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Alenkasestr [34]
3 years ago
6

The amount of mass an object has per its volume is known as ___________. Select one of the options below as your answer:

Physics
2 answers:
Angelina_Jolie [31]3 years ago
7 0
C. Density.
<span>the quantity of mass per unit volume of a substance.</span>
irga5000 [103]3 years ago
5 0
C. Density, Mass/Volume=Density
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A car travels 100 m while decelerating to 8 m/s in 5 s.<br> a) What was its initial speed?
viktelen [127]

Answer:

Vi = 32 [m/s]

Explanation:

In order to solve this problem we must use the following the two following kinematics equations.

v_{f} =v_{i} - (a*t)\\

The negative sign of the second term of the equation means that the velocity decreases, as indicated in the problem.

where:

Vf = final velocity = 8[m/s]

Vi = initial velocity [m/s]

a = acceleration = [m/s^2]

t = time = 5 [s]

Now replacing:

8 = Vi - 5*a

Vi = (8 + 5*a)

As we can see we have two unknowns the initial velocity and the acceleration, so we must use a second kinematics equation.

v_{f}^{2} = v_{i}^{2} - (2*a*d)

where:

d = distance = 100[m]

(8^2) = (8 + 5*a)^2 - (2*a*100)

64 = (64 + 80*a + 25*a^2) - 200*a

0 = 80*a - 200*a + 25*a^2

0 = - 120*a + 25*a^2

0 = 25*a(a - 4.8)

therefore:

a = 0 or a = 4.8 [m/s^2]

We choose the value of 4.8 as the acceleration value, since the zero value would not apply.

Returning to the first equation:

8 = Vi - (4.8*5)

Vi = 32 [m/s]

6 0
4 years ago
Occupants in a single space shuttle in orbit feel weightless. describe a scheme whereby occupants in a pair of shuttles (or even
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3 years ago
If a car has a velocity of 85 km/hr, how long will it take to accelerate to 45 km/hr if the acceleration is -3 km/hr/sec?
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Answer:

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6 0
3 years ago
when a circular plate of metal is heated in an oven, its radius increases at .03 cm/min, at what rate is the area increasing whe
Alinara [238K]

Answer:

Rate of change of area will be 9.796cm^2/min

Explanation:

We have given rate of change of radius \frac{dr}{dt}=0.03cm/min

Radius of the circular plate r = 52 cm

Area is given by A=\pi r^2

So \frac{dA}{dt}=2\pi r\frac{dr}{dt}

Puting the value of r and \frac{dr}{dt}

\frac{dA}{dt}=2\times 3.14\times 52\times 0.03=9.796cm^2/min

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Which clue indicates that a speciation has occurred between two populations?
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I believe the answer is C
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