When you catch a baseball with a glove (instead of your hand), the glove helps by

A 8.0-cm-diameter horizontal pipe gradually narrows to 5.0 cm . When water flows through this pipe at a certain rate, the gauge

adelina 88 [10]

Answer:

A 8.0 cm diameter horizontal pipe gradually narrows to 5.0 cm. The the water flows through this pipe at certain rate, the gauge pressure in these two sections is 31.0 kPa and 24.0 kPa, respectively. What is the volume of rate of flow?

The flow rate is 3.1175×10⁻³ m³/s

Explanation:

To solve the question we rely on Bernoulli's principle as follows $P_{1} +\frac{1}{2}\rho v^{2} _{1} + \rho gz_{1} = P_{2} +\frac{1}{2}\rho v^{2} _{2} + \rho gz_{2}$

thus where the pipe is horizontal we have

z₁ = z₂ hence the above equattion becomes

$P_{1} +\frac{1}{2}\rho v^{2} _{1} = P_{2} +\frac{1}{2}\rho v^{2} _{2}$

since the flow rate is constant then

Q = v₁A₁ = v₂A₂

Where is the area of the two sections given by A₁ = π·D₁²÷4 and

A₂ = π·D₂²÷4

Thereffore A₁ = π·0.08²÷4 = 5.02×10⁻³ m²

and A₂ = π·0.05²÷4 = 1.96×10⁻³ m²

v₁ = v₂A₂/A₁ =0.391×v₂

The given pressures are P₁ = 31.0 kPa and P₂ = 24.0 pKa and

ρ = 1000 kg/m³

Plugging the values into the above equation we get

31.0 kPa +0.5× 1000 kg/m³× (0.391×v₂)² = 24.0 pKa +0.5×1000 kg/m³×v₂²

= 31000+76.3·v₂² =24000+500·v₂²

or 423.706·v₂² = 7000

v₂² = 7000/423.706 = 16.52 or v₂ = 4.065 m/s and v₁ 0.391×4.065 = 1.59 m/s

The flow rate = v₂A₂ = 1.59×1.96×10⁻³ = 3.1175×10⁻³ m³/s

5 0

4 years ago

Is it right? correct me if im wrong

jeka57 [31]

Answer:

i think it is right

Explanation:

8 0

3 years ago

7. A 1300 kg car starts at rest and rolls down a hill from a height of 10.0 m. It then moves across a level (frictionless) surfa

makkiz [27]

Answer:

0.505 m

Explanation:

From the question,

The kinetic energy of the car = energy stored in the spring

1/2mv² = 1/2ke²...................... Equation 1

Where m = mass of the car, v = velocity of the car, k = spring constant of the car, e = extension/compression

make e the subject of the equation

e = v√(m/k)............... Equation 2

We can calculate the value of v, by applying,

v² = u²+2gH...................... Equation 3

Where u = initial velocity of the car, H = height of the car, g = acceleration due to gravity.

Given: u = 0 m/s (from rest), H, 10 m, g = 9.8 m/s²

Substitute into equation 2

v² = 2(10×9.8)

v² = 196

v = √196

v = 14 m/s

Also given: m = 1300 kg, e = 1.0×10⁶ N/m =1000000 N/m

Substitute into equation 2

e = 14√(1300/1000000)

e = 14√(0.00013)

e = 14(0.036)

e = 0.505 m

Hence the maximum distance of the spring is compressed = 0.505 m

5 0

3 years ago

If the dielectric constant is 14.1, calculate the ratio of the charge on the capacitor with the dielectric after it is inserted

Lady bird [3.3K]

Answer:

$\frac{Q}{Q_0}=1$

Explanation:

Capacitance is defined as the charge divided in voltage.

$C=\frac{Q}{V}(1)$

Introducing a dielectric into a parallel plate capacitor decreases its electric field. Therefore, the voltage decreases, as follows:

$V=\frac{V_0}{k}$

Where k is the dielectric constant and $V_0$ the voltage of the capacitor without a dielectric

The capacitance with a dielectric between the capacitor plates is given by:

$C=kC_0$

Where k is the dielectric constant and $C_0$ the capacitance of the capacitor without a dielectric. So, we have:

$Q=CV\\Q=kC_0\frac{V_0}{k}\\Q=C_0V_0\\Q_0=C_0V_0\\Q=Q_0\\\frac{Q}{Q_0}=1$

Therefore, a capacitor with a dielectric stores the same charge as one without a dielectric.

7 0

3 years ago

If the moon is half full on November 2, on approximately what date will it next become full?

anyanavicka [17]

The complete cycle of lunar phases has a period of 29.531 days.

You've described two phases that are separated by 1/4 of the
complete cycle.

If the moon is exactly at First Quarter (half-full and getting bigger)
at, let's say, 12:00 Noon on November 2 where you are, then it will
be Full after

29.531 days / 4 = 7 days 9 hours 11 minutes .

That's 9:11 PM on November 9 .

7 0

3 years ago