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2 years ago

How much heat energy is required to raise the temperature of 1 kilogram of steel by 10°C?

Physics

1 answer:

sammy [17]2 years ago

7 0

from

∆H=MC∆T

but M=1Kg=1000g

∆T=10°C

C=0.466J/g°C

∆H=(1000×0.466×10)Joule

=4660J

explanation

When heat energy is added to a substance, the temperature will change by a certain amount. The relationship between heat energy and temperature is different for every material, and the specific heat is a value that describes how they relate.

heat energy = (mass of substance)(specific heat)(change in temperature)

Q = mc∆T

Q = heat energy (Joules, J)

m = mass of a substance (kg)

c = specific heat (units J/kg∙K)

∆ is a symbol meaning "the change in"

∆T = change in temperature (Kelvins, K

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Position c

Vinil7 [7]

Answer:

B porque tiene energía n..

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3 years ago

A 4 kg car with frictionless wheels is on a ramp that makes a 25° angle with the horizontal. The car is connected to a 1 kg mass

stellarik [79]

Refer to the diagram shown below.

W₁ = (4 kg)*(9.8 m/s²) = 39.2 N
W₂ = (1 kg)*(9.8 m/s²) = 9.8 N

The normal reaction on the 4-kg mass is
N = (39.2 N)*cos(25°) = 35.5273 N
The force acting down the inclined plane due to the weight is
F = (39.2 N)*sin(25°) = 16.5666 N

The net force that accelerates the 4-kg mass at a m/²s down the plane is
F - W₂ = (4 kg)*(a m/s²)
4a = 16.5666 - 9.8
a = 1.6917 m/s²

Answer: 1.69 m/s² (nearest hundredth)

7 0

3 years ago

Select the correct answer.

Murrr4er [49]

Answer:

I will choose the answer A

3 0

3 years ago

Imagine an alternate universe where the value of the Planck constant is . In that universe, which of the following objects would

HACTEHA [7]

Question: The planck constant was not given. In this calculation, planck constant of 6.62607*10^-9 Js is used for the calculation.

Answer:

(a) A virus -------------Classical

(b) A buckyball -----Classical

(d) A turtle ------------Quantum

Explanation:

Calculating the wavelength using the formula;

λ= h/(mv)

where

λ= Wavelength

h = Planck Constant = 6.62607*10^-9 Js

m = mass in kg

v = velocity in m/s

Virus size = 280. nm = 2.80*10⁻⁷ m

A Virus:

m = 9.4 x 10-17 g 9.4*10⁻²⁰ kg

v = 0.50 µm/s = 5 *10⁻⁷ m/s

h = 6.62607*10^-9 Js

Virus size = 280 nm = 2.80*10⁻⁷ m

Substituting into the formula; we have

λ= h/(mv)

λ= 6.62607*10^-9/ (9.4*10⁻²⁰* 5 *10⁻⁷)

= 6.62607*10^-9/4.7*10^-26

= 1.4*10^17 m

Classical : Wavelength is bigger than it's size

(b)

A buckyball

m = 1.2 x 10-21 g = 1.2 *10⁻²⁴ kg

V = 37 m/s

Size = 0.7 nm = 7*10⁻¹⁰ m

Substituting into the formula, we have

λ= h/(mv)

λ= 6.62607*10^-9/ ( 1.2 *10⁻²⁴* 37)

= 6.62607*10^-9/4.44*10^-23

= 1.49 *10^14 m

Classical : Wavelength is bigger than it's size

(c)

A mosquito

Mass = 1.0 mg = 1*10⁻⁶ kg

v = 1.1 m/s

Size = 6.3 mm = 6.3*10⁻³ m

Substituting into the formula, we have

λ= h/(mv)

λ= 6.62607*10^-9/ ( 1*10⁻⁶* 1.1)

= 6.62607*10^-9/1.1*10^-6

= 6.02*10^-3 m

Quantum Approach: The wavelength and the size are comparable

(d)

A turtle

Mass = 710. g = 0.71 kg

Size = 22. cm = 0.22 m

V = 2.8 cm/s. = 0.028 m/s

Substituting into the formula, we have

λ= h/(mv)

λ= 6.62607*10^-9/ ( 0.71* 0.028)

= 6.62607*10^-9/0.01988

= 3.33*10^-7 m

Quantum Approach: The wavelength and the size are comparable

8 0

3 years ago

The acceleration due to gravity is lower on the Moon than on Earth. Which of the following is true about the mass and weight of

Naya [18.7K]

Mass is the same, weight is less

<h3>What is the Weight and mass on Moon ?</h3>

As we know that the mass of the object is the measurement of the quantity of the matter that is present in it

So here we can say that if the mass of the object is m then its total quantity of the matter that is present in it is given as

mass = (density) × (volume)

Now for the weight of the object is defined as the force of gravity due to planet

Fg = mg

so the weight of the object is depending on the acceleration due to gravity of the planet

As we know that the gravity of moon is smaller than the gravity of the earth so here weight on the moon will be smaller than the weight on the Earth

Learn more about Weight on Moon here:

brainly.com/question/4080619

#SPJ4

3 0

2 years ago