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3 years ago

5

Barry wants to convert mechanical energy into electric energy. What can he use?

2 answers:

Sedaia [141]3 years ago

8 0

Answer:

generator

Explanation:

generator converts mechanical energy to electrical

scZoUnD [109]3 years ago

5 0

Answer:

Barry wants to convert mechanical energy into electric energy. What can he use?

Explanation:

You might be interested in

If the specific surface energy for aluminum oxide is 0.90 J/m2 and its modulus of elasticity is (393 GPa), compute the critical

vampirchik [111]

Answer:

critical stress required for the propagation is 27.396615 × $10^{6}$ N/m²

Explanation:

given data

specific surface energy = 0.90 J/m²

modulus of elasticity E = 393 GPa = 393 × $10^{9}$ N/m²

internal crack length = 0.6 mm

to find out

critical stress required for the propagation

solution

we will apply here critical stress formula for propagation of internal crack

( σc ) = $\sqrt{\frac{2E\gamma s}{\pi a}}$ .....................1

here E is modulus of elasticity and γs is specific surface energy and a is half length of crack i.e 0.3 mm = 0.3 × $10^{-3}$ m

so now put value in equation 1 we get

( σc ) = $\sqrt{\frac{2E\gamma s}{\pi a}}$

( σc ) = $\sqrt{\frac{2*393*10^9*0.90}{\pi 0.3*10^{-3}}}$

( σc ) = 27.396615 × $10^{6}$ N/m²

so critical stress required for the propagation is 27.396615 × $10^{6}$ N/m²

6 0

3 years ago

A closed system of mass 10 kg undergoes a process during which there is energy transfer by work from the system of 0.147 kJ per

mr_godi [17]

Answer:

-50.005 KJ

Explanation:

Mass flow rate = 0.147 KJ per kg

mass= 10 kg

Δh= 50 m

Δv= 15 m/s

W= 10×0.147= 1.47 KJ

Δu= -5 kJ/kg

ΔKE + ΔPE+ ΔU= Q-W

0.5×m×(30^2- 15^2)+ mgΔh+mΔu= Q-W

Q= W+ 0.5×m×(30^2- 15^2) +mgΔh+mΔu

= 1.47 +0.5×1/100×(30^2- 15^2)-9.7×50/1000-50

= 1.47 +3.375-4.8450-50

Q=-50.005 KJ

7 0

3 years ago

Read 2 more answers

A charge of +2.00 μC is at the origin and a charge of –3.00 μC is on the y axis at y = 40.0 cm . (a) What is the potential at po

Nimfa-mama [501]

a) Potential in A: -2700 V

b) Potential difference: -26,800 V

c) Work: $4.3\cdot 10^{-15} J$

Explanation:

a)

The electric potential at a distance r from a single-point charge is given by:

$V(r)=\frac{kq}{r}$

where

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have a system of two charges, so the total potential at a certain point will be given by the algebraic sum of the two potentials.

Charge 1 is

$q_1=+2.00\mu C=+2.00\cdot 10^{-6}C$

and is located at the origin (x=0, y=0)

Charge 2 is

$q_2=-3.00 \mu C=-3.00\cdot 10^{-6}C$

and is located at (x=0, y = 0.40 m)

Point A is located at (x = 0.40 m, y = 0)

The distance of point A from charge 1 is

$r_{1A}=0.40 m$

So the potential due to charge 2 is

$V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.40}=+4.50\cdot 10^4 V$

The distance of point A from charge 2 is

$r_{2A}=\sqrt{0.40^2+0.40^2}=0.566 m$

So the potential due to charge 1 is

$V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.566}=-4.77\cdot 10^4 V$

Therefore, the net potential at point A is

$V_A=V_1+V_2=+4.50\cdot 10^4 - 4.77\cdot 10^4=-2700 V$

b)

Here we have to calculate the net potential at point B, located at

(x = 0.40 m, y = 0.30 m)

The distance of charge 1 from point B is

$r_{1B}=\sqrt{(0.40)^2+(0.30)^2}=0.50 m$

So the potential due to charge 1 at point B is

$V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.50}=+3.60\cdot 10^4 V$

The distance of charge 2 from point B is

$r_{2B}=\sqrt{(0.40)^2+(0.40-0.30)^2}=0.412 m$

So the potential due to charge 2 at point B is

$V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.412}=-6.55\cdot 10^4 V$

Therefore, the net potential at point B is

$V_B=V_1+V_2=+3.60\cdot 10^4 -6.55\cdot 10^4 = -29,500 V$

So the potential difference is

$V_B-V_A=-29,500 V-(-2700 V)=-26,800 V$

c)

The work required to move a charged particle across a potential difference is equal to its change of electric potential energy, and it is given by

$W=q\Delta V$

where

q is the charge of the particle

$\Delta V$ is the potential difference

In this problem, we have:

$q=-1.6\cdot 10^{-19}C$ is the charge of the electron

$\Delta V=-26,800 V$ is the potential difference

Therefore, the work required on the electron is

$W=(-1.6\cdot 10^{-19})(-26,800)=4.3\cdot 10^{-15} J$

4 0

3 years ago

Kay [80]

Answer:

A. optical isolation

Explanation:

well I can't really give a good explanation because I also saw the same question in my exams and option A was the correct answer

6 0

3 years ago

True or false. Part of the mission of the NTSB is to determine the probable cause of an accident

eimsori [14]

uniform

welcome 2 Ghana African state western region

6 0

3 years ago