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quester [9]
3 years ago
5

Can electrons travel in speed of light?

Physics
1 answer:
Virty [35]3 years ago
6 0
Yes

Is a pop quiz! :)
:( ;)
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What’s better csp or pv ?
BARSIC [14]

Answer:

CSP

Explanation:

CSP systems store energy through Thermal Energy Storage technologies (TES), so power can be used when there isn't enough sunlight. PV systems, however, can't store thermal energy because they use direct sunlight, rather than heat. For this reason, CSP systems are better for energy storage and efficiency.

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2 years ago
What is an organism's specific role in an ecosystem
Harman [31]
<span>Niche - </span><span>An organism's particular role in an ecosystem, or how it makes its living.</span>
7 0
3 years ago
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The tension in cable da has a magnitude of tda=6.27 lb. find the cartesian components of tension tda, which is directed from d t
Oliga [24]

Complete Question

The Complete Question is attached below

We have that the Cartesian components of tension T_{da} is

T_{DA}=-4.433i-3.49j+2.735k

From the Question we are told that

M_{da}=6.27 lb\\\\w=9.50ft\\\\d=6.60ft\\\\h=4.50ft

\vec {DA}=-4.7i-3.7j+2.9k)ft

\vec {DB}=-1.9i-3.7j+1.9k)ft\\\\\vec {DC}=-1.9i+5.8j-1.6k)ft

Generally the equation for T_{DA}  is mathematically given as

T_{DA}=\phi_{DA}* M_{da}

Where

\phi_{DA}=\frac{-4.7i-3.7j+2.9k}{(-4.7)^2+(-3.7)^2+(2.9)^2}\\\\\phi_{DA}=\frac{-4.7i-3.7j+2.9k}{6.65}

Therefore

T_{DA}=\phi_{DA}* M_{da}

T_{DA}=\frac{-4.7i-3.7j+2.9k}{6.65}* 6.27

T_{DA}=-4.433i-3.49j+2.735k

For more information on this visit

brainly.com/question/20746649?referrer=searchResults

5 0
3 years ago
A nonconducting sphere has radius R = 1.29 cm and uniformly distributed charge q = +3.83 fC. Take the electric potential at the
zalisa [80]

Answer:

a) -2.516 × 10⁻⁴ V

b) -1.33 × 10⁻³ V

Explanation:

The electric field inside the sphere can be expressed as:

E= \frac{kqr}{R^3}

The potential at a distance can be represented as:

V(r) - V(0) = -\int\limits^r_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) = [\frac{qr^2}{8 \pi E_0R^3 }]₀

V(r) =   -[\frac{qr^2}{8 \pi E_0R^3 }]₀

Given that:

q = +3.83 fc = 3.83 × 10⁻¹⁵ C

r = 0.56 cm

 = 0.56 × 10⁻² m

R = 1.29 cm

  =  1.29 × 10⁻² m

E₀ = 8.85 × 10⁻¹² F/m

Substituting our values; we have:

V(r) = -\frac{(3.83*10^{-15}C)(0.560*10^{-2}m)^2}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)^3}

V(r) = -2.15  × 10⁻⁴ V

The difference between the radial distance  and center can be expressed as:

V(r) - V(0) = -\int\limits^R_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) =  [\frac{qr^2}{8 \pi E_0R^3 }]^R

V(r) = -\frac{qR^2}{8 \pi E_0R^3 }

V(r) = -\frac{q}{8 \pi E_0R }

V(r) = -\frac{(3.83*10^{-15}C)}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)}

V(r) = -0.00133

V(r) = - 1.33 × 10⁻³ V

8 0
3 years ago
Two bodies are falling with negligible air resistance, side by side, above a horizontal plane. If one of the bodies is given an
tankabanditka [31]

Answer:4-strikes the plane at same time as the other body

Explanation:

Given

If both bodies is falling on a horizontal plane and second body is given an acceleration in horizontal direction then it does not change the time to reach the Horizontal Plate as there is no change in vertical direction.

Horizontal acceleration will give only horizontal range and horizontal velocity.

8 0
3 years ago
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