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Elis [28]
2 years ago
12

Explain the characteristics of the Gypsum Hills regionIf we have a sample of silicon (Si) atoms that has 14 protons, 14 electron

s, and 18 neutrons
What is the name of this specific silicon isotope?

silicon-14
silicon-32
silicon-46
silicon-153
Physics
1 answer:
amid [387]2 years ago
8 0

Answer:

second one

Explanation:

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A book weighing 12 n is placed on a table. how much support force does a table exert on the book?
Alenkasestr [34]
It should be 12 N. the force of the book on the table should be the same as the force of the table on the book.
7 0
3 years ago
A rigid adiabatic container is divided into two parts containing n1 and n2 mole of ideal gases respectively, by a movable and th
kicyunya [14]

Answer:

Explanation:

Given

Pressure, Temperature, Volume of gases is

P_1, V_1, T_1 & P_2, V_2, T_2

Let P & T be the final Pressure and Temperature

as it is rigid adiabatic container  therefore Q=0 as heat loss by one gas is equal to heat gain by another gas

-Q=W+U_1----1

Q=-W+U_2-----2

where Q=heat loss or gain (- heat loss,+heat gain)

W=work done by gas

U_1 & U_2 change in internal Energy of gas

Thus from 1 & 2 we can say that

U_1+U_2=0

n_1c_v(T-T_1)+n_2c_v(T-T_2)=0

T(n_1+n_2)=n_1T_1+n_2T_2

T=\frac{n_1+T_1+n_2T_2}{n_1+n_2}

where n_1=\frac{P_1V_1}{RT_1}

n_2=\frac{P_2V_2}{RT_2}

T=\frac{\frac{P_1V_1}{RT_1}\times T_1+\frac{P_2V_2}{RT_2}\times T_2}{\frac{P_1V_1}{RT_1}+\frac{P_2V_2}{RT_2}}

T=\frac{P_1V_1+P_2V_2}{\frac{P_1V_1}{T_1}+\frac{P_2V_2}{T_2}}

and P=\frac{P_1V_1+P_2V_2}{V_1+V_2}

6 0
3 years ago
How must a fuse be connected in a circuit to prevent current from flowing when the circuit becomes ""overloaded""?
Lady bird [3.3K]

Answer:

The fuse must be connected between the device and the power intake source.

Explanation:

A fuse is a protective component of electrical appliances that is designed to be sensitive to a particular range of electric current

The fuse is made of a thing metal strip with a known melting point. Once current abive its carrying capacity flows through it, large heat is generated in the metal strip which melts it and causes the metal strip to cut int two protecting the device from the power spike.

3 0
2 years ago
An engine absorbs 1749 J from a hot reservoir and expels 539 J to a cold reservoir in each cycle. a. What is the engine’s effici
Masja [62]

Answer:

a)η = 69.18 %

b)W= 1210 J

c)P=3967.21 W

Explanation:

Given that

Q₁ = 1749 J

Q₂ = 539  J

From first law of thermodynamics

Q₁   = Q₂ +W

W=Work out put

Q₂=Heat rejected to the cold reservoir

Q₁ =heat absorb by hot  reservoir

W= Q₁- Q₂

W= 1210 J

The efficiency given as

\eta=\dfrac{W}{Q_1}

\eta=\dfrac{1210}{1749}

\eta=0.6918

η = 69.18 %

We know that rate of work done is known as power

P=\dfrac{W}{t}

P=\dfrac{1210}{0.305}\ W

P=3967.21 W

8 0
3 years ago
A 10-kg package drops from chute into a 25-kg cart with a velocity of 3 m/s. The cart is initially at rest and can roll freely w
amid [387]

Answer:

(a) the final velocity of the cart is 0.857 m/s

(b) the impulse experienced by the package is 21.43 kg.m/s

(c) the fraction of the initial energy lost is 0.71

Explanation:

Given;

mass of the package, m₁ = 10 kg

mass of the cart, m₂ = 25 kg

initial velocity of the package, u₁ = 3 m/s

initial velocity of the cart, u₂ = 0

let the final velocity of the cart = v

(a) Apply the principle of conservation of linear momentum to determine common final velocity for ineleastic collision;

m₁u₁  + m₂u₂ = v(m₁  +  m₂)

10 x 3   + 25 x 0   = v(10  +  25)

30  = 35v

v = 30 / 35

v = 0.857 m/s

(b) the impulse experienced by the package;

The impulse = change in momentum of the package

J = ΔP = m₁v - m₁u₁

J = m₁(v - u₁)

J = 10(0.857 - 3)

J = -21.43 kg.m/s

the magnitude of the impulse experienced by the package = 21.43 kg.m/s

(c)

the initial kinetic energy of the package is calculated as;

K.E_i = \frac{1}{2} mu_1^2\\\\K.E_i = \frac{1}{2} \times 10 \times (3)^2\\\\K.E_i = 45 \ J\\\\

the final kinetic energy of the package;

K.E_f = \frac{1}{2} (m_1 + m_2)v^2\\\\K.E_f = \frac{1}{2} \times (10 + 25) \times 0.857^2\\\\K.E_f = 12.85 \ J

the fraction of the initial energy lost;

= \frac{\Delta K.E}{K.E_i} = \frac{45 -12.85}{45} = 0.71

7 0
3 years ago
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