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Len [333]
3 years ago
9

If energy is transferred spontaneously as heat from a substance with a temperature of T1 to a substance with a temperature of T2

, which of the following statements must be true?
1-T1 < T2

2-T1 = T2

3-T1 > T2

4-more information is needed
Physics
1 answer:
LuckyWell [14K]3 years ago
6 0

Answer: The statement T_{1} > T_{2} must be true.

Explanation:

As it is given that heat is being transferred from a substance with temperature T_{1} to a substance with temperature T_{2}.

It is known that heat will always being transferred from a higher temperature to a lower temperature. Because at higher temperature the molecules of a substance acquire more energy and when they lose energy then a decrease in temperature occurs.

Hence, in the given situation T_{1} > T_{2}.

Thus, we can conclude that the statement T_{1} > T_{2} must be true.

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A toroid with a square cross section 3.0 cm ✕ 3.0 cm has an inner radius of 25.1 cm. It is wound with 600 turns of wire, and it
coldgirl [10]

Answer:

B = 1.353 x 10⁻³ T

Explanation:

The Magnetic field within a toroid is given by

B = μ₀ NI/2πr, where N is the number of turns of the wire, μ₀ is the permeability of free space, I is the current in each turn and r is the distance at which the magnetic field is to be determined from the center of the toroid.

To find r we need to add the inner radius and outer radius and divide the value by 2. Hence,

r = (a + b)/2, where a is the inner radius and b is the outer radius which can be found by adding the length of a square section to the inner radius.

b = 25.1 + 3 = 28.1 cm

a = 25.1 cm

r = (25.1 + 28.1)/2 = 26.6 cm = 0.266m

B = 4π x 10⁻⁷ x 600 x 3/2π x 0.266

B = 1.353 x 10⁻³ T

The strength of the magnetic field at the center of the square cross section is 1.3 x 10⁻³ T

5 0
3 years ago
A scuba diver's tank contains 0.240 kg of o2 compressed into a volume of 3.10 l. what volume (in liters) would this oxygen occup
ohaa [14]
Given:
m = 0.240 kg = 240 g, the mass of O₂
V = 3.10 L = 3.10 x 10⁻³ m³, the volume

Because the molar mass of oxygen is 16, the number of moles of O₂ is
n = (240 g)/(2*16 g/mol) =  7.5 mol

As an ideal gas,
p*V = nRT
or
V = (nRT)/p
where R = 8.314 J/(mol-K)

When
p = 0.910 atm = (0.910 atm) * (101325Pa/atm) = 92205.75 Pa
T = 27 °C = (27 + 273) K = 300 K
then the volume is
V= \frac{(7.5 \, mol)*(8.314 \, J/(mol-K))*(300 \, K)}{(92205.75 \, Pa)} =0.2029 \, m^{3}

V = (0.2029 m³)*(10³ L/m³) = 202.9 L

Answer: 203 liters
3 0
3 years ago
A 5000 kg open train car is rolling on frictionless rails at 22 m/s when it starts pouring rain. A few minutes later, the car’s
Phantasy [73]

Answer:

500 kg

Explanation:

It is given that,

The mass of a open train car, M = 5000 kg

Speed of open train car, V = 22 m/s

A few minutes later, the car’s speed is 20 m/s

We need to find the mass of water collected in the car. It is based on the conservation of momentum as follows :

initial momentum = final momentum

Let m is final mass

MV=mv

m=\dfrac{MV}{v}\\\\m=\dfrac{5000\times 22}{20}\\\\=5500\ kg

Water collected = After mass of train - before mass of train

= 5500 - 5000

= 500 kg

So, 500 kg of water has collected in the car.

3 0
3 years ago
Wire sizes are often reported using the AWG (American Wire Gauge) system in which smaller diameter wires are said to be of highe
NISA [10]

Answer:

Explanation:

24 - gauge wire , diameter = .51 mm .

Resistivity of copper ρ = 1.72 x 10⁻⁸ ohm-m

R = ρ l / s

1.72x 10⁻⁸ / [3.14 x( .51/2)² x 10⁻⁶ ]

= 8.42 x 10⁻² ohm

= .084 ohm

B )  Current required through this wire

= 12 / .084 A

= 142.85 A

C )

Let required length be l

resistance = .084 l

2 = 12 / .084 l

l = 12 / (2 x .084)

= 71.42 m

6 0
3 years ago
Read 2 more answers
Two point charges 3q and −8q (with q &gt; 0) are at x = 0 and x = L, respectively, and free to move. A third charge is placed so
riadik2000 [5.3K]

Answer:

Explanation:

The unknown charge can not remain in between the charge given because force on the middle charge will act in the same direction due to both the remaining charges.

So the unknown charge is somewhere on negative side of x axis . Its charge will be negative . Let it be - Q and let it be at distance - x on x axis.

force on it due to rest of the charges will be equal and opposite so

k3q Q / x² =k 8q Q / (L+x)²

8x² = 3 (L+x)²

2√2 x = √3 (L+x)

2√2 x - √3 x = √3 L

x(2√2 - √3 ) = √3 L

x = √3 L / (2√2 - √3 )

Let us consider the balancing force on 3q

force on it due to -Q and -8q will be equal

kQ . 3q / x² = k3q  8q / L²

Q = 8q  (x² / L²)

so charge required = - 8q  (x² / L²)

and its distance from x on negative x side = √3 L / (2√2 - √3 )

3 0
3 years ago
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