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Mazyrski [523]
4 years ago
13

A piece of bismuth with a mass of 4.06 g 4.06 g gains 423 J 423 J of heat. If the specific heat of bismuth is 0.123 J / ( g ° C

) 0.123 J/(g°C) , what is the change in temperature of the sample?
Physics
1 answer:
Sholpan [36]4 years ago
4 0

Answer: 846°C

Explanation:

The quantity of Heat Energy (Q) required to heat bismuth depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Given that:

Q = 423 joules

Mass of bismuth = 4.06g

C = 0.123 J/(g°C)

Φ = ?

Then, Q = MCΦ

423 J = 4.06g x 0.123 J/(g°C) x Φ

423 J = 0.5J/°C x Φ

Φ = (423J/ 0.5g°C)

Φ = 846°C

Thus, the change in temperature of the sample is 846°C

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A man is flying in a hot-air balloon in a straight line at a constant rate of 5 feet per second, while keeping it at a constant
icang [17]

Answer:

x = 220.85 ft

Explanation:

Let at any moment of time the friend's car is at some horizontal distance "x" from the position of balloon.

Now if the altitude of the balloon is fixed and it is at height "h"

so here we will have

tan \theta = \frac{h}{x}

now we know that

initially the angle of the friend's car is 35 degree

so the horizontal distance will be

x_1 = h cot35

similarly if the angle after passing the car position is 36 degree

then we have

x_2 = h cot36

now the speed of the balloon is constant

so we have

v = \frac{x_1 + x_2}{\Delta t}

5 ft/s = \frac{h cot35 + h cot36}{90 s}

5 ft/s = \frac{2.8h}{90}

h = 160.45 ft

so the final position of friend when the angle is 36 degree

x = \frac{h}{tan36}

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Read 2 more answers
A 1400.0 kg car crests a 3200.0 m pass in the mountains and briefly comes to rest. The car descends 1000 m before climbing and c
rjkz [21]

Answer:

a) v = 88.54 m/s

b) vf = 26.4 m/s

Explanation:

Given that;

m = 1400.0 kg

a)

by using the energy conservation

loss in potential energy is equal to gain in kinetic energy

mg × ( 3200-2800) = 1/2 ×m×v²

so

1400 × 9.8 × 400 = 0.5 × 1400 × v²

5488000 = 700v²

v² = 5488000 / 700

v² = 7840

v = √7840

v = 88.54 m/s

b)

Work done by all forces is equal to change in KE

W_gravity + W_non - conservative = 1/2×m×(vf² - vi²)

we substitute

1400 × 9.8 × ( 3200-2800) - (5 × 10⁶) = 1/2 × 1400 × (vf²  -0 )

488000 = 700 vf²

vf² = 488000 / 700

vf² = 697.1428

vf = √697.1428

vf = 26.4 m/s

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3 years ago
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