In which object more force has to be applied to move in between two objects of mas 11

Which of the following types of light microscopy improves the resolution of thick specimens by illuminating one plane of the spe

Vilka [71]

Answer:

confocal microscopy

Explanation:

According to my research on different types of microscopes, I can say that based on the information provided within the question the tool being mentioned in this situation is a confocal microscopy. This is an extremely powerful microscope used to develop extremely sharp images of cells and tissues by viewing one plane of the specimen at a given time.

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

7 0

3 years ago

A cardboard box has four forces acting upon it, as shown in the diagram below. What is the magnitude of the unknown force, Funkn

shutvik [7]

D. 11N
Nznsnksksksksmsmsms

4 0

3 years ago

The per-unit impedance of a single-phase electric load is 0.3. The base power is 500 kVA, and the base voltage is 13.8 kV. a. Fi

Leto [7]

Answer:

114.26

Explanation:

a)Formula for per unit impedance for change of base is

Zpu2= Zpu1×(kV1/kV2)²×(kVA2/kVA1)

Zpu2: New per unit impedance

Zpu1: given per unit impedance

kV1: give base voltage

kV2: New bas votlage

kVA1: given bas power

kVA2: new base power

In the question

Zpu2=??

Zpu1= 0.3

kV2=24kV

kV1= 13.8 kV

kVA2= 1MVA ×1000= 1000 kVA

kVA1=500kVA

Zpu2= 0.3(13.8/24)²×(1000/500)

Zpu2= 0.198

b) to find ohmic impedance we will first calculate base value of impedance(Zbase). So,

Zbase= kV²/MVA

Zbase= 13.8²/(500/1000)

Zbase=380.88

Now that we have base value of impedance, Zbase, we can calculate actual ohmic value of impedance(Zactual) by using the following formula:

Zpu=Zactual/Zbase

0.3= Zactual/380.88

Zactual= 114.26 ohms

8 0

3 years ago

A thin, metallic spherical shell of radius 0.347 m0.347 m has a total charge of 7.53×10−6 C7.53×10−6 C placed on it. A point cha

USPshnik [31]

Answer:

E = 12640.78 N/C

Explanation:

In order to calculate the electric field you can use the Gaussian theorem.

Thus, you have:

$\Phi_E=\frac{Q}{\epsilon_o}$

ФE: electric flux trough the Gaussian surface

Q: net charge inside the Gaussian surface

εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2

If you take the Gaussian surface as a spherical surface, with radius r, the electric field is parallel to the surface anywhere. Then, you have:

$\Phi_E=EA=E(4\pi r^2)=\frac{Q}{\epsilon_o}\\\\E=\frac{Q}{4\pi \epsilon_o r^2}$

r can be taken as the distance in which you want to calculate the electric field, that is, 0.795m

Next, you replace the values of the parameters in the last expression, by taking into account that the net charge inside the Gaussian surface is:

$Q=7.53*10^{-6}C+3.65*10^{-6}C=1.115*10^{-5}C$

Finally, you obtain for E:

$E=\frac{1.118*10^{-5}C}{4\pi (8.85*10^{-12C^2/Nm^2})(0.795m)^2}=12640.78\frac{N}{C}$

hence, the electric field at 0.795m from the center of the spherical shell is 12640.78 N/C

3 0

3 years ago

Find the total electric charge of 1.7 kg of electrons. me=9.11×10−31kg, e=1.60×10−19C.

Gelneren [198K]

Answer:

$2.99\cdot 10^{11}C$

Explanation:

The mass of one electron is

$m_e = 9.11\cdot 10^{-31}kg$

So the number of electrons contained in M=1.7 kg of mass is

$N=\frac{M}{m_e}=\frac{1.7 kg}{9.11\cdot 10^{-31}kg}=1.87\cdot 10^{30}$

The charge of one electron is

$e=1.60\cdot 10^{-19} C$

So, the total charge of these electrons is equal to the charge of one electron times the number of electrons:

$Q=Ne=(1.87\cdot 10^{30})(1.6\cdot 10^{-19}C)=2.99\cdot 10^{11}C$

8 0

3 years ago