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2 years ago

What is the volume of an irregularly shaped object that has a mass 3.0 grams and a density of 6.0 g/mL

Physics

1 answer:

pogonyaev2 years ago

4 0

Answer: The volume of an irregularly shaped object is 0.50 ml

Explanation:

To calculate the volume, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of object = $6.0g/ml$

mass of object = 3.0 g

Volume of object = ?

Putting in the values we get:

$6.0g/ml=\frac{3.0g}{\text{Volume of substance}}$

${\text{Volume of substance}}=0.50ml$

Thus the volume of an irregularly shaped object is 0.50 ml

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While playing basketball in PE class, Logan lost his balance after making a lay-up and colliding with the padded wall behind the

olga2289 [7]

Answer:

a.) F = 3515 N

b.) F = 140600 N

Explanation: given that the

Mass M = 74kg

Initial velocity U = 7.6 m/s

Time t = 0.16 s

Force F = change in momentum ÷ time

F = (74×7.6)/0.16

F = 3515 N

b.) If Logan had hit the concrete wall moving at the same speed, his momentum would have been reduced to zero in 0.0080 seconds

Change in momentum = 74×7.6 + 74×7.6

Change in momentum = 562.4 + 562.4 = 1124.8 kgm/s

F = 1124.8/0.0080 = 140600 N

6 0

2 years ago

Calculate the distance d from the center of the sun at which a particle experiences equal attractions from the earth and the sun

fredd [130]

Answer:

149.34 Giga meter is the distance d from the center of the sun at which a particle experiences equal attractions from the earth and the sun.

Explanation:

Mass of earth = m = $5.976\times 10^{24} kg$

Mass of Sun = M = 333,000 m

Distance between Earth and Sun = r = 149.6 gm = 1.496\times 10^{11} m[/tex]

1 giga meter = $10^{9} meter$

Let the mass of the particle be m' which x distance from Sun.

Distance of the particle from Earth = (r-x)

Force between Sun and particle:

$F=G\frac{M\times m'}{x^2}=G\frac{333,000 m\times m'}{x^2}$

Force between Sun and particle:

$F'=G\frac{mm'}{(r-x)^2}$

Force on particle is equal:

F = F'

$G\frac{333,000 m\times m'}{x^2}=G\frac{mm'}{(r-x)^2}$

$\frac{x}{r-x}=\sqrt{333,000}$ = ±577.06

Case 1:

$\frac{x}{r-x}=577.06$

x = $1.49\times 10^{11} m=149.34 Gm$

Acceptable as the particle will lie in between the straight line joining Earth and Sun.

Case 2:

$\frac{x}{r-x}=-577.06$

x = $1.49\times 10^{11} m=149.86 Gm$

Not acceptable as the particle will lie beyond on line extending straight from the Earth and Sun.

3 0

2 years ago

When Kevin pulls his cotton shirt off his body, the electrons get transferred from the (shirt or body) to the (shirt or body) .

Masja [62]

<span>When Kevin pulls his cotton shirt off his body, the electrons get transferred from the shirt (in form of static charges i.e. electrons to the body. So, the shirt becomes positively charged and Kevin’s body becomes negatively charged.

As a result of charge transfer from the shirt to the body, we can hear a crackling sound. or if observed in dark, a sparkle can be seen.</span>

6 0

2 years ago

A man runs at a velocity of 4.5 m/s for 15.0 min. When going up an increasingly steep hill, he slows down at a constant rate of

madreJ [45]

The man ran <u>4252.5 meters.</u>

Why?

To solve the problem, we need to divide the exercise into two movements, the first on while the was running at 4.5 m/s for 15 min, and then, while he was slowing down (going up because of the hill).

First movement: Running at 4.5 m/s for 15 min.

We need convert from minutes to seconds,

$1min=60seconds\\\\15min*\frac{60seconds}{1min}=900seconds$

Now, calculating the distance covered for the first movement, we have:

$x_{1}=0+v_{1}*t_{1}\\\\x_{1}=4.5\frac{m}{s}*900s=4050m$

So, we know that the man covered 4050m for the first movement, it will be our initial position for the second movement.

Second movement: acceleration -0.05m/s^2 (because he's slowing down) for 90 seconds, at 4.5m/s.

$x_{2}=x_{1}+v_{1}*t+\frac{1}{2}at^{2}\\\\x_{2}=4050m+4.5m\frac{m}{s}*90seconds-\frac{1}{2}*(0.05\frac{m}{s^{2}})*(90s)^{2}\\\\x_{2}=4050m+405m-(0.5*0.05\frac{m}{s^{2}}*8100s^{2})=4050m+405m-202.5m\\\\x_{2}=4252.5m$

Hence, we have that he ran 4252.5 m.

Have a nice day!

4 0

2 years ago

(HELP!!! 30 pts if answered right. )What formula gives the strength of an electric field, E, at a distance from a known source c

umka2103 [35]

Answer:

$E=\frac{k\,Q}{d^2}$

Explanation:

The strength of an electric field E produced by a single charge Q at a distance d from it is given by the formula: $E=\frac{k\,Q}{d^2}$ , where K represents the Coulomb constant.

Since the electric field E is derived from the Coulomb Force per unit charge using a positive test charge, the field's units will be in units of Newtons/Coulomb, and be the formula for the Coulomb electric force between to charges (Q1 and Q2),

$F_C=k\frac{Q_1\,Q_2}{d^2}$

but modified with only one charge showing in the numerator of the expression.

8 0

2 years ago