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3 years ago

15

There is a 200n boulder stuck in ur yard. You want to use 2 meter long lever to pry it loose from the ground. If you are able to

generate 50n of force to the other end of the lever, where should the fulcrum be placed? How far from the boulder? How far from u?

1 answer:

Kaylis [27]3 years ago

7 0

40cm from boulder and 160cm from you.

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If the pressing force on the slidding surfaces is greater then friction will be

Tema [17]

Answer:

Kinetic friction is lesser than limiting friction. Two surfaces are rubbed together, first with a smaller force and then with a greater force.

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3 years ago

2. Two identical conducting spheres are placed with their centers 0.30 m apart. One is given a charge of 12 x 10-9 C and the oth

Maru [420]

Answer:

A. -2.16 * 10^(-5) N

B. 9 * 10^(-7) N

Explanation:

Parameters given:

Distance between their centres, r = 0.3 m

Charge in first sphere, Q1 = 12 * 10^(-9) C

Charge in second sphere, Q2 = -18 * 10^(-9) C

A. Electrostatic force exerted on one sphere by the other is:

F = (k * Q1 * Q2) / r²

F = (9 * 10^9 * 12 * 10^(-9) * -18 * 10^(-9)) / 0.3²

F = -2.16 * 10^(-5) N

B. When they are brought in contact by a wire and are then in equilibrium, it means they have the same final charge. That means if we add the charges of both spheres and divided by two, we'll have the final charge of each sphere:

Q1 + Q2 = 12 * 10^(-9) + (-18 * 10^(-9))

= - 6 * 10^(-9) C

Dividing by two, we have that each sphere has a charge of -3 * 10^(-9) C

Hence the electrostatic force between them is:

F = [9 * 10^9 * (-3 * 10^(-9)) * (-3 * 10^(-9)] / 0.3²

F = 9 * 10^(-7) N

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3 years ago

Find the quantity of heat needed

krok68 [10]

Answer:

Approximately $3.99\times 10^{4}\; \rm J$ (assuming that the melting point of ice is $0\; \rm ^\circ C$ .)

Explanation:

Convert the unit of mass to kilograms, so as to match the unit of the specific heat capacity of ice and of water.

$\begin{aligned}m&= 100\; \rm g \times \frac{1\; \rm kg}{1000\; \rm g} \\ &= 0.100\; \rm kg\end{aligned}$

The energy required comes in three parts:

Energy required to raise the temperature of that $0.100\; \rm kg$ of ice from $(-10\; \rm ^\circ C)$ to $0\; \rm ^\circ C$ (the melting point of ice.)
Energy required to turn $0.100\; \rm kg$ of ice into water while temperature stayed constant.
Energy required to raise the temperature of that newly-formed $0.100\; \rm kg$ of water from $0\; \rm ^\circ C$ to $10\;\ rm ^\circ C$ .

The following equation gives the amount of energy $Q$ required to raise the temperature of a sample of mass $m$ and specific heat capacity $c$ by $\Delta T$ :

$Q = c \cdot m \cdot \Delta T$ ,

where

$c$ is the specific heat capacity of the material,
$m$ is the mass of the sample, and
$\Delta T$ is the change in the temperature of this sample.

For the first part of energy input, $c(\text{ice}) = 2100\; \rm J \cdot kg \cdot K^{-1}$ whereas $m = 0.100\; \rm kg$ . Calculate the change in the temperature:

$\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (0\; \rm ^\circ C) - (-10\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}$ .

Calculate the energy required to achieve that temperature change:

$\begin{aligned}Q_1 &= c(\text{ice}) \cdot m(\text{ice}) \cdot \Delta T\\ &= 2100\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 2.10\times 10^{3}\; \rm J\end{aligned}$ .

Similarly, for the third part of energy input, $c(\text{water}) = 4200\; \rm J \cdot kg \cdot K^{-1}$ whereas $m = 0.100\; \rm kg$ . Calculate the change in the temperature:

$\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (10\; \rm ^\circ C) - (0\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}$ .

Calculate the energy required to achieve that temperature change:

$\begin{aligned}Q_3&= c(\text{water}) \cdot m(\text{water}) \cdot \Delta T\\ &= 4200\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 4.20\times 10^{3}\; \rm J\end{aligned}$ .

The second part of energy input requires a different equation. The energy $Q$ required to melt a sample of mass $m$ and latent heat of fusion $L_\text{f}$ is:

$Q = m \cdot L_\text{f}$ .

Apply this equation to find the size of the second part of energy input:

$\begin{aligned}Q_2&= m \cdot L_\text{f}\\&= 0.100\; \rm kg \times 3.36\times 10^{5}\; \rm J\cdot kg^{-1} \\ &= 3.36\times 10^{4}\; \rm J\end{aligned}$ .

Find the sum of these three parts of energy:

$\begin{aligned}Q &= Q_1 + Q_2 + Q_3 = 3.99\times 10^{4}\; \rm J\end{aligned}$ .

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3 years ago

The diffusion rate for a solute is 4.0 x 10^-11 kg/s in a solvent- filled channel that has a cross-sectional area of 0.50 cm^2 a

zlopas [31]

Answer:

$s = 9.6\times 10^{-12}kg/s$

Explanation:

Given:

Solute Diffusion rate = 4.0 × 10⁻¹¹ kg/s

Area of cross-section = 0.50 cm²

Length of channel =0.25 cm

Now for the new channel

Area of cross-section = 0.30 cm²

Length of channel =0.10 cm

let the Solute Diffusion rate of new channel = s

now equating the diffusion rate per unit volume for both the channels

$\frac{4\times 10^{-11}}{0.50\times 0.25}=\frac{s}{0.30\times 0.10}$

thus,

$s = 9.6\times 10^{-12}kg/s$

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4 years ago

Explain why urban sprawl occurs even though it negatively impacts the environment.

Grace [21]

Urban sprawl occurs when housing is filled in one location and car dependent communities are forced to moved away from the central urban areas where the population is too vast, even though this impacts the environment by increasing pollution and causing environmental degradation.

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4 years ago

Read 2 more answers