Le Which describes how light waves can interact with wood?

A photon with a frequency of 5.48 × 1014 hertz is emitted when an electron in a mercury atom falls to a lower energy level. Iden

ludmilkaskok [199]

The color should be green.

3 0

3 years ago

The radius of a sphere is increasing at a rate of 4 mm/s. how fast is the volume increasing when the diameter is 40 mm?

marin [14]

Using r to represent the radius and t for time, you can write the first rate as:

drdt=4mms

or

r=r(t)=4t

The formula for a solid sphere's volume is:

V=V(r)=43πr3

When you take the derivative of both sides with respect to time...

dVdt=43π(3r2)(drdt)

...remember the Chain Rule for implicit differentiation. The general format for this is:

dV(r)dt=dV(r)dr(t)⋅dr(t)dt with V=V(r) and r=r(t) .

So, when you take the derivative of the volume, it is with respect to its variable r (dV(r)dr(t)) , but we want to do it with respect to t (dV(r)dt) . Since r=r(t) and r(t) is implicitly a function of t , to make the equality work, you have to multiply by the derivative of the function r(t) with respect to t (dr(t)dt) as well. That way, you're taking a derivative along a chain of functions, so to speak (V→r→t ).

Now what you can do is simply plug in what r is (note you were given diameter) and what drdt is, because dVdt describes the rate of change of the volume over time, of a sphere.

dVdt=43π(3(20mm)2)(4mms) =6400πmm3s

Since time just increases, and the radius increases as a function of time, and the volume increases as a function of a constant times the radius cubed, the volume increases faster than the radius increases, so we can't just say the two rates are the same.

7 0

3 years ago

Which of the following is not an example of a physical change?

Mazyrski [523]

Answer:

Explanation:

Cutting a string in half because

b is irreversible

c is a cheical and d is also a chemical change

8 0

3 years ago

A block attached to a spring with an unknown spring constant oscillates with a period of 2.0 s. What is the period if

Zigmanuir [339]

Answer:

a) If the mass is doubled, then the period is increased by $\sqrt{2}$ . Hence, the period of the system is 2.828 seconds.

b) If the mass is halved, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.

d) If the spring constant is doubled, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

Explanation:

The statement is incomplete. We proceed to present the complete statement: A block attached to a spring with unknown spring constant oscillates with a period of 2.00 s. What is the period if a. The mass is doubled? b. The mass is halved? c. The amplitude is doubled? d. The spring constant is doubled?

We have a block-spring system, whose angular frequency ( $\omega$ ) is defined by the following formula:

$\omega = \sqrt{\frac{k}{m} }$ (1)

Where:

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass, measured in kilograms.

And the period ( $T$ ), measured in seconds, is determined by the following expression:

$T = \frac{2\pi}{\omega}$ (2)

By applying (1) in (2), we get the following formula:

$T = 2\pi\cdot \sqrt{\frac{m}{k} }$

a) If the mass is doubled, then the period is increased by $\sqrt{2}$ . Hence, the period of the system is 2.828 seconds.

b) If the mass is halved, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.

d) If the spring constant is doubled, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

8 0

3 years ago

How much heat does it take to raise the temperature of 10.0 kg of water by 1.0 C?

fomenos

Answer:The specific heat capacity of water is 4,200 joules per kilogram per degree Celsius (J/kg°C). This means that it takes 4,200 J to raise the temperature of 1 kg of water by 1°C.

Explanation:

7 0

2 years ago