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2 years ago

11

Two bowling balls each have a mass of 6.8 kg. They are located next to each other with their centers 21.8 cm apart. What gravita

tional force do they exert on each other?

1 answer:

12345 [234]2 years ago

3 0

Answer:

6.49 x 10^-8 N

Explanation:

formula is

F= G * ((m1 * m2)/r^2)

F = 6.67x10^-11 * ((6.8*6.8/.218)

F = 6.49 x 10^-8 Newtons

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A block is at rest on the incline shown in the figure. The coefficients of static and ki- netic friction are μs = 0.62 and μk =

GREYUIT [131]

The frictional force is 218.6 N

Explanation:

The block in the problem is at rest along the inclined surface: this means that the net force acting along the direction parallel to the incline must be zero.

There are two forces acting along this direction:

- The component of the weight parallel to the incline, downward along the plane, of magnitude

$mg sin \theta$

where

m = 46 kg is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

$\theta=29^{\circ}$ is the angle of the incline

- The (static) frictional force, acting upward, of magnitude $F_f$

Since the block is in equilibrium, we can write

$mg sin \theta - F_f = 0$

And substituting, we find the force of friction:

$F_f = mg sin \theta = (46)(9.8)(sin 29^{\circ})=218.6 N$

Learn more about frictional force along an inclined plane:

brainly.com/question/5884009

#LearnwithBrainly

7 0

2 years ago

The weighted average of the masses of the isotopes of an element

solmaris [256]

Answer:

Atomic mass

Explanation:

The weighted average of the masses of the various isotopes of an element makes the atomic mass. This is why most atomic mass values usually have decimals.

In order to calculate the atomic mass of an element using the weighted average masses, we use the expression below:

Atomic mass = ∑ $m_{i}$ $α_{i}$

Where $m_{i}$ is the mass of isotope i

$α_{i}$ is the abundance of isotope i

The abundance or geonormal abundance is the proportion by which each of the isotope occurs in nature.

4 0

2 years ago

A 70 kg man's arm, including the hand, can be modeled as BIO 75-cm-long uniform rod with a mass of 3.5 kg. When the man raises b

Butoxors [25]

Answer:

Δy = 7.1 cm

Explanation:

The center of mass of a body is defined

$y_{cm}$ = 1 /M ∑ $m_{i} y_{i}$ i

Where M is the total mass of the body, m mass of each part and ‘y’ height

Let's apply this equation to our case

We locate the reference system on the shoulders

The height of the arms is at its midpoint

y = -75/2 = 37.5 cm

With arms down

$y_{cm}$ = 1/70 (63 y₀ - 3.5 37.5 - 3.5 37.5)

$y_{cm}$ = 1/70 (63 y)₀ - 7 37.5)

With arms up

$y_{cm}$ ’= 1/70 (63 y₀ + 3.5 y + 3.5 y)

$y_{cm}$ ’= 1/70 (63y₀ + 7 35.5)

let's subtract the two equations

$y_{cm}$ ’ - $y_{cm}$ = 1/70 2 (7 35.5)

Δy = $y_{cm}$ ’ - $y_{cm}$ = 2 7 35.5 / 70

ΔY = 7.1 cm

7 0

2 years ago

A transformer is used to light a lamp rated 40w, 240v from a 400v A.C supply. Calculate:

zubka84 [21]

Answer:1.81

(a) Explanation:the turn ratio= input voltage÷output voltage.

400÷220=1.81.

Don't know how to solve b part...

5 0

2 years ago

A spectral line from a star is observed to have a wavelength of 656.5 nm. The rest wavelength of this line is 656.8 nm. (a) What

nignag [31]

Answer:

speed of star is 1.37 × $10^{5}$ m/s

it is approaching to earth because wavelength of star is decreasing than rest

if emit same wavelength it does not move anywhere

it will remain steady condition with respect earth

Explanation:

given data

wavelength = 656.5 nm

rest wavelength = 656.8 nm

to find out

the speed of the star , is it approaching

solution

we know here equation that is

wavelength shift / wavelength at rest = velocity of source / speed of light

so put all value and find

velocity of source = 3 × $10^{-9}$ × 3 × $10^{8}$ / 656.8 × $10^{-9}$

velocity of source star = 1.37 × $10^{5}$ m/s

and

it is approaching to earth because wavelength of star is decreasing than rest

and

if emit same wavelength it does not move anywhere

it will remain steady condition with respect earth

6 0

3 years ago