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olya-2409 [2.1K]

3 years ago

10

The power dissipated in each of two resistors is the same. The current across resistor A is triple that across resistor B. If th

e resistance of resistor B is R, what is the resistance of A?

1 answer:

Svetllana [295]3 years ago

5 0

Answer:

Explanation:

this is the answer to your question

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You are a member of an alpine rescue team and must project a box of supplies, with mass m, up an incline of constant slope angle

AlekseyPX

Answer:

$v = \sqrt{2gh + \frac{2\mu_kgh}{tan\theta}}$

Explanation:

When we push the box from the bottom of the incline towards the top then by work energy theorem we can say that

Work done by all the forces = change in kinetic energy of the system

$- mgh - F_f (s) = 0 - \frac{1}{2}mv^2$

here we know that

$F_f = \mu_k mg cos\theta$

also we know that the length of the incline is given as

$s = \frac{h}{sin\theta}$

now we have

$- mgh - \mu_k mgcos\theta(\frac{h}{sin\theta}) = -\frac{1}{2}mv^2$

so we have

$v = \sqrt{2gh + \frac{2\mu_kgh}{tan\theta}}$

3 0

3 years ago

A red laser from a physics lab is marked as producing 632.8 nm light. When light from this laser falls on two closely spaced sli

goblinko [34]

Given Information:

Wavelength of the red laser = λr = 632.8 nm

Distance between bright fringes due to red laser = yr = 5 mm

Distance between bright fringes due to laser pointer = yp = 5.14 mm

Required Information:

Wavelength of the laser pointer = λp = ?

Answer:

Wavelength of the laser pointer = λp = ?

Explanation:

The wavelength of the monochromatic light can be found using young's double slits formula,

y = Dλ/d

y/λ = D/d

Where

λ is the wavelength

y is the distance between bright fringes.

d is the double slit separation distance

D is the distance from the slits to the screen

For the red laser,

yr/λr = D/d

For the laser pointer,

yp/λp = D/d

Equating both equations yields,

yr/λr = yp/λp

Re-arrange for λp

λp = yp*λr/yr

λp = (5*632.8)/5.14

λp = 615.56 nm

Therefore, the wavelength of the small laser pointer is 615.56 nm.

3 0

3 years ago

What is the momentum of an 18-kg object moving at 0.5 m/s?

Otrada [13]

sorry Idk the answer ..???

6 0

3 years ago

You illuminate the grating in a spectrometer at normal incidence θi=0° with a beam of light that has a wavelength of 6562.8 Å. T

monitta

Answer:

a) θ₁ = 23.14 ° , b) θ₂ = 51.81 °

Explanation:

An address network is described by the expression

d sin θ = m λ

Where is the distance between lines, λ is the wavelength and m is the order of the spectrum

The distance between one lines, we can find used a rule of proportions

d = 1/600

d = 1.67 10⁻³ mm

d = 1-67 10⁻³ m

Let's calculate the angle

sin θ = m λ / d

θ = sin⁻¹ (m λ / d)

First order

θ₁ = sin⁻¹ (1 6.5628 10⁻⁷ / 1.67 10⁻⁶)

θ₁ = sin⁻¹ (3.93 10⁻¹)

θ₁ = 23.14 °

Second order

θ₂ = sin⁻¹ (2 6.5628 10⁻⁷ / 1.67 10⁻⁶)

θ₂ = sin⁻¹ (0.786)

θ₂ = 51.81 °

3 0

3 years ago

.Find the uncertainty in a calculated electrical potential difference from the measurements of current and resistance. Electric

maw [93]

Answer:

a) The uncertainty in calculated V, ΔV = 25.3

b) The uncertainty in calculated v, Δv = 0.41 m/s

c) The uncertainty in calculated V, ΔV = 22.2 V

Explanation:

We'll use Upper-Lower Bounds method of uncertainty to estimate the uncertainties.

a) I = 5.1 A, ΔI = 0.3 A

I = (5.1 ± 0.3) A

R = 77.5 ohms, ΔR = 0.4 ohms

R = (77.5 ± 0.4) ohms

V = IR = 5.1 × 77.5 = 395.25 V

The lower bound for the voltage will be calculated using the lower bounds for the current and resistance

Iₗ = 5.1 - 0.3 = 4.8 A

Rₗ = 77.5 - 0.4 = 77.1 ohms

Vₗ = 4.8 × 77.1 = 370.08 V

The upper bound for the voltage will be calculated using the upper bounds for the current and resistance

Iᵤ = 5.1 + 0.3 = 5.4 A

Rᵤ = 77.5 + 0.4 = 77.9 ohms

Vᵤ = 5.4 × 77.9 = 420.66 V

The average of the differences from the mean voltage/true value is 25.3 V

V = 395.25 V, Δ = 25.3V

V = (395.25 ± 25.3) V

b) x = 2.9 m, Δx = 0.3 m

x = (2.9 ± 0.3) m

t = 4.4 s, Δt = 1.8 s

t = (4.4 ± 1.8) ohms

v = x/t = 2.9/4.4 = 0.659 m/s

The lower bound for average speed will be calculated using the lower bounds for distance and upper bounds for time.

xₗ = 2.9 - 0.3 = 2.6 m

tᵤ = 4.4 + 1.8 = 6.2 s

vₗ = 2.6/6.2 = 0.419 m/s

The upper bound for the average speed will be calculated using the upper bound for the distance and lower bound for time

xᵤ = 2.9 + 0.3 = 3.2 m

tₗ = 4.4 - 1.8 = 2.6 s

vᵤ = 3.2/2.6 = 1.231 m/s

The average of the differences from the mean average speed/true value is 0.41 m/s

v = 0.659 m/s, Δv = 0.41 m/s

v = (0.659 ± 0.41) m/s

c) ) I = 9.8 A, ΔI = 0.5 A

I = (9.8 ± 0.5) A

R = 40.5 ohms, ΔR = 0.2 ohms

R = (40.5 ± 0.2) ohms

V = IR = 9.8 × 40.5 = 396.9 V

The lower bound for the voltage will be calculated using the lower bounds for the current and resistance

Iₗ = 9.8 - 0.5 = 9.3 A

Rₗ = 40.5 - 0.2 = 40.3 ohms

Vₗ = 9.3 × 40.3 = 374.79 V

The upper bound for the voltage will be calculated using the upper bounds for the current and resistance

Iᵤ = 9.8 + 0.5 = 10.3 A

Rᵤ = 40.5 + 0.2 = 40.7 ohms

Vᵤ = 10.3 × 40.7 = 419.21 V

The average of the differences from the mean voltage/true value is 22.2 V

V = 396.9 V, Δ = 22.2 V

V = (396.9 ± 22.2) V

7 0

3 years ago