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3 years ago

It requires 350 joules to raise a certain amount of a substance from 10.0°C to 30.0°

Physics

2 answers:

Hitman42 [59]3 years ago

5 0

The answer is 15g, you would use the specific heat formaula

malfutka [58]3 years ago

3 0

Answer:

$m=14.6g$

Explanation:

You can use the equation $Q=cm(T_{2}-T_{1})$ where Q is the heat, c is the specific heat of the substance, $T_{2}$ and $T_{1}$ are temperature and m is the mass of the substance.

Solving for m we have:

$m=\frac{Q}{c(T_{2}-T_{1})}$

The problem gives you the information about the variables:

$Q=350J$

$c=1.2\frac{J}{g^{o}C}$

$T_{1}=10.0^{o}C$

$T_{2}=30.0^{o}C$

Replacing that values:

$m=\frac{350J}{1.2\frac{J}{g^{o}C}(30.0^{o}C-10.0^{o}C)}$

$m=\frac{350J}{1.2\frac{J}{g^{o}C}(20.0^{o}C)}$

$m=14.6g$

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A 5-m steel beam is lowered by means of two cables unwinding at the same speed from overhead cranes. As the beam approaches the

ArbitrLikvidat [17]

Answer:

a) The angular acceleration of the beam is 0.5 rad/s²CW (direction clockwise due the tangential acceleration is positive)

b) The acceleration of point A is 3.25 m/s²

The acceleration of point E is 0.75 m/s²

Explanation:

a) The relative acceleration of B with respect to D is equal:

$a_{B} =a_{D} +(a_{B/D} )_{n} +(a_{B/D} )_{t}$

Where

aB = absolute acceleration of point B = 2.5 j (m/s²)

aD = absolute acceleration of point D = 1.5 j (m/s²)

(aB/D)n = relative acceleration of point B respect to D (normal direction BD) = 0, no angular velocity of the beam

(aB/D)t = relative acceleration of point B respect to D (tangential direction BD)

$a_{B} =a_{D} +(a_{B/D} )_{t}$

$2.5j=1.5j +(a_{B/D} )_{t}\\(a_{B/D} )_{t}=j=1m/s^{2}$

We have that

(aB/D)t = BDα

Where α = acceleration of the beam

BDα = 1 m/s²

Where

BD = 2

$2\alpha =1\\\alpha =0.5rad/s^{2}CW$

b) The acceleration of point A is:

$a_{A} =a_{D} +(a_{A/D} )_{t}$

(aA/D)t = ADαj

$a_{A} =a_{D} +AD\alpha j\\a_{A}=1.5j+(3.5*0.5)j\\a_{A}=3.25jm/s^{2}$

The acceleration of point E is:

(aE/D)t = -EDαj

$a_{E} =a_{D} -ED\alpha j\\a_{E}=1.5j-(1.5*0.5)j\\a_{E}=0.75jm/s^{2}$

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3 years ago

The theory of plate tectonics describes how the Earth’s lithosphere is broken into plates and these plates move over the molten

Mumz [18]

Answer:

Explanation:

A theory can be changed when new evidence is found. A law doesn't change because it is universally a fact. It doesn't need new evidence to support it.

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3 years ago

A rectangular pane of glass is 91.1 cm wide and 155.9 cm long, and its area is equal to the length multiplied by the width. Usin

Kamila [148]

911.1×155.9 = 14,202.49
14,202.49 in 3 significant figures would be either 14,200 or 1.42×10^4

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3 years ago

Read 2 more answers

The electric force on object A by object B is F. If the charge of object B were only half as large as it is, but everything else

nalin [4]

Answer:

The new electric force on object B by object A is one half of the magnitude of the original force F, in the opposite direction. Equals -F/2

Explanation:

We are asked to demonstrate what happens when you reduce the magnitude of the charge of an object in an interactive system.

Assumptions.

The new force is F'

The system is isolated from other external interferences

The distance between objects A and B remains unchanged

Analysis.

The force F on object A by object B is given by the formula

F=(k*q,a*q,b)/d^2

where k is the Coulomb's Law constant, and equals 9x10^9 N/m s

q,a and q,b equals the original charges of objects A and B

d is the distance between objects A and B

Now, we are to calculate the new force, let's call it F', and following the same reasoning.

F'=(k*q',a*q',b)/d'^2

Given that only q', b is different to the original, we have that

F'=(k*q,a*q',b)/d^2

Now, q',b=q,b/2 and replacing

F'=(k*q,a*q,b)/(2*d^2)

Now we find the relation between F' and F

F'/F=[(k*q,a*q,b)/(2*d^2)]/[(k*q,a*q,b)/d^2]

Reducing similar terms

F'/F=1/2

However, we should also consider that the forces F and F' are in oposite directions, so simplifying the vectorial notation,

F'/F=-1/2, meaning that F'=-F/2

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3 years ago

An electron in the n=7 level of the hydrogen atom relaxes to a lower energy level, emitting light of 1005 nm. Part A What is the

matrenka [14]

Answer:

rom level n = 7 to level n = 3

Explanation:

Bohr's model describes the energy levels for the hydrogen atom

En = -13.606 / n²

Where n is an integer with values of 1, 2, 3

An electronic transition occurs between two permitted levels of energy

ΔE = $E_{nf}$ - $E_{no}$

Let's apply these relationships our problem.

Let's start by knowing the energy of level n = 7

E₇ = - 13.606 / 7²

E₇ = - 0.27767 eV

Now let's see what the energy of the emitted photon

E = h f

c = λ f

f = c / λ

E = h c / λ

E = 6.63 10⁻³⁴ 3 10⁸/1005 10⁻⁹

E = 19,791 10⁻²⁰ J

Let's reduce to eV

E = 19,791 10⁻²⁰ (1 eV / 1.6 10⁻¹⁹)

E = 1,237 eV

The possible transitions from this level are towards n = 6, 5, 4,3,2, 1

We must test the different values until we find the right one

Energy of the states

n $E_{n}$

6 -0.378

5 -0.544

4 -0.850

3 -1,512

2 -3,402

1 -13,606

Let's examine the transition n = 7 to n = 6

ΔE = - 0.27767 - (-0.3779)

ΔE = 0.10023 eV

n = 7 to n = 5

ΔE = -0.27767 - (-0.5442)

ΔE = 0.267 eV

n = 7 a n = 4

ΔE = -0.27767- (-0.8504)

ΔE = 0.573 eV

n = 7 a n = 3

ΔE = -0.27767 - (- 1.5118)

ΔE = 1.234 Ev

This is the transition sought, so that the electron goes from level n = 7 to level n = 3

7 0

3 years ago