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jeyben [28]
2 years ago
13

The magnetic lines of force always travel from north to south true or false

Physics
1 answer:
kumpel [21]2 years ago
7 0

Answer:

True

Explanation:

Magnetic field lines outside of a permanent magnet always run from the north magnetic pole to the south magnetic pole. Therefore, the magnetic field lines of the earth run from the southern geographic hemisphere towards the northern geographic hemisphere.

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A manufacturer selected a metal to use in producing a lightweight button for clothing. A metal that has a density of 2.71 g/cm3
Natali5045456 [20]

Just find the density of every metal and select the one with a density of 2.71 g/cm³ . This is:

Metal 1

ρ = m/V

ρ = 22.1 g / 3 cm³

ρ = 7.367 g / cm³

Metal 2

ρ = m/V

ρ = 42 g / 4 cm³

ρ = 10.5 g / cm³

Metal 3

ρ = m/V

ρ = 9.32 g / 5 cm³

ρ = 1.864 g / cm³

Metal 4

ρ = m/V

ρ = 8.13 g / 3 cm³

ρ = 2.71 g / cm³

<h2>R / Metal 4 was selected.</h2>
4 0
3 years ago
The Carson family's pancake recipe uses 2 teaspoons of baking powder for every 1/3 of a teaspoon of salt. How much baking powder
melomori [17]

Answer:

6 teaspoons of baking powder required.

Explanation:

Given that

According to the recipe of pancake,

For every \frac{1}{3} teaspoon of salt, 2 teaspoons of baking baking powder is required.

To find:

How much baking powder will be needed, if 1 teaspoon of salt was used ?

Solution:

This problem can be solved using ratio.

\frac{1}3 teaspoon of salt : 2 teaspoons of baking powder

Let us multiply the above ratio with 3.

\frac{1}{3}\times 3 teaspoon of salt : 2 \times 3 teaspoons of baking powder

OR

1 teaspoon of salt : 6 teaspoons of baking powder

So, answer is <em>6 teaspoons </em>of baking powder required.

Also, we can use the unitary method:

\frac{1}3 teaspoon of salt needs =  2 teaspoons of baking powder

1  teaspoon of salt needs =  \frac{2}{\frac{1}3} teaspoons of baking powder

1  teaspoon of salt needs = 2 \times 3 = <em>6</em>  teaspoons of baking powder needed

So, the answer is:

<em>6 teaspoons of baking powder </em>required.

5 0
3 years ago
Read 2 more answers
Heat is transferred at a rate of 2 kW from a hot reservoir at 800 K to a cold reservoir at 300 K. Calculate the rate at which th
vekshin1

Answer:

0.00417 kW/K or 4.17 W/K

Second law is satisfied.

Explanation:

Parameters given:

Rate of heat transfer, Q = 2kW

Temperature of hot reservoir, Th = 800K

Temperature of cold reservoir, Tc = 300K

The rate of entropy change is given as:

ΔS = Q * [(1/Tc) - (1/Th)]

ΔS = 2 * (1/300 - 1/800)

ΔS = 2 * 0.002085

ΔS = 0.00417 kW/K or 4.17 W/K

Since ΔS is greater than 0, te the second law of thermodynamics is satisfied.

6 0
3 years ago
Help pls sand quickly plssssss
tamaranim1 [39]

Answer:

The magnet produces an electric current in the wire

Explanation:

6 0
3 years ago
Read 2 more answers
A glass flask whose volume is 1000 cm^3 at a temperature of 1.00°C is completely filled with mercury at the same temperature. W
Marat540 [252]

Answer:

the coefficient of volume expansion of the glass is \mathbf{  ( \beta_{glass} )= 1.333 *10^{-5} / K}

Explanation:

Given that:

Initial volume of the glass flask = 1000 cm³ = 10⁻³ m³

temperature of the glass flask and mercury= 1.00° C

After heat is applied ; the final temperature = 52.00° C

Temperature change ΔT = 52.00° C - 1.00° C = 51.00° C

Volume of the mercury overflow = 8.50 cm^3 = 8.50 ×  10⁻⁶ m³

the coefficient of volume expansion of mercury is 1.80 × 10⁻⁴ / K

The increase in the volume of the mercury =  10⁻³ m³ ×  51.00 × 1.80 × 10⁻⁴

The increase in the volume of the mercury = 9.18*10^{-6} \ m^3

Increase in volume of the glass =  10⁻³ × 51.00 × \beta _{glass}

Now; the mercury overflow = Increase in volume of the mercury - increase in the volume of the flask

the mercury overflow = (9.18*10^{-6}  -  51.00* \beta_{glass}*10^{-3})\ m^3

8.50*10^{-6} = (9.18*10^{-6}  -51.00* \beta_{glass}* 10^{-3} )\ m^3

8.50*10^{-6} - 9.18*10^{-6} = ( -51.00* \beta_{glass}* 10^{-3} )\ m^3

-6.8*10^{-7} =  ( -51.00* \beta_{glass}* 10^{-3} )\ m^3

6.8*10^{-7} =  ( 51.00* \beta_{glass}* 10^{-3} )\ m^3

\dfrac{6.8*10^{-7}}{51.00 * 10^{-3}}=  ( \beta_{glass} )

\mathbf{  ( \beta_{glass} )= 1.333 *10^{-5} / K}

Thus; the coefficient of volume expansion of the glass is \mathbf{  ( \beta_{glass} )= 1.333 *10^{-5} / K}

5 0
3 years ago
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