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Arada [10]
3 years ago
15

Complete the statement below with all that apply. Average speed is what?

Physics
2 answers:
iVinArrow [24]3 years ago
6 0

Answer:

a, b, d

Explanation:

Dmitrij [34]3 years ago
3 0

Answer:

i think its "A"

Explanation:

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PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!<br><br> Which number marks a crest?
lyudmila [28]

number 2 because the curve demstrates the crest GOOD LUCK i hope i got you the correct answer if not sorry

5 0
3 years ago
An average hole drift velocity of 103 cm/sec results when 2 V is applied across a 1 cm long semiconductor bar. What is the hole
Helga [31]

Answer:

ε = 2 V/cm

Explanation:

To calculate the mobility inside this bar, we just need to apply the expression that let us determine the mobility. This expression is the following:

ε = ΔV / L

Where:

ε: Hole mobility inside the bar

ΔV: voltage applied in the bar

L: Length of the bar

We already have the voltage and the length so replacing in the above expression we have:

ε = 2 V / 1 cm

<h2>ε = 2 V/cm</h2><h2></h2>

The data of the speed can be used for further calculations, but in this part its not necessary.

Hope this helps

8 0
3 years ago
Factors that affect pressure in fluid​
satela [25.4K]

Answer:

The factors that affect are depth of the fluid and its density

5 0
3 years ago
A point charge q1 = 1.0 µC is at the origin and a point charge q2 = 6.0 µC is on the x axis at x = 1 m.
iris [78.8K]

To solve this problem we will apply the concepts related to the Electrostatic Force given by Coulomb's law. This force can be mathematically described as

F = \frac{kq_1q_2}{d^2}

Here

k = Coulomb's Constant

q_{1,2} = Charge of each object

d = Distance

Our values are given as,

q_1 = 1 \mu C

q_2 = 6 \mu C

d = 1 m

k =  9*10^9 Nm^2/C^2

a) The electric force on charge q_2 is

F_{12} = \frac{ (9*10^9 Nm^2/C^2)(1*10^{-6} C)(6*10^{-6} C)}{(1 m)^2}

F_{12} = 54 mN

Force is positive i.e. repulsive

b) As the force exerted on q_2 will be equal to that act on q_1,

F_{21} = F_{12}

F_{21} = 54 mN

Force is positive i.e. repulsive

c) If q_2 = -6 \mu C, a negative sign will be introduced into the expression above i.e.

F_{12} = \frac{(9*10^9 Nm^2/C^2)(1*10^{-6} C)(-6*10^{-6} C)}{(1 m)^{2}}

F_{12} = F_{21} = -54 mN

Force is negative i.e. attractive

6 0
3 years ago
A brass rod with a length of 1.22 m and a cross-sectional area of 2.19 cm2 is fastened end to end to a nickel rod with length L
GaryK [48]

Answer:

a) L₂ = 0.676 m

b) σ₁ = 2.28*10⁸ N/m²

σ₂ = 9.62*10⁸ N/m²

c) ε₁ = 0.00253678

ε₂ = 0.00457875

Explanation:

Given info

L₁ = 1.22 m

A₁ = 2.19 cm² = 2.19*10⁻⁴ m²

L₂ = ?

A₂ = 0.52 cm² = 0.52*10⁻⁴ m²

P = 5.00*10⁴ N

E₁ = 9*10¹⁰ N/m²

E₂ = 2.1*10¹¹ N/m²

In order to get the length L of the nickel rod if the elongations of the two rods are equal, we can say that

ΔL₁ = ΔL₂   ⇒  P*L₁/(A₁*E₁) = P*L₂/(A₂*E₂)

⇒  L₂ = A₂*E₂*L₁ / (A₁*E₁)

⇒  L₂ = (0.52*10⁻⁴ m²)*(2.1*10¹¹ N/m²)*(1.22 m) / (2.19*10⁻⁴ m²*9*10¹⁰ N/m²)

⇒  L₂ = 0.676 m

The stress in the brass rod is obtained as follows

σ₁ = P/A₁ ⇒ σ = 5.00*10⁴ N / 2.19*10⁻⁴ m² = 2.28*10⁸ N/m²

The stress in the niquel rod is obtained as follows

σ₂ = P/A₂ ⇒ σ = 5.00*10⁴ N / 0.52*10⁻⁴ m² = 9.62*10⁸ N/m²

The strain in the brass rod is obtained as follows

σ₁ = E₁*ε₁    ⇒   ε₁ = σ₁ / E₁

⇒   ε₁ = 2.28*10⁸ N/m² / 9*10¹⁰ N/m² = 0.00253678

The strain in the niquel rod is obtained as follows

σ₂ = E₂*ε₂    ⇒   ε₂ = σ₂ / E₂

⇒   ε₂ = 9.62*10⁸ N/m² / 2.1*10¹¹ N/m² = 0.00457875

3 0
3 years ago
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