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mylen [45]
3 years ago
11

How much heat do you need to raise the temperature of 150 g of oxygen from -30c to -15c?

Physics
2 answers:
Ludmilka [50]3 years ago
6 0

Answer:

2.1 kJ

Explanation:

The heat (q) required to raise the temperature of oxygen can be calculated using the following expression.

q = c × m × ΔT

where,

c: specific heat capacity (c(O₂): 0.913 J/g.°C)

m: mass

ΔT: change in the temperature

q = c × m × ΔT

q = (0.913 J/g.°C) × 150 g × [-15°C - (-30°C)] = 2.1 × 10³ J = 2.1 kJ

Naddika [18.5K]3 years ago
4 0
The amount of heat needed to raise the temperature of a substance by \Delta T is given by
Q=m C_s \Delta T
where
m is the mass of the substance
Cs is its specific heat capacity
\Delta T is the increase in temperature

For oxygen, the specific heat capacity is approximately 
C_s = 0.92 J/(g K)
The variation of temperature for the sample in our problem is 
\Delta T= -15^{\circ}C-(-30^{\circ} C)=+15^{\circ}C=15 K
while the mass is m=150 g, so the amount of heat needed is
Q=m C_s \Delta T=(150 g)(0.92 J/g K)(15 K)=2070 J
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Chứng minh V=kq/r từ mối liên hệ giữ E và V
Marat540 [252]

Answer:V=Aq=KQr

Explanation:

Không biết V bạn kí hiệu ở đây là gì nhỉ? Có phải là điện thế?

Điện thế tại 1 điểm trong điện trường được định nghĩa là công làm vật dịch chuyển từ vị trí đó đến vô cùng. V = A/q

Chứng minh thì được, nhưng chỉ e bạn không có hiểu biết về nguyên hàm, tích phân nên không hiểu.

- Xét tại vị trí cách điện tích Q một đoạn x, khi đó điện tích q sẽ chịu 1 lực: dF=KQ.qx2

Điện tích q dịch chuyển 1 khoảng dx rất nhỏ. Khi đó công do lực điện trường gây ra là:

dA=dF.dx=KQ.qx2dx

Công để dịch chuyển điện tích q từ vị trí r đến vô cùng là:

A=∫∞rdA=∫∞rKQ.qx2dx=KQ.qr−KQ.q∞=KQ.qr

Theo đúng định nghĩa: V=Aq=KQr

4 0
3 years ago
A ball is launched with initial speed v from the ground level up a frictionless hill. The hill becomes steeper as the ball slide
Triss [41]

Answer:

H(max) = (v²/2g)

Explanation:

The maximum height the ball will climb will be when there is no friction at all on the surface of the hill.

Normally, the conservation of kinetic energy (specifically, the work-energy theorem) states that, the change in kinetic energy of a body between two points is equal to the work done in moving the body between the two points.

With no frictional force to do work, all of the initial kinetic emergy is used to climb to the maximum height.

ΔK.E = W

ΔK.E = (final kinetic energy) - (initial kinetic energy)

Final kinetic energy = 0 J, (since the body comes to rest at the height reached)

Initial kinetic energy = (1/2)(m)(v²)

Workdone in moving the body up to the height is done by gravity

W = - mgH

ΔK.E = W

0 - (1/2)(m)(v²) = - mgH

mgH = mv²/2

gH = v²/2

H = v²/2g.

7 0
3 years ago
A horizontal uniform board of weight 125N and length 4 m is supported by vertical chains at each end. A person weighing 500N sit
Misha Larkins [42]

Let's apply an equation of equilibrium to the situation: The sum of the moments about the left end of the board must equal 0.

We have three moments to add. Positive force values indicate upward direction and negative values indicate downward direction. All distances given below are measured to the right side of the left end of the board:

  1. The weight of the board, -125N, located at 2m (center of the board due to its uniform density)
  2. The tension in the right chain, +250N, located at 4m
  3. The weight of the person, -500N, located at a distance "x"

The sum of the moments must equal 0 and is given by:

ΣFx = 0

F is the magnitude of force, x = distance from the left end of the board

Plug in all of the force and distance values and solve for x:

ΣFx = 250(4) - 125(2) - 500x = 0

500x = 750

x = 1.5m

7 0
3 years ago
How does the antenna work to detect the electromagnetic signal produced when radio stations broadcast
galina1969 [7]

Answer:

The antenna which is a transmitting and receiving device emits energy from current as radio waves, it does this by attracting the radio waves which are a form of EMWs and converts it to small voltages which are amplified to the final voltage signal which hear

3 0
3 years ago
Please help with physics 2 part question giving 50 points
AveGali [126]

Answer:

2.88×10⁻⁹ s

2.40×10¹⁵ m/s²

Explanation:

Given:

v₀ = 12300 m/s

v = 6.92×10⁶ m/s

Δx = 0.997 cm = 0.00997 m

Part 1) Find: t

Δx = ½ (v + v₀)t

0.00997 m = ½ (6.92×10⁶ m/s + 12300 m/s) t

t = 2.88×10⁻⁹ s

Part 2) Find: a

(6.92×10⁶ m/s)² = (12300 m/s)² + 2a (0.00997 m)

a = 2.40×10¹⁵ m/s²

5 0
3 years ago
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