Ask question

Ask question

All categories

2 years ago

9

Laura adds 50mL of boiling water to 100mL of ice water. If the 150 mL of water is then put into a freezer, at what temperature w

ill the water freeze?

1 answer:

Dennis_Churaev [7]2 years ago

8 0

Water that is pure, will always freeze at 0 degrees.

You might be interested in

3N<br> 3 N<br> What is the net force of the box?

zepelin [54]

6N I think I’m pretty sure

3 0

3 years ago

Which of the following are true statements about the electric field? a. A moving charge produces a magnet anomaly. b. The electr

mars1129 [50]

Answer:

a,b,c,d,,f, g, j

Explanation:

e) equipotential lines are lines connecting points of equal potential

h) electric field inside the conductor is non-zero even when there is net movement of charge or non-zero current.

i) capacitors' plates are charged and an electric field exists between the plates.

4 0

2 years ago

If the force on the tympanic membrane (eardrum) increases by about 1.4 N above the force from the atmospheric pressure, the memb

Sveta_85 [38]

Answer:

10.58 ft

Explanation:

Force, F = 1.4 N

Diameter of membrane = 7.4 mm

radius of membrane, r = 7.4 / 2 = 3.7 mm = 3.7 x 10^-3 m

Area, A = 3.14 x (3.7 x 10^-3)^2 = 4.3 x 10^-5 m^2

Density, d = 1.03 x 10^3 kg/m^3

Pressure at depth, P = h x d x g

Let h be the depth.

Pressure = force / Area

h x 1.03 x 10^3 x 9.8 = 1.4 / (4.3 x 10^-5)

h = 3.225 m = 10.58 ft

Thus, the depth of water is 10.58 ft.

6 0

3 years ago

A 6cm diameter horizontal pipe gradually narrows to 4cm.

Rom4ik [11]

Answer: $Q=0.5612 m^3/s$

Explanation:

Given

diameter of pipe( $d_1$ )=6 cm

diameter of pipe( $d_2$ )=4 cm

$P_1=32 kPa$

$P_2=24 kPa$

$A_1=\frac{\pi }{4}6^2=9\pi cm^2$

$A_2=\frac{\pi }{4}4^2=4\pi cm^2$

$v_1=\frac{Q}{A_1}$

Applying bernoulli's equation

$\frac{P_1}{\rho g}+\frac{v^2_1}{2g}+z_1=\frac{P_2}{\rho g}+\frac{v^2_2}{2g}+z_2$

$\frac{P_1}{\rho g}+\frac{\frac{Q^2}{A_1^2}}{2g}+z_1=\frac{P_2}{\rho g}+\frac{\frac{Q^2}{A_2^2}}{2g}+z_2$

since $z_1=z_2$

$\frac{32\times 10^3}{10^3\times 9.81}+\frac{Q^2}{2A_1^2g}=\frac{24\times 10^3}{10^3\times 9.81}+\frac{Q^2}{2A_2^2g}$

$Q^2=\frac{8\times 2\times 81\pi ^2\times 16\pi ^2\times 10^{-4}}{65\pi ^2}$

$Q^2=3149.3722\times 10^{-4}$

$Q=\sqrt{3149.3722\times 10^{-4}}$

$Q=0.5612 m^3/s$

4 0

2 years ago

Please help me with this.

Stells [14]

The answer is D, only B is a mineral. I always look at it as mineral's atoms are usually closer together but always check your answers.

7 0

3 years ago