4.4 m

The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 7.7 s.

We need to find the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding. For this the centripetal force is balanced by the frictional force.

$\mu mg=\dfrac{mv^2}{r}$

v is the speed of cat, $v=\dfrac{2\pi r}{t}$

$\mu=\dfrac{4\pi^2r}{gt^2}\\\\\mu=\dfrac{4\pi^2\times 4.4}{9.8\times (7.7)^2}\\\\\mu=0.29$

So, the least coefficient of static friction between the cat and the merry-go-round is 0.29.

4 0

3 years ago

after singing for a few minutes the soloist rejoins the choir a second choir, consist of 90 additional singers then joins in eac

vredina [299]

If the soloist produces "x" decibels and the 10-person choir produces "y" decibels, combined they will produce "x+y" decibels.

The second choir has 90 additional singers, we base our description on the first choir. If a 10-person choir produces "x+y" decibels, then the 90 person choir produces 10 (x+y) decibels.

7 0

3 years ago

Enter the expression

vlabodo [156]

I see the word "when..." kind of fading out at the end of the first line.
Whatever comes after it may be important.

If you're just supposed to copy the expression into the box,
then the problem is that you left the 'e' out of it.

I'm guessing that you're supposed to enter whatever the expression becomes
when either N₀ or ' t ' has some special value that's in the first line.

Just taking a wild guess here . . . . .

If it's "Enter the expression ..... , when t=0 ." ,
then the correct answer in the box is N₀ .

But that's just a wild guess. As I pointed out, you cut off
the picture in the middle of the word 'when', and I've got
a hunch that there's something important after it.

6 0

3 years ago

Determine the magnitude of the effective value of g⃗ at a latitude of 60 ∘ on the earth. assume the earth is a rotating sphere.

dezoksy [38]

In addition to acceleration of gravity we experience centrifugal acceleration away from the axis of rotation of the earth. this additional acceleration has value ac = r w^2 where w = angular velocity and r is distance from your spot on earth to the earth's axis of rotation so r = R cos(l) where l = 60 deg is the lattitude and R the earth's radius and w = 1 / (24hr x 3600sec/hr)
<span>now you look up R and calculate ac then you combine the centrifugal acc. vector ac with the gravitational acceleration vector ag = G Me/R^2 to get effective ag' = ag -</span>

5 0

3 years ago

Car A has a mass of 1,200 kg and is traveling at a rate of 22 km/hr. It collides with car B. Car B has a mass of 1,900 kg and is

anastassius [24]

The car A has a mass of 1200 kg.

The car B has the mass of 1900 kg.

It is given that velocity of car A is given as 22 Km/hr

The car B has the velocity of 25 Km/hr.

Let the mass of two bodies are denoted as $m_{1} \ and\ m_{2}$

Let the velocity of cars A and B are denoted as $v_{1} \ and\ v_{2}$

The momentum before collision is-

$p_{i} =m_{1} v_{1} +m_{2} v_{2}$

[Here p stand for momentum.]

We are asked to calculate the final momentum of the system after collision.

The answer of the question is based law of conservation of linear momentum.

As per law of conservation of linear momentum the sum total linear momentum for an isolated system is always constant.Hence irrespective of the type of collision[elastic and inelastic],the momentum of the system is always constant which is a universal truth.

Let after the collision the velocity of A and B are $v'_{1} \ and\ v'_{2}$

Hence the final momentum of the system is-

$p_{f} = m_{1} v'_{1} +m_{2} v'_{2}$

As per the law of conservation of linear momentum, the initial and final momentum must be equal i.e

$p_{i} =p_{f}$

$m_{1} v_{1} +m_{2}v_{2} =m_{1} v'_{1} +m_{2} v'_{2}$

Hence the option A is right.

7 0

3 years ago

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