Answer:
521 nm
Explanation:
Given the values and units we are given, I'm assuming 5.76*10^14 Hz is frequency.
The formula to use here is λ * υ = c, where λ is wavelength, υ is frequency, and c is the speed of light.
λ = ![\frac{3*10^8\frac{m}{s} }{5.76*10^{14}Hz} = {5.20833*10^{-7} m}\approx{521 *10^{-9}m}={521 nm}](https://tex.z-dn.net/?f=%5Cfrac%7B3%2A10%5E8%5Cfrac%7Bm%7D%7Bs%7D%20%7D%7B5.76%2A10%5E%7B14%7DHz%7D%20%3D%20%7B5.20833%2A10%5E%7B-7%7D%20m%7D%5Capprox%7B521%20%2A10%5E%7B-9%7Dm%7D%3D%7B521%20nm%7D)
HI buddy!
An integrated circuit(IC) is a semiconductor water on which thousands of tiny resistors, transistors are fabricated.
Remember that an integrated circuit are the heart and brains of most circuits. They are the keystone of modern electronics.
I hope this helps!
The EMT must assume that any unwitnessed water-related incident is accompanied by potential spinal damage.
<h3>What is spinal damage?</h3>
- Nerves or the spinal cord in any way damaged at the end of the spinal canal.
- A rapid strike or cut to the spine can cause a traumatic spinal cord damage.
- Below the damage site, a spinal cord injury frequently results in a lifelong loss of strength, feeling, and function.
- A lot of people with spinal cord injuries may lead productive, independent lives with the help of rehabilitation and assistive technology.
- Symptom-reducing medications and spinal stabilisation surgery are used as treatments.
- Herniated discs are among the common injuries and diseases of the spine. Stenosis of the lower back and Scoliosis are others.
- After taking part in a rehabilitation programme, over 80% of people with incomplete spinal cord injury (SCI) can walk again.
Learn more about spinal cord here:
brainly.com/question/23916836
#SPJ4
Answer:0.0704 kg
Explanation:
Given
initial Absolute pressure
=210+101.325=311.325
![T_1=25^{\circ}\approx 298 K](https://tex.z-dn.net/?f=T_1%3D25%5E%7B%5Ccirc%7D%5Capprox%20298%20K)
![V=0.025 m^3](https://tex.z-dn.net/?f=V%3D0.025%20m%5E3)
![T_2=50^{\circ}\approx 323 K](https://tex.z-dn.net/?f=T_2%3D50%5E%7B%5Ccirc%7D%5Capprox%20323%20K)
as the volume remains constant therefore
![\frac{P_1}{T_1}=\frac{P_2}{T_2}](https://tex.z-dn.net/?f=%5Cfrac%7BP_1%7D%7BT_1%7D%3D%5Cfrac%7BP_2%7D%7BT_2%7D)
![\frac{311.325}{298}=\frac{P_2}{323}](https://tex.z-dn.net/?f=%5Cfrac%7B311.325%7D%7B298%7D%3D%5Cfrac%7BP_2%7D%7B323%7D)
![P_2=337.44 KPa](https://tex.z-dn.net/?f=P_2%3D337.44%20KPa)
therefore Gauge pressure is 337.44-101.325=236.117 KPa
Initial mass ![m_1=\frac{P_1V}{RT_1}=\frac{311.325\times 0.025}{0.0287\times 298}](https://tex.z-dn.net/?f=m_1%3D%5Cfrac%7BP_1V%7D%7BRT_1%7D%3D%5Cfrac%7B311.325%5Ctimes%200.025%7D%7B0.0287%5Ctimes%20298%7D)
![m_1=0.91 kg](https://tex.z-dn.net/?f=m_1%3D0.91%20kg)
Final mass ![m_2=\frac{P_2V}{RT_2}=\frac{311.325\times 0.025}{0.0287\times 323}](https://tex.z-dn.net/?f=m_2%3D%5Cfrac%7BP_2V%7D%7BRT_2%7D%3D%5Cfrac%7B311.325%5Ctimes%200.025%7D%7B0.0287%5Ctimes%20323%7D)
![m_2=0.839](https://tex.z-dn.net/?f=m_2%3D0.839)
Therefore
=0.91-0.839=0.0704 kg of air needs to be removed to get initial pressure back
Answer:
the rate of acceleration of the train is 4 m/s²
Explanation:
Given;
initial velocity of the train, u = 10 m/s
change in time of motion, dt = 5 s
final velocity of the train, v = 30 m/s
The rate of acceleration of the train is calculated as;
![a = \frac{dv}{dt} = \frac{v-u}{dt} = \frac{30-10}{5} = \frac{20}{5} = 4 \ m/s^2](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7Bdv%7D%7Bdt%7D%20%3D%20%5Cfrac%7Bv-u%7D%7Bdt%7D%20%3D%20%5Cfrac%7B30-10%7D%7B5%7D%20%3D%20%5Cfrac%7B20%7D%7B5%7D%20%3D%204%20%5C%20m%2Fs%5E2)
Therefore, the rate of acceleration of the train is 4 m/s²