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3 years ago

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what's a tensor quantity ? ~

[Explain, in your own words]

[With an example]

Physics

2 answers:

Juli2301 [7.4K]3 years ago

8 0

TENSOR QUANTITY ~ 

this is a physical change
tensor quantity is a magnitude aa well as multiple direction
it is neither victor nor scanler
vector :.  which have only magnitude is known as victor
scaler :.  which has magnitude as well as a direction is know as scanler

<h3>EXAMPLE : </h3>

tensor : construction beam in a bridge 
vector : the car is driving velocity of 75.3 km 
scaler : the climate temperature was decreased

Liula [17]3 years ago

4 0

Answer:

A tensor is a quantity for example stress or strain which has magnitude , direction and a plane in which it acts . stress and strain are both tensor quantities , in real engineering components , stress and strain are 3-D tensors

Explanation:

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There is a parallel plate capacitor. Both plates are 4x2 cm and are 10 cm apart. The top plate has surface charge density of 10C

liberstina [14]

Answer:

1) The total charge of the top plate is 0.008 C

b) The total charge of the bottom plate is -0.008 C

2) The electric field at the point exactly midway between the plates is 0

3) The electric field between plates is approximately 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates is approximately 1.807 × 10⁻⁷ N

Explanation:

The given parameters of the parallel plate capacitor are;

The dimensions of the plates = 4 × 2 cm

The distance between the plates = 10 cm

The surface charge density of the top plate, σ₁ = 10 C/m²

The surface charge density of the bottom plate, σ₂ = -10 C/m²

The surface area, A = 0.04 m × 0.02 m = 0.0008 m²

1) The total charge of the top plate, Q = σ₁ × A = 0.0008 m² × 10 C/m² = 0.008 C

b) The total charge of the bottom plate, Q = σ₂ × A = 0.0008 m² × -10 C/m² = -0.008 C

2) The electrical field at the point exactly midway between the plates is given as follows;

$V_{tot} = V_{q1} + V_{q2}$

$V_q = \dfrac{k \cdot q}{r}$

Therefore, we have;

The distance to the midpoint between the two plates = 10 cm/2 = 5 cm = 0.05 m

$V_{tot} = \dfrac{k \cdot q}{0.05} + \dfrac{k \cdot (-q)}{0.05} = \dfrac{k \cdot q}{0.05} - \dfrac{k \cdot q}{0.05} = 0$

The electric field at the point exactly midway between the plates, $V_{tot}$ = 0

3) The electric field, 'E', between plates is given as follows;

$E =\dfrac{\sigma }{\epsilon_0 } = \dfrac{10 \ C/m^2}{8.854 \times 10^{-12} \ C^2/(N\cdot m^2)} \approx 1.1294 \times 10^{12}\ N/C$

E ≈ 1.1294 × 10¹² N/C

The electric field between plates, E ≈ 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates

The charge on an electron, e = -1.6 × 10⁻¹⁹ C

The force on an electron in the middle of the two plates, $F_e$ = E × e

∴ $F_e$ = 1.1294 × 10¹² N/C × -1.6 × 10⁻¹⁹ C ≈ 1.807 × 10⁻⁷ N

The force on an electron in the middle of the two plates, $F_e$ ≈ 1.807 × 10⁻⁷ N

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Answer:

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Comparing Wave A (black) to Wave B (green), Wave A has a

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I think it's C, longer wave length.

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A massless spring with spring constant 16.4 N/m hangs vertically. A body of mass 0.193 kg is attached to its free end and then r

svet-max [94.6K]

Answer:

(A) 0.2306 m

(B) 1.467 Hz

(C) 0.1152 m

Explanation:

spring constant (K) = 16.4 N/m

mass (m) = 0.193 kg

acceleration due to gravity (g) = 9.8 m/s^{2}

(A) force = Kx, where x = extension

mg = Kx

0.193 x 9.8 = 16.4x

x = 0.1153 m

now the mass actually falls two times this value before it gets to its equilibrium position ( turning point ) and oscillates about this point

therefore

2x = 0.2306 m

(B) frequency (f) = \frac{1}{2π} x $\sqrt{\frac{k}{m}}$

frequency (f) = \frac{1}{2π} x $\sqrt{\frac{16.4}{0.193}}$

frequency = 1.467 Hz

(C) the amplitude is the maximum position of the mass from the equilibrium position, which is half the distance the mass falls below the initial length of the spring

= \frac{0.2306}{2} = 0.1152 m

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3 years ago