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3 years ago

6

A ride at an amusement park moves the riders in a circle at a rate of 6.0 m/s. If the radius of the ride is 9.0 meters, what is

the acceleration of the ride?
4.0 m/s2
0.67 m/s2
0.075 m/s2
54 m/s2

2 answers:

NikAS [45]3 years ago

8 0

<span>4.0 m/s2

it's 9 squared divided by 6</span>

yulyashka [42]3 years ago

4 0

The answer is B 4.0 m/s2

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A storage tank containing oil (SG=0.92) is 10.0 meters high and 16.0 meters in diameter. The tank is closed, but the amount of o

blagie [28]

Answer:

a-1 Graph is attached. The relation is linear.

a-2 The corresponding height for 68 kPa Pressure is 7.54 m

a-3 The corresponding weight for 68 kPa Pressure is 1394726kg

b The original height of the column is 5.98 m

Explanation:

Part a

a-1

The graph is attached with the solution. The relation is linear as indicated by the line.

a-2

By the equation

$P=\rho \times g \times h$

Here

P is the pressure which is given as 68 kPa.
ρ is the density of the oil whose SG is 0.92. It is calculated as

$\rho=S.G \times \rho_{water}\\\rho=0.92 \times 1000 kg/m^3\\\rho=920 kg/m^3\\$

g is the gravitational constant whose value is 9.8 m/s^2
h is the height which is to be calculated

$P=\rho \times g \times h\\h=\frac{P}{\rho \times g}\\h=\frac{68 \times 10^3}{920 \times 9.8}\\h=7.54m$

So the height of column is 7.54m

a-3

By the relation of volume and density

$M=\rho \times V$

Here

ρ is the density of the oil which is 920 kg/m^3
V is the volume of cylinder with diameter 16m calculated as follows

$V=\pi r^2h\\V=3.14\times (8)^2 \times 7.54\\V=1515.23 m^3$

Mass is given as

$M=\rho \times V\\M=920 \times 1515.23\\M=1394726kg$

So the mass of oil leading to 68kPa is 1394726kg

Part b

Pressure variation is given as

$\Delta P=P_{obs}-P_{atm}\\\Delta P=115-101 kPa\\\Delta P=14 kPa\\$

Now corrected pressure is as

$P_c=P_g-\Delta P\\P_c=68-14 kPa\\P_c=54 kPa$

Finding the value of height for this corrected pressure as

$P_c=\rho \times g \times h\\h=\frac{P_c}{\rho \times g}\\h=\frac{54 \times 10^3}{920 \times 9.8}\\h=5.98m$

The original height of column is 5.98m

4 0

3 years ago

What do electrons in the same shell have in common? AThey have the same amount of energy incorrect answer BThey are all positive

Keith_Richards [23]

Answer:

They have the same amount of energy

Explanation:

Electrons are said to be the subatomic particles that move around the nucleus of an atom. These electrons are negatively charged particles that are seen to be quite smaller than the nucleus of an atom.

The electron shells of these atoms are usually being filled from the inside out with the low-energy shells closer to the nucleus being filled before they can go into the much higher-energy shells that are a bit out

8 0

3 years ago

Help awnser need experts

pishuonlain [190]

12. The answer would be C. 1.50 s. This is because if you divide 60 by 40, you will get 1.5.

13. For this one I'm not sure, but what I can tell you is that the heavier something is the faster it will sink, the lighter it is, it will float.

5 0

3 years ago

0.16 mol of argon gas is admitted to an evacuated 70 cm^3 container at 30°C. The gas then undergoes an isothermal expansion to a

Semmy [17]

Answer:

The final pressure of the gas is 9.94 atm.

Explanation:

Given that,

Weight of argon = 0.16 mol

Initial volume = 70 cm³

Angle = 30°C

Final volume = 400 cm³

We need to calculate the initial pressure of gas

Using equation of ideal gas

$PV=nRT$

$P_{i}=\dfrac{nRT}{V}$

Where, P = pressure

R = gas constant

T = temperature

Put the value in the equation

$P_{i}=\dfrac{0.16\times8.314\times(30+273)}{70\times10^{-6}}$

$P_{i}=5.75\times10^{6}\ Pa$

$P_{i}=56.827\ atm$

We need to calculate the final temperature

Using relation pressure and volume

$P_{2}=\dfrac{P_{1}V_{1}}{V_{2}}$

$P_{2}=\dfrac{56.827\times70}{400}$

$P_{2}=9.94\ atm$

Hence, The final pressure of the gas is 9.94 atm.

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2 years ago

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svlad2 [7]

Answer:

The coin is less dense than the water thefore it can float.

Explanation:

5 0

3 years ago

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