Which of the following items is an example of a symbol Bernice bobs her hair

A granite monument has a volume of 25,365.4 cm3. The density of granite is 2.7 g/cm3. Use this information to calculate the mass

Nana76 [90]

V = 25,364.4 cm^3 Is volumer = 2.7g/cm^3 Is density
To calculate mass you use formula:m= V*rTo avoid remembering this formula you can see the type of unit on each given variable. We can see that we have g/cm^3 and cm^3. If we multiply them, we negate cm^3 and cm^3 and we are left with g which is unit for mass.
the answer is :
m = 68,486,6 g

6 0

2 years ago

A stone is thrown vertically into the air at an initial velocity of 96 ft/s. On Mars, the height s (in feet) of the stone above

vladimir1956 [14]

Answer:

240 ft

Explanation:

t = Time taken

u = Initial velocity = 96 ft/s

v = Final velocity

s = Displacement

a = Acceleration = 12 m/s² on Mars 32 ft/s² on Earth negative due to upward direction

Mars

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=96\times t+\frac{1}{2}\times -12\times t^2\\\Rightarrow s=96t-6t^2\ ft$

Earth

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=96\times t+\frac{1}{2}\times -32\times t^2\\\Rightarrow s=96t-16t^2\ ft$

Differentiating the first equation with respect to time we get

$\frac{ds}{dt}=96-12t$

Equating with zero

$0=96-12t\\\Rightarrow t=\frac{96}{12}=8\ s$

Differentiating the second equation with respect to time we get

$\frac{ds}{dt}=96-32t$

Equating with zero

$0=96-32t\\\Rightarrow t=\frac{96}{32}=3\ s$

Applying the time taken to the above equations, we get

$s=96t-6t^2\ ft\\\Rightarrow s=96\times 8-6\times 8^2\\\Rightarrow s=384$

$s=96t-16t^2\\\Rightarrow s=96\times 3-16\times 3^2\\\Rightarrow s=144$

Difference in height = 384-144 = 240 ft

The stone will travel 240 ft higher on Mars

6 0

2 years ago

A 10n force is applied to a 25kg mass to slide it across a frictional surface. What is the acceleration of the mass?

klasskru [66]

Answer: a = 0.4m/s^2 - 9.8*c where c is the coefficient of kinetic friction of the surface

Explanation: We know that, by the second Newton's law, a = F/m

where a is the acceleration, F is the net force and m is the mass of the object.

Then, if the surface is frictionless, the total force applied in the object is 10N, and the mass of the object is 25kg, so the acceleration is:

a =10N/25kg = 0.4m/s^2.

But if the surface is frictional, there will be a force of friction applied in the mass (this depends on the coefficient of friction and the weight of the mass), this means that the acceleration will be reduced.

If = -(9.8*25)*c

where c is a number that is bigger than 0 and smaller than 1, is called the coefficient of kinetic friction.

So the total force is now:

F = (10 - 9.8*25*c)

Then, the acceleration in a frictional surface is equal to:

a = (10 - 9.8*25*c)/25 = 0.4m/s^2 - 9.8*c

6 0

3 years ago

Read 2 more answers

A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0

otez555 [7]

The given question is incomplete. The complete question is as follows.

A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 m/s each second. The push force has a horizontal component of 20 N and a vertical component of 25 N downward. Calculate the coefficient of kinetic friction between the box and the floor.

Explanation:

The given data is as follows.

$F_{1}$ = 20 N, $F_{2}$ = 25 N, a = -0.9 $m/s^{2}$

W = 83 N

m = $\frac{83}{9.81}$

= 8.46

Now, we will balance the forces along the y-component as follows.

N = W + $F_{2}$

= 83 + 25 = 108 N

Now, balancing the forces along the x component as follows.

$F_{1} - F_{r}$ = ma

$20 - F_{r} = 8.46 \times (-0.9)$

$F_{r}$ = 7.614 N

Also, we know that relation between force and coefficient of friction is as follows.

$F_{r} = \mu \times N$

$\mu = \frac{F_{r}}{N}$

= $\frac{7.614}{108}$

= 0.0705

Thus, we can conclude that the coefficient of kinetic friction between the box and the floor is 0.0705.

7 0

3 years ago

Read 2 more answers

As an interstellar cloud of hydrogen gas shrinks in size, its rate of rotation

FinnZ [79.3K]

Answer:

INCREASES, BECAUSE ITS ANGULAR MOMENTUM IS CONSERVED.

Explanation: Interstellar cloud of Hydrogen is an accumulation of Hydrogen gas in the cloud.

As the Interstellar cloud of Hydrogen shrinks (reduces) in size,the rate of rotation of the shrinked Interstellar cloud Increases because its angular momentum is conserved. GASEOUS MOLECULES MAKE UP ABOUT 99% OF THE INTERSTELLAR CLOUD WITH HYDROGEN HAVING ABOUT 90% OF THE VOLUME OF GASES IN THE INTERSTELLAR CLOUD.

4 0

3 years ago