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3 years ago

1.1 A book lies on a table. This book is not moving because:

Physics

1 answer:

Mandarinka [93]3 years ago

6 0

Answer:

B The forces acting on the book are in equilibrium

Explanation:

The book that lies on the table and does not move according to Newton's First Law of motion, because there are no net forces acting on it, therefore the net force (the sum of forces acting on the book) is zero

Therefore, the magnitude of the forces acting on the book and the direction in which the forces acts on the balance each other such that the book is in an equilibrium state because the forces acting on the book (which can cause the book to move) are in equilibrium

Therefore, the book is not moving because the forces on the book are in equilibrium.

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If a body p with a positive charge is placed in contact with a body q (initially uncharged), what will be the nature of the char

uysha [10]

If a body p with a positive charge is placed in contact with a body q (initially uncharged), then the nature of charge gained by q must be positive, because rubbing an uncharged body with a charged body or placed in contact with a positive charged body, helps gain a charge to the uncharged body.

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Let's know, how a element gain positive charge?

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2 years ago

How to understand that an inductor behaves like short circuit and capacitor like an open circuit in steady state in a network?

garik1379 [7]

It depends on the steady-state frequency. At zero frequency an inductor behaves like an open circuit. As the frequency increases, the inductor acts more like an open circuit and a capacitator acts more like a short circuit

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3 years ago

Please help i don't understand. will mark brainliest

soldier1979 [14.2K]

Answer:

marblebrainiest plz\c cmarble

Explanation:

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3 years ago

Read 2 more answers

Three resistors, each having a resistance, R, are connected in parallel to a 1.50 V battery. If the resistors dissipate a total

ivanzaharov [21]

Answer:

The value of resistance of each resistor, R is 2.25 Ω

Explanation:

Given;

voltage across the three resistor, V = 1.5 V

power dissipated by the resistors, P = 3.00 W

the resistance of each resistor, = R

The effective resistance of the three resistors is given by;

R(effective) = R/3

Apply ohms law to determine the current delivered by the source;

V = IR

I = V/R

I = 3V/R

Also, power is calculated as;

P = IV

P = (3V/R) x V

P = 3V²/R

R = 3V² / P

R = (3 x 1.5²) / 3

R = 2.25 Ω

Therefore, the value of resistance of each resistor, R is 2.25 Ω

4 0

3 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