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3 years ago

Look at the triangle below.

Physics

1 answer:

slega [8]3 years ago

7 0

Answer:

a = 32.48538491

b = 25.59883264

Step-by-step explanation:

To find the length of a, use the sine function:

$sin\alpha = \frac{opposite}{hypotenuse}$

$sin 38 = \frac{20}{a} \\a = \frac{20}{sin38}\\a = 32.48538491$

To find the length of b, use tangent:

$tan\alpha = \frac{opposite}{adjacent}$

$tan 38 = \frac{20}{b} \\b = \frac{20}{sin38}\\b = 25.59883264$

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Answer:

Explanation:

Component of force perpendicular to stick

= F Sin 60°

=√3 / 2 F.

Taking torque about the other end

= √3 / 2 F x 1 Nm

Weight of stick = 60 gm

= 60 x 10⁻³ kg

= 60 x 10⁻³ x 9.8 N

= .588 N

This weight will act from the middle point of stick so torque about the

other end

= .588 x 1 Nm

Balancing these two torques we have

.588 = √3 /2 F

$F=\frac{2\times0.588}{\sqrt{3} }$

F = 0.679 N

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3 years ago

When light propagates from a material with a given index of refraction into a material with a smaller index of refraction, the s

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3 years ago

The circuit shown in the figure (Figure 1) uses a neon-filled tube. This neon lamp has a threshold voltage V0 for conduction, be

Svetllana [295]

Part 1)

Answer:

Explanation:

As we know by equation of charging of the capacitor we will have

$V = E(1 - e^{-t/RC})$

so we will have

$87 = 105(1 - e^{-t/RC})$

here we know that

$R = 3.00 \times 10^6 ohm$

$C = 0.250 \mu F$

so we have

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Part b)

Answer:

The time will increase.

Explanation:

As we know that on increasing the value of the resistance the the product of the resistance and capacitance will increase so the time will increase to get the above voltage.

Part c)

Answer:

The capacitor discharges through a very low resistance (the lamp filled with ionized gas), and so the discharge time constant is very short. Thus the flash is very brief.

Explanation:

Since the lamp resistance is very small so the energy across the lamp will totally lost in very short interval of time

Part d)

Answer:

Once the lamp has flashed, the stored energy in the capacitor is gone, and there is no source of charge to maintain the lamp current. The lamp "goes out", the lamp resistance increases, and the capacitor starts to recharge. It charges again and the process will repeat.

Explanation:

Since we know that the battery is connected to the given system so after whole energy of capacitor is flashed out it is again charged by the battery and the process will continue

3 0

3 years ago

What does "Position (m)” represent in the graph?

SCORPION-xisa [38]

Position (m)” represent the dependent variable in the graph.

5 0

3 years ago

Read 2 more answers

A heavy rope, 60 ft long, weighs 0.5 lb/ft and hangs over the edge of a building 130 ft high. (Let x be the distance in feet bel

Nataliya [291]

Answer:

Riemann sum

W = lim n→∞ Σ 0.5xᵢΔx (with the summation done from i = 1 to n)

Integral = W = ∫⁶⁰₀ 0.5x dx

Workdone in pulling the entire rope to the top of the building = 900 lb.ft

Riemann sum for pulling half the length of the rope to the top of the building

W = lim n→∞ Σ 0.5xᵢΔx (but the sum is from i = 1 to n/2)

Integral = W = ∫⁶⁰₃₀ 0.5x dx

Work done in pulling half the rope to the top of the building = 675 lb.ft

Step-by-step explanation:

Using Riemann sum which is an estimation of area under a curve

The portion of the rope below the top of the building from x to (x+Δx) ft is Δx.

The weight of rope in that part would be 0.5Δx.

Then workdone in lifting this portion through a length xᵢ ft would be 0.5xᵢΔx

So, the Riemann sum for this total work done would be

W = lim n→∞ Σ 0.5xᵢΔx (with the summation done from i = 1 to n)

The Riemann sum can easily be translated to integral form.

In integral form, with the rope being 60 ft long, we have

W = ∫⁶⁰₀ 0.5x dx

W = [0.25x²]⁶⁰₀ = 0.25 (60²) = 900 lb.ft

b) When half the rope is pulled to the top of the building, 60 ft is pulled up until the length remaining is 30 ft

Just like in (a)

But the Riemann sum will now be from the start of the curve, to it's middle

Still W = lim n→∞ Σ 0.5xᵢΔx (but the sum is from i = 1 to n/2)

W = ∫⁶⁰₃₀ 0.5x dx

W = [0.25x²]⁶⁰₃₀ = 0.25 (60² - 30²) = 675 lb.ft

Hope this Helps!!!

7 0

3 years ago