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2 years ago

5

When the power in their house unexpectedly went off, Radhika’s mother said that she would check and replace the fuse. What exact

ly would she replace ?
A) a piece of wire B) a bulb. C) an electric meter. D) a switch

2 answers:

Genrish500 [490]2 years ago

4 0

Answer:

A) a piece of wire

Explanation:

A simple fuse consists of a small piece of wire which has a high resistance and low melting point so that when a high current flows through it , it melts breaking the electric circuit and saving the electrical appliance from any danger.

AVprozaik [17]2 years ago

3 0

A. A piece of wire

Explaintion

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Positive ions from a base and negative ions from an acid form a .

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Positive ions from a base and negative ion from an acid form salt.

<span>To add, table salt or common salt is a mineral composed primarily of sodium chloride, a chemical compound belonging to the larger class of salts. Rock salt or halite is also the common term for salt in its natural form.</span>

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2 years ago

Read 2 more answers

An automobile tire is inflated with air originally at 10.0°C and normal atmospheric pressure. During the process, the air is com

solong [7]

Answer:

(a) $3.81\times 10^5\ Pa$

(b) $4.19\times 1065\ Pa$

Explanation:

<u>Given:</u>

$T_1$ = The first temperature of air inside the tire = $10^\circ C =(273+10)\ K =283\ K$

$T_2$ = The second temperature of air inside the tire = $46^\circ C =(273+46)\ K= 319\ K$

$T_3$ = The third temperature of air inside the tire = $85^\circ C =(273+85)\ K=358 \ K$

$V_1$ = The first volume of air inside the tire

$V_2$ = The second volume of air inside the tire = $30\% V_1 = 0.3V_1$

$V_3$ = The third volume of air inside the tire = $2\%V_2+V_2= 102\%V_2=1.02V_2$

$P_1$ = The first pressure of air inside the tire = $1.01325\times 10^5\ Pa$

<u>Assume:</u>

$P_2$ = The second pressure of air inside the tire

$P_3$ = The third pressure of air inside the tire
n = number of moles of air

Since the amount pof air inside the tire remains the same, this means the number of moles of air in the tire will remain constant.

Using ideal gas equation, we have

$PV = nRT\\\Rightarrow \dfrac{PV}{T}=nR = constant\,\,\,(\because n,\ R\ are\ constants)$

Part (a):

Using the above equation for this part of compression in the air, we have

$\therefore \dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow P_2 = \dfrac{V_1}{V_2}\times \dfrac{T_2}{T_1}\times P_1\\\Rightarrow P_2 = \dfrac{V_1}{0.3V_1}\times \dfrac{319}{283}\times 1.01325\times 10^5\\\Rightarrow P_2 =3.81\times 10^5\ Pa$

Hence, the pressure in the tire after the compression is $3.81\times 10^5\ Pa$ .

Part (b):

Again using the equation for this part for the air, we have

$\therefore \dfrac{P_2V_2}{T_2}=\dfrac{P_3V_3}{T_3}\\\Rightarrow P_3 = \dfrac{V_2}{V_3}\times \dfrac{T_3}{T_2}\times P_2\\\Rightarrow P_3 = \dfrac{V_2}{1.02V_2}\times \dfrac{358}{319}\times 3.81\times 10^5\\\Rightarrow P_3 =4.19\times 10^5\ Pa$

Hence, the pressure in the tire after the car i driven at high speed is $4.19\times 10^5\ Pa$ .

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2 years ago

The magnitude of the weight of a 3.0 kg object on the surface of the earth is 29 N. True False

madreJ [45]

True

In fact, the weight of an object on the surface of the Earth is given by:
$F=mg$
where m is the mass of the object and $g=9.81 m/s^2$ is the gravitational acceleration on Earth's surface. If we use the mass of the object, m=3.0 kg, we find
$F=mg=(3.0 kg)(9.81 m/s^2)=29 N$

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2 years ago

An 100 kg object traveling at 50 m/s collides (perfectly inelastic) with a 50 kg object initially at rest.

qaws [65]

Answer:

Option C. 5,000 kg m/s

Explanation:

<u>Linear Momentum on a System of Particles </u>

Is defined as the sum of the momenta of each particles in a determined moment. The individual momentum is the product of the mass of the particle by its speed

P=mv

The question refers to an 100 kg object traveling at 50 m/s who collides with another object of 50 kg object initially at rest. We compute the moments of each object

$m_1=(100\ kg)(50\ m/s)=5,000\ kg\ m/s$

$m_2=(50\ kg)(0\ m/s) = 0$

The sum of the momenta of both objects prior to the collision is

$P=5,000\ kg\ m/s+0\ kg\ m/s$

$\boxed{ P=5,000\ kg\ m/s}$

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3 years ago

An element's atomic number is 58. How many protons would an atom of this element have?

agasfer [191]

58 the number of protons are the same as your atomic number<span />

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2 years ago