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maks197457 [2]
2 years ago
7

The electrical resistance of an element in a platinum resistance thermometer at 100°c, 0°c and room temperature are 75.00Ω, 63.0

0Ω and 64.992Ω respectively. Use the data to calculate room temperature
Physics
1 answer:
Dominik [7]2 years ago
6 0

Answer:

16.6 °C

Explanation:

From the question given above, the following data were obtained:

Temperature at upper fixed point (Tᵤ) = 100 °C

Resistance at upper fixed point (Rᵤ) = 75 Ω

Temperature at lower fixed point (Tₗ) = 0 °C

Resistance at lower fixed point (Rₗ) = 63.00Ω

Resistance at room temperature (R) = 64.992 Ω

Room temperature (T) =?

T – Tₗ / Tᵤ – Tₗ = R – Rₗ / Rᵤ – Rₗ

T – 0 / 100 – 0 = 64.992 – 63  / 75 – 63

T / 100 = 1.992 / 12

Cross multiply

T × 12 = 100 × 1.992

T × 12 = 199.2

Divide both side by 12

T = 199.2 / 12

T = 16.6 °C

Thus, the room temperature is 16.6 °C

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A) Show that the surface temperature of a star can be inferred from measurements of blackbodyfluxes at two different frequencies
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The answers to the questions have been solved in the attachment.

Explanation:

The answers to part a to e are all contained in the attachment. For answer part b, temperature and frequency were assumed to be fixed or constant. V² is directly proportional to T telling us that variation in T gives us a square in the frequency variation. This tells us why it is difficult when both frequencies are on this side of the black body.

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3 years ago
A wheel rotating about a fixed axis has a constant angular acceleration of 4.0 rad/s2. In a 4.0-s interval the wheel turns throu
anastassius [24]

Answer:

a. 3 s.

Explanation:

Given;

angular acceleration of the wheel, α = 4 rad/s²

time of wheel rotation, t = 4 s

angle of rotation, θ = 80 radians

Apply the kinematic equation below,

\theta = \omega_1 t \ + \ \frac{1}{2} \alpha t^2\\\\80 = 4\omega_1 + \frac{1}{2}*4*4^2\\\\80 = 4\omega_1 + 32\\\\ 4\omega_1 = 48\\\\ \omega_1 = \frac{48}{4}\\\\ \omega_1 = 12 \ rad/s

Given initial angular velocity, ω₀ = 0

Apply the kinematic equation below;

\omega_1 = \omega_o + \alpha t_1\\\\12 = 0 + 4t\\\\4t = 12\\\\t = \frac{12}{4}\\\\t = 3 \ s

Therefore, the wheel had been in motion for 3 seconds.

a. 3 s.

8 0
2 years ago
A uniform, solid sphere of radius 2.50 cm and mass 4.75 kg starts with a purely translational speed of 3.00 m/s at the top of an
allsm [11]

Answer:

The final translational seed at the bottom of the ramp is approximately 4.84 m/s

Explanation:

The given parameters are;

The radius of the sphere, R = 2.50 cm

The mass of the sphere, m = 4.75 kg

The translational speed at the top of the inclined plane, v = 3.00 m/s

The length of the inclined plane, l = 2.75 m

The angle at which the plane is tilted, θ = 22.0°

We have;

K_i + U_i = K_f + U_f

K = (1/2)×m×v²×(1 + I/(m·r²))

I = (2/5)·m·r²

K =  (1/2)×m×v²×(1 + 2/5) = 7/10 × m×v²

U = m·g·h

h = l×sin(θ)

h = 2.75×sin(22.0°)

∴ 7/10×4.75×3.00² + 4.75×9.81×2.75×sin(22.0°) = 7/10 × 4.75×v_f² + 0

7/10×4.75×3.00² + 4.75×9.81×2.75×sin(22.0°) ≈ 77.93

∴ 77.93 ≈ 7/10 × 4.75×v_f²

v_f² = 77.93/(7/10 × 4.75)

v_f ≈ √(77.93/(7/10 × 4.75)) ≈ 4.84

The final translational seed at the bottom of the ramp, v_f ≈ 4.84 m/s.

3 0
2 years ago
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