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kenny6666 [7]
3 years ago
8

The amount of energy that must be absorbed or lost to raise or lower the temperature of 1 g of liquid water by 1°c _____.

Physics
1 answer:
Fiesta28 [93]3 years ago
6 0

Answer:

4.2 J

Explanation:

Specific heat capacity: This is defined as the amount of a heat required to rise a unit mass of a substance through a temperature of 1 K

From specific heat capacity,

Q = cmΔt.............................. Equation 1

Where Q = amount of energy absorbed or lost, c = specific heat capacity of water, m = mass of water, Δt = Temperature rise.

Given: m = 1 g = 0.001 kg, Δt = 1 °C

Constant : c = 4200 J/kg.°C

Substitute into equation 1

Q = 0.001×4200(1)

Q = 4.2 J.

Hence the energy absorbed or lost = 4.2 J

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WARRIOR [948]

The frequency of the light observed from the Earth is 4.945\cdot 10^{14}Hz

Explanation:

First of all, we start by noticing that the galaxy is receding from Earth (moving away): this means that according to the Doppler effect, the frequency of the light as seen from the Earth must be shorter than the real frequency of the light emitted by the galaxy.

Furthermore, we can quantify the change in frequency of the light using the following equation:

\frac{\Delta f}{f}=\frac{v}{c}

where

\Delta f is the change in frequency

f is the real frequency

v is the velocity of recession of the galaxy (negative if the galaxy is moving away)

c is the speed of light

In this problem, we have:

f=5.00\cdot 10^{14} Hz

v=-3325 km/s = -3.325\cdot 10^6 m/s

c=3\cdot 10^8 m/s

Substituting and solving for \Delta f, we find

\Delta f = \frac{v}{c}f=\frac{-3.325\cdot 10^6}{3\cdot 10^8}(5.00\cdot 10^{14})=-5.542\cdot 10^{12} Hz

And therefore, the frequency of the light observed from the Earth is

f'=f+\Delta f = 5.00\cdot 10^{14} +(-5.52\cdot 10^{12})=4.945\cdot 10^{14}Hz

Learn more about frequency and waves:

brainly.com/question/5354733

brainly.com/question/9077368

#LearnwithBrainly

4 0
3 years ago
Find the work done in pumping gasoline that weighs 6600 newtons per cubic meter. A cylindrical gasoline tank 3 meters in diamete
Sonbull [250]

Answer:

<em>work done in pumping the entire fuel is 466587 J</em>

<em></em>

Explanation:

weight of the gasoline per volume = 6600 N/m^3

diameter of the tank = 3 m

length of the tank = 2 m

height of the tractor tank above the top of the tank = 5 m

work done in pumping fuel to this height = ?

First, we find the volume of the fuel

since the tank is cylindrical,<em> we assume that the fuel within also takes the cylindrical shape.</em>

<em>Also, we assume that the fuel completely fills the tank.</em>

volume of a cylinder = \pi r^{2}l

where r = radius = diameter ÷ 2 = 3/2 = 1.5 m

volume of the cylinder = 3.142 x 1.5^{2} x 2 = 14.139 m^3

we then find the total weight of the fuel in Newton

total weight = (weight per volume) x volume

total weight = 6600 x 14.139 = 93317.4 N

work done = (total weight of the fuel) x (height through which the fuel is pumped)

work done in pumping = 93317.4 x 5 = <em>466587 J</em>

8 0
2 years ago
A 100-n object and a 50-n object are placed on scales a and b respectively inside an elevator ascending with constant velocity
german

Answer: b

Explanation:

8 0
2 years ago
The number of wave troughs that pass by every second is a wave's _________.
professor190 [17]
A waves frequency (in Hertz) is how many crests pass by a point per second. easily confused with period, which is the amount of time it takes for a full wave to pass by a certain point 
6 0
3 years ago
An 85-kg man plans to tow a 109 000-kg airplane along a runway by pulling horizontally on a cable attached to it. Suppose that h
Lelu [443]

Answer:

The greatest acceleration the man can give the airplane is 0.0059 m/s².

Explanation:

Given that,

Mass of man = 85 kg

Mass of airplane = 109000 kg

Distance = 9.08

Coefficient of static friction = 0.77

We need to calculate the greatest friction force

Using formula of friction

F=\mu mg

Where, m = mass of man

g = acceleration due to gravity

Put the value into the formula

F = 0.77\times85\times9.8

F= 641.41\ N

We need to calculate the acceleration

Using formula of newton's second law

F = ma

a=\dfrac{F}{m}

Put the value into the formula

a=\dfrac{ 641.41}{109000}

a=0.0059\ m/s^2

Hence, The greatest acceleration the man can give the airplane is 0.0059 m/s².

3 0
3 years ago
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