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3 years ago

Which statement about a flat magnet that can stick onto a refrigerator is true

Physics

1 answer:

zubka84 [21]3 years ago

7 0

The magnetic particles are mixed with other substances to make it flexible.

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61. A physics student has a single-occupancy dorm room. The student has a small refrigerator that runs with a current of 3.00 A

Mademuasel [1]

Answer:

Part a)

$percentage = 21.3$ %

Part b)

$percentage = 2.13 \times 10^{-5}$ %

Explanation:

As we know that total power used in the room is given as

$P = P_1 + P_2 + P_3 + P_4$

here we have

$P_1 = (110)(3) = 330 W$

$P_2 = 100 W$

$P_3 = 60 W$

$P_4 = 3 W$

$P = 330 + 100 + 60 + 3$

$P = 493 W$

Part a)

Since power supply is at 110 Volt so the current obtained from this supply is given as

$110\times i = 493$

$i = 4.48 A$

now resistance of transmission line

$R = \frac{\rho L}{A}$

$R = \frac{(2.8 \times 10^{-8})(10\times 10^3)}{\pi(4.126\times 10^{-3})^2}$

$R = 5.23 \ohm$

now power loss in line is given as

$P = i^2 R$

$P = (4.48)^2(5.23)$

$P = 105 W$

Now percentage loss is given as

$percentage = \frac{loss}{supply} \times 100$

$percentage = \frac{105}{493} \times 100$

$percentage = 21.3$ %

Part b)

now same power must have been supplied from the supply station at 110 kV, so we have

$110 \times 10^3 (i ) = 493$

$i = 4.48\times 10^{-3} A$

now power loss in line is given as

$P = i^2 R$

$P = (4.48 \times 10^{-3})^2(5.23)$

$P = 1.05 \times 10^{-4} W$

Now percentage loss is given as

$percentage = \frac{loss}{supply} \times 100$

$percentage = \frac{1.05 \times 10^{-4}}{493} \times 100$

$percentage = 2.13 \times 10^{-5}$ %

6 0

3 years ago

Explain why the model of this chemical reaction obeys the Law of Conservation of Matter. A) because there are the same number of

Sav [38]

Answer:

(A) because there are the same number of atoms of each element shown on both sides

Explanation:

The Law of conservation of mass says that in a reaction the matter of the products should be equivalent to the matter of the reactants and the mass of the system should remain constant over time.

In a chemical reaction, while atoms bond is breaking of 1 substance than new bonds are formed in another substance and new substances are formed. However, in the overall reaction, they keep the same elements, no new elements can go and come from the outside. For example:

HCl + NaOH -----> NaCl + H2O

In this reaction, on both sides the same number of atoms of each element are present.

5 0

3 years ago

1. An EM wave representing 650 nm laser light is traveling in the +zˆ direction. At one point in space and time, the electric fi

ivann1987 [24]

Answer:

Explanation:

a ) Direction of the magnetic field will be in positive x direction.

The direction of the vector E X B gives the direction of motion of wave.

b ) Magnitude of magnetic field is given by the relation

E₀ / B₀ = c , c is velocity of light

B₀ = E₀ / c

= 20 / (3 x 10⁸)

= 6.67 x 10⁻⁸ T

c ) Average power flowing per unit area by this wave is called Poynting vector

c ε₀E₀² , ε₀ = 8.85X10⁻¹²

= 3 X 10⁸ X 8.85 X 10⁻¹² X 20²

= 1.062 W m⁻²

5 0

3 years ago

Does water expand more as the temperature drops even more below 0 deg? I am worried about the coolant in my rad. as the temp.is

alex41 [277]

Water is, indeed, an exception. Normally when temperature drops, material shrinks. Water doesn't, water expands instead.

4 0

4 years ago

A single Oreo cookie provides 53 kcal of energy. An athlete does an exercise that involves repeatedly lifting (without accelerat

Sever21 [200]

Answer:

Approximately $325$ (rounded down,) assuming that $g = 9.81\; {\rm N \cdot kg^{-1}}$ .

The number of repetitions would increase if efficiency increases.

Explanation:

Ensure that all quantities involved are in standard units:

Energy from the cookie (should be in joules, ${\rm J}$ ):

$\begin{aligned} & 53\; {\rm kCal} \times \frac{1\; {\rm kJ}}{4.184\; {\rm kCal}} \times \frac{1000\; {\rm J}}{1\; {\rm kJ}} \approx 2.551 \times 10^{5}\; {\rm J} \end{aligned}$ .

Height of the weight (should be in meters, ${\rm m}$ ):

$\begin{aligned} h &= 2\; {\rm dm} \times \frac{1\; {\rm m}}{10\; {\rm dm}} = 0.2\; {\rm m}\end{aligned}$ .

Energy required to lift the weight by $\Delta h = 0.2\; {\rm m}$ without acceleration:

$\begin{aligned} W &= m\, g\, \Delta h \\ &= 100\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times 0.2\; {\rm m} \\ &= 196\; {\rm N \cdot m} \\ &= 196\; {\rm J} \end{aligned}$ .

At an efficiency of $0.25$ , the actual amount of energy required to raise this weight to that height would be:

$\begin{aligned} \text{Energy Input} &= \frac{\text{Useful Work Output}}{\text{Efficiency}} \\ &= \frac{196\; {\rm J}}{0.25} \\ &=784\; {\rm J}\end{aligned}$ .

Divide $2.551 \times 10^{5}\; {\rm J}$ by $784\; {\rm J}$ to find the number of times this weight could be lifted up within that energy budget:

$\begin{aligned} \frac{2.551 \times 10^{5}\; {\rm J}}{784\; {\rm J}} &\approx 325 \end{aligned}$ .

Increasing the efficiency (the denominator) would reduce the amount of energy input required to achieve the same amount of useful work. Thus, the same energy budget would allow this weight to be lifted up for more times.

4 0

2 years ago