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2 years ago

Describe how private citizens have a voice in which projects the federal government will fund

1 answer:

3 0

<span>Two main ways, free elections, casting your vote for the candidate supporting your goals. Second, Lobbying the sitting gov.</span>

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Consider a steel guitar string of initial length l=1.00m and cross-sectional area a=0.500mm2. the young's modulus of the steel i

laiz [17]

L = 1.00 m, the original length
A = 0.5 mm² = 0.5 x 10⁻⁶ m², the cross sectional area
E = 2.0 x 10¹¹ n/m², Young's modulus
P = 1500 N, the applied tension

Calculate the stress.
σ = P/A = (1500 N)/(0.5 x 10⁻⁶ m²) = 3 x 10⁹ N/m²

Let δ = the stretch of the string.
Then the strain is
ε = δ/L

By definition, the strain is
ε = σ/E = (3 x 10⁹ N/m²)/(2 x 10¹¹ N/m²) = 0.015
Therefore
δ/(1 m) = 0.015
δ = 0.015 m = 15 mm

Answer: 15 mm

4 0

2 years ago

An engineer is designing a process for a new transistor. She uses a vacuum chamber to bombard a thin layer of silicon with ions

adelina 88 [10]

Answer:

$E=5.7\times 10^{-3}\ V/m$

Explanation:

Given that

$m_p=5.18\times 10^{-26}\ kg$

re= 46 cm

Vp= 180 m/s

We know that

$E=\dfrac{\Delta V}{r}$

$\Delta V=\dfrac{1}{4}\dfrac{m_pv_p^2}{e}$

$E=\dfrac{1}{4}\dfrac{m_pv_p^2}{e.r_e}$

Now by putting the all given values in the questions

$E=\dfrac{1}{4}\times \dfrac{5.18\times 10^{-26}\times 180^2}{1.6\times 10^{-19}\times 0.46}$

$E=5.7\times 10^{-3}\ V/m$

So the average electric field is $E=5.7\times 10^{-3}\ V/m$ .

6 0

2 years ago

A force of 200 N is applied on small piston of a pascal press. What would be the

VladimirAG [237]

Answer:

The force applied on the big piston is 1306.67 N

Explanation:

Given;

force applied on small piston, F₁ = 200 N

diameter of the small piston, d₁ = 4.37 cm

radius of the small piston, r₁ = d₁/2 = 2.185 cm

Area of the small piston, A₁ = πr₁² = π(2.185 cm)² = 15 cm²

Area of the big piston, A₂ = 98 cm²

The pressure of the piston is given by;

$P = \frac{F}{A} \\\\\frac{F_1}{A_1} = \frac{F_2}{A_2}\\\\ F_2 = \frac{F_1A_2}{A_1}$

Where;

F₂ is the force on big piston

$F_2 = \frac{200*98}{15} \\\\F_2 = 1306.67 \ N$

Therefore, the force applied on the big piston is 1306.67 N

3 0

1 year ago

Force F acts between two charges, q1 and q2, separated by a distance d. If q1 is increased to twice its original value and the d

Step2247 [10]

Okay, haven't done physics in years, let's see if I remember this.

So Coulomb's Law states that $F = k \frac{Q_1Q_2}{d^2}$ so if we double the charge on $Q_1$ and double the distance to $(2d)$ we plug these into the equation to find

<span> $F_{new} = k \frac{2Q_1Q_2}{(2d)^2}=k \frac{2Q_1Q_2}{4d^2} = \frac{2}{4} \cdot k \frac{Q_1Q_2}{d^2} = \frac{1}{2} \cdot F_{old}$ </span>

So we see the new force is exactly 1/2 of the old force so your answer should be $\frac{1}{2}F$ if I can remember my physics correctly.

9 0

2 years ago

Read 2 more answers

A quarterback throws a football at 40km/hr to a receiver 50yd away. How much time does it take the ball to reach the receiver

Akimi4 [234]

Given:

Distance = 50 yard = 45.72 meter

Speed = 40 km/hr = 11.11 m/s

To find:

Time required by ball to reach the receiver = ?

Formula used:

speed = $\frac{distance}{time}$

Solution:

The speed of the ball is given by,

speed = $\frac{distance}{time}$

Thus,

Time = $\frac{distance}{speed}$

Distance = 50 yard = 45.72 meter

Speed = 40 km/hr = 11.11 m/s

Time = 4.12 second

Hence, ball reaches the receiver in 4.12 second.

3 0

2 years ago