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ziro4ka [17]
3 years ago
7

A kettle heats 1.75 kg of water. The specific latent heat of vaporisation of water is 3.34 x 106 J/kg. How much energy would be

needed to boil off the water? Write the answer out in full (i.e. not in standard form).
Physics
1 answer:
labwork [276]3 years ago
7 0

The amount of energy needed to boil off the water is 5,845,000 J

Explanation:

The amount of thermal energy needed to completely boil a certain amount of a liquid substance already at boiling temperature is given by

Q=m\lambda_v

where

m is the amount of the substance

\lambda_v is the specific latent heat of vaporization of the substance

In this problem, the substance is water, and its mass is

m = 1.75 kg

Its specific latent heat of vaporization is

\lambda_v = 3.34\cdot 10^6 J/kg

Assuming that the water is already at boiling temperature, therefore, the thermal energy needed to boil off the water is

Q=(1.75)(3.34\cdot 10^6)=5,845,000 J

Learn more about specific heat:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

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What is the magnitude of the force that is exerted on a 20 kg mass to give it an acceleration of 10.0
ANEK [815]

Answer:Mass of the body = 20 kg.

Final Velocity = 5.8 m/s.

Initial velocity = 0

Time = 3 seconds.  

Using the Formula,  

Acceleration = (v - u)/ t

= (5.8 - 0)/ 3

= 1.6 m /s².

Now, Using the Formula,  

Force = mass × acceleration

= 20 × 1.6

=

Explanation: I REALLY  HOPE THIS HELPS I'M KINDA NEW AT THIS :] :]

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3 years ago
How many seconds of space should you keep between you if you're driving a 100 foot truck 30 mph?
Marrrta [24]

Answer:

<em> The space in seconds that will be kept = 2.27 seconds</em>

Explanation:

S = d/t..................... Equation 1

making t the subject of formula in the equation above,

t = d/S.................... Equation 2

Where S = speed, d = distance, t = time.

<em>Conversion: (i)if 1 mph = 0.44704 m/s,</em>

<em>                 then, 30 mph = 30×0.44704    </em>

<em>                = 13.41 m/s</em>

<em>               (ii) If 1 foot = 0.3048 m</em>

<em>            then, 100 foot = 30.48 m.</em>

<em>Given: S = 30 mph = 13.41 m/s, d = 100 foot = 30.48 m</em>

<em>Substituting these values into equation 2</em>

<em>t = 30.48/13.41</em>

<em>t = 2.27 seconds.</em>

<em>Therefore the space in seconds that will be kept = 2.27 seconds</em>

7 0
3 years ago
How do you find a controlled variable
adelina 88 [10]

Hello there!


Essentially, a control variable is what is kept the same throughout the experiment, and it is not of primary concern in the experimental outcome. Any change in a control variable in an experiment would invalidate the correlation of dependent variables (DV) to the independent variable (IV), thus skewing the results.

7 0
3 years ago
vector u has a magnitude of 20 and direction of 0°.vector v has amagnitude of 40and a direction of 60°.find the magnitude and di
pantera1 [17]

Addition of vectors:

vector u

has a magnitude of 20 and a direction of 0º with respect to the horizontal, vector v has a magnitude of 40 and a direction of 60º with respect to the horizontal.

a) Find the magnitude and direction of the resultant to the nearest whole

number.

Vector Sum:

The resultant of two vectors is simply the vector sum of the vectors. There are a handful of ways to present the resultant factor; the notation that shows the vector magnitude and direction is called the polar vector notation. An example of a vector presented in polar vector notation is

a∠θ where a is the magnitude and θis the angle that the vector makes with the horizontal axis.

Answer and Explanation:

Let's first present the vectors in rectangular vector notation.

For the vector →u of magnitude 20 and direction 0∘ to the horizontal axis, the vector is →u=^i20.

For the vector →v

of magnitude 40 and direction 60∘ to the horizontal axis, the vector is →v=^i40cos60∘+^j40sin60∘.

The resultant vector →w is the vector sum of the vectors, i.e.

→w=→u+→v

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=^i(20+40cos60∘)+^j(40sin60∘)

=^i40+^j20√3

For a vector ^ix+^jy, the magnitude of the vector is √x2+y2 and the direction above the horizontal axis is θ=tan−1(yx).

Let's use the formulas:

|→w|∠θ=√(40)2+(20√3)2∠tan−1(20√340)≈52.9∠40.9∘

The magnitude of the vector is about 53 units in the direction 41-degrees above the horizontal axis.

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How to tell how much work gravity does on something?
Grace [21]
By dropping a ball and seeing how long it takes to hit the ground or throw a ball up and time it as well
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