8750 J of heat are applied to a piece of aluminium causing a 56C increase in its

Light with a frequency of 7.30 x 1014 hz lies in the violet region of the visible spectrum. What is the wavelength of this frequ

Leona [35]

From the theory we know that:

c = λ / T

f = 1 / T

Where:

c = 3. $10^{8}$ / m (the speed of light)

λ is the wavelengh (in meters)

T is the period (in seconds)

f is the frequency (in Hz)

We were told that:

f = 7.30 . $10^{14}$

And we want to find out the value of λ.

c = λ / T

c = λ . 1/T

Swaping 1/T = f

c = λ . f

λ = c / f

λ = 3 . $10^{8}$ / 7.30 . $10^{14}$

λ = 4.12 $10^{-7}$ m

Response: 4.12 $10^{-7}$ m = 412 nm

:-)

6 0

3 years ago

1. After a rabbit population reaches the carrying capacity of its habitat, what will most likely happen to

kifflom [539]

C) alternately increase and decrease

6 0

3 years ago

The area of the effort and load of the Piston of a hydrolic are 0.5 and 5m respectively. If a force of 100 Newton is applied on

Yanka [14]

Answer:

Explanation:

Using Pascal laws, which states that pressure are the input equals the pressure at the output.

Pressure is given as force/area

P1=P2

Then,

F1/A1=F2/A2

Cross multiply

F1A2=F2A1

Given that

Ae=0.5m² area of effort

Al=5m² area of load

Fl=? Force if load

Fe= 100N. Force of effort

Then applying pascal

Fl/Al=Fe/Ae

Fl/5=100/0.5

FL/5=200

Fl=200×5

Fl=1000N

The first safety load is 1000N

6 0

3 years ago

A small metal bar whose initial temperature was 20 degrees Celsius is dropped into a large container of boiling water. How long

Vesnalui [34]

Below is the answer. I hope it help.
T ( t ) = C e k t + T m where Tm is the temperature of the surroundings
T ( t ) = C e k t + T m
T ( 0 ) = 20
T ( 1 ) = T ( 0 ) + 2 = 22
C + T m = 20 C+Tm=20
C e k + T m = 22

4 0

3 years ago

A spring hangs from the ceiling with an unstretched length of x 0 = 0.35 m . A m 1 = 6.3 kg block is hung from the spring, causi

Jlenok [28]

Answer:

x₂=0.44m

Explanation:

First, we calculate the length the spring is stretch when the first block is hung from it:

$\Delta x_1=0.50m-0.35m=0.15m$

Now, since the stretched spring is in equilibrium, we have that the spring restoring force must be equal to the weight of the block:

$k\Delta x_1=m_1g$

Solving for the spring constant k, we get:

$k=\frac{m_1g}{\Delta x_1}\\\\k=\frac{(6.3kg)(9.8m/s^{2})}{0.15m}=410\frac{N}{m}$

Next, we use the same relationship, but for the second block, to find the value of the stretched length:

$k\Delta x_2=m_2g\\\\\Delta x_2=\frac{m_2g}{k}\\\\\implies \Delta x_2=\frac{(3.7kg)(9.8m/s^{2})}{410N/m}=0.088m$

Finally, we sum this to the unstretched length to obtain the length of the spring:

$x_2=0.35m+0.088m=0.44m$

In words, the length of the spring when the second block is hung from it, is 0.44m.

3 0

3 years ago