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murzikaleks [220]
3 years ago
13

Charles designed a system to pull sleds uphill, but he needs to know the angle of the slope. He knows that the hill is 13 m high

and that the distance from the base of the hill to the top is 20 m. calculate the angle of the slope. Hint: You may assume that the hill forms a right triangle
Physics
1 answer:
timama [110]3 years ago
6 0

Answer:

Explanation:

It is given that,

Height of the hill, h = 13 m

The distance from the base of the hill to the top is 20 m, d = 20 m

We need to find the angle of the slope. It is already mentioned that the the hill forms a right triangle. For a right angled triangle the angle is given by :

sin\theta=\dfrac{perpendicular}{hypotenuse}

sin\theta=\dfrac{13}{20}

\theta=sin^{-1}(\dfrac{13}{20})

\theta=40.54^{\circ}

So, the angle of the slope is 40.54 degrees. Hence, this is the required solution.

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3 years ago
1) Consider a source particle of charge qS=9 C located at (−9,5) [distances in meters]. Find the electric field vector at the ta
xeze [42]

Answer:

1)  E = 2.25 i^+ 0.809j^)  10⁹ N / C , 2)   E = 2.39 10⁹ N / C , 3)    θ = 19.8º , 4)   F = 19.12 10⁹ N , 5)  E = (1.32 i^+ 3.56 j^) 109 N/C

Explanation:

1) The equation for the electric field is

       E = k q / r²

Where K is the Coulomb constant that is worth 8.99 10⁹ N m² /C², q is the load and r is the distance of the load to the test point

Since the electric field is a vector magnitude, we can find its component

X axis

     Ex = k q / x²

where the distance on the axis is

      x = √ (X₂-x₁)²

      x = √ (-15 + 9)² = 6 m

      Eₓ = 8.99 10⁹ 9/6²

      Eₓ = 2.25 10⁹ N /C

Y Axis

     y = √ (y₂-y₁)² = √ (15-5)² = 10 m

     E_{y} = 8.99 10⁹ 9/10²

      E_{y}  = 0.809 10⁹ N / C

     

     E = Eₓ i^+   E_{y}  j^

     E = 2.25 i^+ 0.809j^)  10⁹ N / C

2) the magnitude can be found using the Pythagorean triangle

        E = √ (Eₓ² +  E_{y}²)

        E = √ (2.25² + 0.809²) 10⁹

        E = 2.39 10⁹ N / C

3) to find the angle let's use trigonometry

       tan θ = E_{y} / Eₓ

       θ = tan⁻¹ E_{y} / Eₓ

       θ = tan⁻¹ (0.809 / 2.25)

       θ = 19.8º

Regarding the positive side of the x axis

4) a charge  q2 = 8C is placed, let's calculate the force

      F = q E

      F = 8 2.39 10⁹

      F = 19.12 10⁹ N

5) The total electric field at the origin, let's look for its components

     q₁ = 9C

     r₁ = -9 i ^ + 5 j ^

     q₂ = 8 C

     r₂ = -15 i ^ + 15 j ^

X axis

     Eₓ = E₁ₓ + E₂ₓ

     Eₓ = k q₁ / Dx₁² + k q₂ / Dx₂²

     Eₓ = 8.99 10⁹ (9 / (9-0)² + 8 / (15-0)²)

     Eₓ = 1.32 109 N / A

Y Axis

    E_{y} =  E_{1y} + E_{2y}

      E_{y} = k q₁ / Δy₁² + k q₂ / Δy₂²

      E_{y} = 8.99 109 (9/5² + 8/15²)

      E_{y} = 3.56 109 N / A

     E = (1.32 i^+ 3.56 j^) 109 N/C

7 0
4 years ago
48 You are given two amplifiers, A and B, to connect in cascade between a 10-mV, 100-k source and a 100- load. The amplifiers ha
Sedbober [7]

Answer:

SABL

Explanation:

The best amplifier will be the one that gives us a bigger gain. In each stage will be a load factor that will reduce the gain, that is defined as:

Fp=\frac{R_{in}}{R_{in}+R_{out}}\\

where Rin is the input resistance of the next stage and Rout the output resistance of the previous stage.

Analyzing SABL:

Fp_1=\frac{100K}{100K+100K}=0.5\\\\Fp_2=\frac{10K}{10K+10K}=0.5\\\\Fp_3=\frac{100}{100+1K}=0.0909

the total gain will be the total gain of each stage multiplied by the load factor.

SABL_{gain}=0.5*100*0.5*10*0.0909\\SABL_{gain}=22.73\\SABL_{gain_{db}}=20*\log{22.73}=27.13db

Analyzing SBAL:

Fp_1=\frac{10K}{10K+100K}=0.0909\\\\Fp_2=\frac{100K}{100K+10K}=0.909\\\\Fp_3=\frac{100}{100+10K}=0.0099

the total gain will be the total gain of each stage multiplied by the load factor.

SABL_{gain}=0.0909*10*0.99*100*0.0099\\SABL_{gain}=0.89\\SABL_{gain_{db}}=20*\log{0.89}=-1.01db

So the best amplifier arrangement is SABL.

4 0
3 years ago
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