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3 years ago

Multiple Response: Newton's second law (select 3)

2 answers:

8 0

The BIG Equation. Newton's second law of motion can be formally stated as follows: The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.

12345 [234]3 years ago

8 0

D i believe is one and c maybe

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A disk between vertebrae in the spine is subjected to a shearing force of 600 N. Find its shear deformation, taking it to have t

Art [367]

Answer:

3.34×10^-6m

Explanation:

The shear modulus can also be regarded as the rigidity. It is the ratio of shear stress and shear strain

can be expressed as

shear stress/(shear strain)

= (F/A)/(Lo/ . Δx)

Stress=Force/Area

The sheear stress can be expressed below as

F Lo /(A *Δx)

Where A=area of the disk= πd^2/4

F=shearing force force= 600N

Δx= distance

S= shear modulus= 1 x 109 N/m2

Lo= Lenght of the cylinder= 0.700 cm=7×10^-2m

If we make Δx subject of the formula we have

Δx= FLo/(SA)

If we substitute the Area A we have

Δx= FLo/[S(πd^2/4]

Δx=4FLo/(πd^2 *S)

If we input the values we have

(4×600×0.7×10^-2)/10^9 × 3.14 ×(4×10^-2)^2

= 3.35×10^-6m

Therefore, its shear deformation is 3.35×10^-6m

A=area of the disk= πd^2/4

= [3.142×(4×10^-2)^2]/4

7 0

4 years ago

Consider five atoms from the second period : lithium , beryllium , boron , carbon , and nitrogen , which of these elements has t

kap26 [50]

Answer:

Lithium has the lowest electronegativity.

Explanation:

Electronegativity measures the tendency of an atom to attract the bonding pair of electrons. As we move left to right in a period, the number of shell remains same but the number of electron increases(negative charge increases. Hence, electronegativity also increases.

As Lithium is the left most element in this period. It has the lowest electronegativity value which is equal to 0.98.

6 0

3 years ago

A thin uniform cylindrical turntable of radius 2.2 m and mass 35 kg rotates in a horizontal plane with an initial angular speed

fenix001 [56]

Explanation:

The given data is as follows.

M = 35 kg, radius (r) = 2.2 m,

m = 17 kg, = 11 rad/s

We assume that will be the final angular speed.

Now, according to the conservation of angular momentum.

$L_{1} = L_{2}$

or, $I_{1} \times \omega_{1} = I_{2} \times \omega_{2}$

Putting the given values into the above formula as follows.

$I_{1} \times \omega_{1} = I_{2} \times \omega_{2}$

or,

= $\frac{(0.5 \times 35 \times (2.2)^{2}) \times 11}{(0.5 \times 35 \times (2.2)^{2} + 17 \times (1.5)^{2})}$

= 7.58 rad/s

Thus, we can conclude that the angular speed of the clay and turntable is 7.58 rad/s.

4 0

4 years ago

Read 2 more answers

A point source at the origin emits sound of frequency 175 Hz uniformly in all directions. On the x-axis at x=100 m, the sound in

hammer [34]

Answer:

Multiple answers:

1. Power output P=17.59W

2.Intensity 160m I=17.6W/ $m^{2}$

3. dB = 77.3

4. f=178.5 Hz

Explanation:

First one comes from the expression

$I=\frac{P}{4\pi r^{2} }$

where I is the intensity, P is the power and r is the radio of the spherical wave, or in this case, the distance x. I solved for the Power by multiplying Intensity with the area (4 $\pi x^{2}$

Second one is done with:

$\frac{I_{2} }{I_{1} } =\frac{x^{2}_{1} }{x^{2} _{2}}$

Solving for Intensity 2, the result mentioned.

The third is simply computed with

$dB=10*log\frac{I}{10^{-12} }$

And finally the last one is done with doppler effect, taking into account the speed of the air as in 10ºC 337m/s.

$f=f_{initial} *(\frac{s+v_{receiver} }{s+v_{source} } )$

Where Finitial is the frequency emitted and s is the speed of the sound. The wind blowing in positive is, in principle, going away of the observer.

8 0

3 years ago

A child is riding a merry-go-round that has an instantaneous angular speed of 1.25 rad/s and an angular acceleration of 0.745 ra

skelet666 [1.2K]

Answer:

So the acceleration of the child will be $8.05m/sec^2$