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Luda [366]
2 years ago
7

Calculate the calories lost when 95 g of water cools from 45 ∘C to 29 ∘C. Express your answer to two significant figures and inc

lude the appropriate units. Use abbreviations for the units.
Physics
1 answer:
lubasha [3.4K]2 years ago
8 0

Answer:

1,520.00 calories

Explanation:

Water molecules are linked by hydrogen bonds that require a lot of heat (energy) to break, which is released when the temperature drops. That energy is called specific heat or thermal capacity (ĉ) when it is enough to change the temperature of 1g of the substance (in this case water) by 1°C. Water ĉ equals 1 cal/(g.°C).

Given that ĉ = Q / (m.ΔT),

where Q= calories transferred between the system and its environment or another system (unity: calorie or cal) (what we are trying to find out),

m= mass of the substance (unity: grams or g), and

ΔT= difference of temperature (unity: Celsius degrees or °C); and

m= 95g and ΔT= 16°C:

Q= 1 cal/(g.°C).95g.16°C =<u> 1,520.00 cal </u>

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melamori03 [73]
If you subscribe I’ll answer QF Aotrx
8 0
2 years ago
In the sport of parasailing, a person is attached to a rope being pulled by a boat while hanging from a parachute-like sail. A r
scoray [572]

Answer:

W = 506.75 N

Explanation:

tension = 2300 N

Rider is towed at a constant speed means there no net force acting on the rider.

hence taking all the horizontal force and vertical force in consideration.

net horizontal  force:

F cos 30° - T cos 19° = 0

F cos 30° = 2300 × cos 19°

F = 2511.12 N

net vertical force:

F sin 30° - T sin 19°- W = 0

W = F sin 30° - T sin 19°

W =  2511.12 sin 30° - 2300 sin 19°

W = 506.75 N

8 0
2 years ago
A solid sphere of brass (bulk modulus of 14.0 ✕ 1010 N/m2) with a diameter of 2.20 m is thrown into the ocean. By how much does
astra-53 [7]

Answer:

Diameter decreases by the diameter of 0.0312 m.

Explanation:

Given that,

Bulk modulus =  14.0 × 10¹⁰ N/m²

Diameter d = 2.20 m

Depth = 2.40 km

Pressure = ρ g h = 1030 × 9.81 × 2.4 × 1000

               = 24.25 × 10⁶  N/m²

Volume = \dfrac{4}{3} \pi r^3

         \dfrac{\Delta V}{V}=\dfrac{(\Delta r)^3}{r^3}

Bulk modulus is equal to

B = -\dfrac{\Delta P}{\dfrac{\Delta V}{V} }

now

B = -\dfrac{24.25 \times 10^6}{\dfrac{(\Delta r)^3}{r^3} }

B = -\dfrac{24.25 \times 10^6}{\dfrac{(\Delta r)^3}{1.1^3} }

(\Delta r)^3 = \dfrac{24.25 \times 10^6 \times 1.1^3}{14.0 \times 10^{10}}

Δ r = -0.0156 m

change in diameter

Δ d = -2 × 0.0156

Δ d = -0.0312 m

Diameter decreases by the diameter of 0.0312 m.

7 0
2 years ago
13. An aircraft heads North at 320 km/h rel:
AURORKA [14]

The velocity of the aircraft relative to the ground is 240 km/h North

Explanation:

We can solve this problem by using vector addition. In fact, the velocity of the aircraft relative to the ground is the (vector) sum between the velocity of the aircraft relative to the air and the velocity of the air relative to the ground.

Mathematically:

v' = v + v_a

where

v' is the velocity of the aircraft relative to the ground

v is the velocity of the aircraft relative to the air

v_a is the velocity of the air relative to the ground.

Taking north as positive direction, we have:

v = +320 km/h

v_a = -80 km/h (since the air is moving from North)

Therefore, we find

v'=+320 + (-80) = +240 km/h (north)

Learn more about vector addition:

brainly.com/question/4945130

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7 0
2 years ago
A wire with a length of 150 m and a radius of 0.15 mm carries a current with a uniform current density of 2.8 x 10^7A/m^2. The c
Mrac [35]

Answer:

The current is 2.0 A.

(A) is correct option.

Explanation:

Given that,

Length = 150 m

Radius = 0.15 mm

Current densityJ=2.8\times10^{7}\ A/m^2

We need to calculate the current

Using formula of current density

J = \dfrac{I}{A}

I=J\timesA

Where, J = current density

A = area

I = current

Put the value into the formula

I=2.8\times10^{7}\times\pi\times(0.15\times10^{-3})^2

I=1.97=2.0\ A

Hence, The current is 2.0 A.

7 0
2 years ago
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