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3 years ago

10

According to Newton’s first law of motion, what must happen to move an object at rest? It must move by its own inertia. It must

gain potential energy. A force must act on it. Gravity must act on it.

2 answers:

Nesterboy [21]3 years ago

6 0

I think it is a force must act on it.

Jet001 [13]3 years ago

4 0

a force must act on it

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A 50 mm diameter steel shaft and a 100 mm long steel cylindrical bushing with an outer diameter of 70 mm have been incorrectly s

BARSIC [14]

Answer:

The axial force is $P = 15.93 k N$

Explanation:

From the question we are told that

The diameter of the shaft steel is $d = 50mm$

The length of the cylindrical bushing $L =100mm$

The outer diameter of the cylindrical bushing is $D = 70 \ mm$

The diametral interference is $\delta _d = 0.005 mm$

The coefficient of friction is $\mu = 0.2$

The Young modulus of steel is $207 *10^{3} MPa (N/mm^2)$

The diametral interference is mathematically represented as

$\delta_d = \frac{2 *d * P_B * D^2}{E (D^2 -d^2)}$

Where $P_B$ is the pressure (stress) on the two object held together

So making $P_B$ the subject

$P_B = \frac{\delta _d E (D^2 - d^2)}{2 * d* D^2}$

Substituting values

$P_B = \frac{(0.005) (207 *10^{3} ) * (70^2 - 50^2))}{2 * (50) (70) ^2 }$

$P_B = 5.069 MPa$

Now he axial force required is

$P = \mu * P_B * A$

Where A is the area which is mathematically evaluated as

$\pi d l$

So $P = \mu P_B \pi d l$

Substituting values

$P = 0.2 * 5.069 * 3.142 * 50 * 100$

$P = 15.93 k N$

8 0

3 years ago

a kid on a playground swing makes 6 complete to-and-from swings each 30 seconds (a) the frequency of the swinging is? (b) the pe

Mariana [72]

The frequency is how many per second:

(6 swings)/(30 sec) = (6/30) swing/sec = 0.2 per sec = 0.2 Hz .

The period is how long each one takes, or seconds per swing.
It's exactly the flip of frequency.
So we could just take the frequency, flip it, and find 1 / 0.2 ,
but let's do it the long way:

(30 sec) / (6 swings) = (30/6) sec/swing = 5 seconds .

4 0

3 years ago

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Newton's third law states that for every action force there is an equal and opposite reaction force. An idiot in your class says

Otrada [13]

A. Because the third laws say that for every action force the is an equal and opposite reaction force

4 0

3 years ago

Read 2 more answers

This group of elements is in group 2?

asambeis [7]

Answer:

If you meant the Periodic Table it's the alkaline-earth metal

beryllium (Be)

magnesium (Mg)

calcium (Ca)

strontium (Sr)

barium (Ba)

radium (Ra)

Explanation:

5 0

2 years ago

a body of radius R and mass m is rolling horizontally without slipping with speed v. it then rolls us a hill to a maximum height

ki77a [65]

Answer:

mR²/2

Explanation:

Here is the complete question

An object of radius′

R′ and mass ′

M′ is rolling horizontally without slipping with speed ′

V′

. It then rolls up the hill to a maximum height h = 3v²/4g. The moment of inertia of the object is (g= acceleration due to gravity)

Solution

Since it rolls without slipping, there is no friction. So, its initial mechanical energy at the horizontal surface equals its final mechanical energy at the top of the hill.

Since the object is rolling initially, and on horizontal ground, it initial energy is kinetic and made up of rotational and translational kinetic energy.

So, E = K + K'

E = 1/2mv² + 1/2Iω² where m = mass of object, v = speed of object, I = moment of inertia of object and ω = angular speed of object = v/r where v = speed of object and R = radius of object.

Also, the final mechanical energy of the object, E' is its potential energy at the top of the hill. So, E' = mgh.

Since E = E',

1/2mv² + 1/2Iω² = mgh

substituting the values of ω and h into the equation, we have

1/2mv² + 1/2Iω² = mgh

1/2mv² + 1/2I(v/R)²= mg(3v²/4g)

Expanding the brackets, we have

1/2mv² + 1/2Iv²/R²= 3mv²/4

Dividing through by v², we have

1/2m + I/2R²= 3m/4

Subtracting m/2 from both sides, we have

I/2R² = 3m/4 - m/2

Simplifying, we have

I/2R² = m/4

Multiplying through by 2R², we have

I = m/4 × 2R²

I = mR²/2

6 0

3 years ago