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2 years ago

When a solid compound dissolves in water,

1 answer:

6 0

The answer is d. Each of its particles becomes surrounded by water molecules.

Notice that when you dissolve a solid compound in the water, it is no longer visible in or eyes, it mixes in the water, it is because it is being divided into smaller particles and each particles were surrounded by the water molecules.

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A block of unknown mass is attached to a spring with a spring constant of 7.00 N/m 2 and undergoes simple harmonic motion with a

KatRina [158]

Answers:

a) $0.80 kg$

b) $2.12 s$

c) $1.093 m/s^{2}$

Explanation:

We have the following data:

$k=7 N/m$ is the spring constant

$A=12.5 cm \frac{1 m}{100 cm}=0.125 m$ is the amplitude of oscillation

$V=32 cm/s=0.32 m/s$ is the velocity of the block when $x=\frac{A}{2}=0.0625 m$

Now let's begin with the answers:

<h3>a) Mass of the block</h3>

We can solve this by the conservation of energy principle:

$U_{o}+K_{o}=U_{f}+K_{f}$ (1)

Where:

$U_{o}=k\frac{A^{2}}{2}$ is the initial potential energy

$K_{o}=0$ is the initial kinetic energy

$U_{f}=k\frac{x^{2}}{2}$ is the final potential energy

$K_{f}=\frac{1}{2} m V^{2}$ is the final kinetic energy

Then:

$k\frac{A^{2}}{2}=k\frac{x^{2}}{2}+\frac{1}{2} m V^{2}$ (2)

Isolating $m$ :

$m=\frac{k(A^{2}-x^{2})}{V^{2}}$ (3)

$m=\frac{7 N/m((0.125 m)^{2}-(0.0625 m)^{2})}{(0.32 m/s)^{2}}$ (4)

$m=0.80 kg$ (5)

<h3>b) Period</h3>

The period $T$ is given by:

$T=2 \pi \sqrt{\frac{m}{k}}$ (6)

Substituting (5) in (6):

$T=2 \pi \sqrt{\frac{0.80 kg}{7 N/m}}$ (7)

$T=2.12 s$ (8)

<h3>c) Maximum acceleration</h3>

The maximum acceleration $a_{max}$ is when the force is maximum $F_{max}$ , as well :

$F_{max}=m.a_{max}=k.x_{max}$ (9)

Being $x_{max}=A$

Hence:

$m.a_{max}=kA$ (10)

Finding $a_{max}$ :

$a_{max}=\frac{kA}{m}$ (11)

$a_{max}=\frac{(7 N/m)(0.125 m)}{0.80 kg}$ (12)

Finally:

$a_{max}=1.093 m/s^{2}$

5 0

3 years ago

Carbon is allowed to diffuse through a steel plate 15 mm thick. The concentrations of carbon at the two faces are 0.65 and 0.30

beks73 [17]

Answer:

T=575.16K

Explanation:

To solve the problem we proceed to use the 1 law of diffusion of flow,

Here,

$J=-D\frac{\Delta C}{\Delta x}$

$\Delta C$ is the rate in concentration

$\Delta x$ is the rate in thickness

D is the diffusion coefficient, where,

$D= D_0 exp(\frac{Q_d}{RT})$

Replacing D in the first law,

$J=-(D_0 exp(\frac{-Q_D}{RT}))\frac{\Delta }{\Delta x}$

clearing T,

$T=\frac{Q_d}{R*ln(\frac{J*\Delta x}{D_0*\Delta C})}$

Replacing our values

$T=-\frac{80000}{8.31*ln(\frac{(6.2*10^{-7})(-15*10^{-3})}{(1.43*10^{-9})(0.65-0.30)})}$

$T=-\frac{80000}{-138.09}$

$T=575.16K$

4 0

3 years ago

1.A Radio station broadcasts modern song on medium wave 350 Hz every day at ten o’clock in the morning. The velocity of radio wa

love history [14]

Answer:

$ans \: = \boxed{{4.8 \times 10}^{ - 4} Hz}$

Explanation:

$given \to \\ f_{r} = 350 \: \\ v_{r} = {3 \times 10}^{8} \\ but \to \\ v = f \gamma \to \: \gamma = \frac{v}{f} : hence \to \\ \gamma _{r} = \frac{v_{r}}{f_{r}} = \frac{3 \times 10^{8} }{350} = \boxed{857,142.85714 \: m}\\ therefore \to \\ given \to \\ f_{w} = water \: frequency = \: \boxed{ ?}\: \\ v_{w} = 14 50 \\ but \to \\ v = f \gamma \to \: \gamma = \frac{v}{f} : hence \to \\ \gamma _{w} = \frac{v_{w}}{f_{w}} = \frac{1}{100} \times \gamma _{r} = \frac{1}{100} \times 857,142.85714 \\\gamma _{w} = \boxed{8,571.4285714 \: m} : hence \to \: \\ f_{w} = \frac{v_{w}}{ \gamma _{w}} = \frac{1450}{8,571.4285714} = \boxed{0.1691666667} \\ if \: the \: number \: of \: times = \boxed{ x} \\ f_{r} (x)=f_{w} \\ (x) = \frac{f_{w}}{f_{r}} = \frac{0.1691666667}{350} = 0.0004833333 \\ hence \to \\ the \: frequency \: of \: the \: radio \: wave \: is \to \: \boxed{{4.8 \times 10}^{ - 4} }\: \\ that \: of \: the \: wave \: created \: in \: the \: water.$

♨Rage♨

8 0

3 years ago

If the back of the truck is 1.3 m above the ground and the ramp is inclined at 22 ∘ , how much time do the workers have to get t

solong [7]

Refer to the diagram shown below.

Assume that
(a) The piano rolls down on frictionless wheels,
(b) Wind resistance is negligible.

The distance along the ramp is
d = (1.3 m)/sin(22°) = 3.4703 m

The component of the piano's weight along the ramp is
mg sin(22°)
If the acceleration down the ramp is a, then
ma = mg sin(22°)
a = g sin(22°) = (9.8 m/s²) sin(22°) = 3.671 m/s²

The time, t, to travel down the ramp from rest is given by
(3.4703 m) = 0.5*(3.671 m/s²)*(t s)²
t² = 3.4703/1.8355 = 1.8907
t = 1.375 s

Answer: 1.375 s