Help please am stuck

A test rocket is launched vertically from ground level (y=0), at time t=0.0s. The rocket engine provides constant upward acceler

Luden [163]

From the definition of average velocity,

$\bar v=\dfrac{\Delta y}{\Delta t}=\dfrac{49\,\mathrm m}t$ ,

and the fact that constant acceleration means

$\bar v=\dfrac{v_f+v_0}2=\dfrac{30\,\frac{\mathrm m}{\mathrm s}}2=15\,\dfrac{\mathrm m}{\mathrm s}$

we can solve for the time $t$ :

$15\,\dfrac{\mathrm m}{\mathrm s}=\dfrac{49\,\mathrm m}t\implies t=3.3\,\mathrm s$

7 0

2 years ago

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

Solid barium sulfate is placed into a beaker to form a saturated solution of barium sulfate. the solution has a barium concentra

borishaifa [10]

Ksp = [Ba⁺²][SO₄⁻²]

[Ba⁺²] = [SO₄⁻²] for barium sulfate
Thus,
Ksp = (1 x 10⁻⁵)²
Ksp = 1 x 10⁻¹⁰

5 0

2 years ago

Read 2 more answers

The slope of the following graph represents

guajiro [1.7K]

Answer:

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4 0

3 years ago

If a fuse melts, does it create an open circuit, a closed circuit, or a short circuit?

Andrew [12]

Answer:

short circuit

7 0

3 years ago

Read 2 more answers