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3 years ago

9

Calculate the pressure on a man’s foot when a woman who weighs 520 N steps on his foot with her heel which has an area of 0.001

m 2with all of her weight.

1 answer:

Vinvika [58]3 years ago

6 0

Answer:

520000 or 520000 pa

Force = 520N

Area of contact = 0.001

Pressure: 520000 or 520000

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The average intensity of light emerging from a polarizing sheet is 0.708 W/m2, and that of the horizontally polarized light inci

Pachacha [2.7K]

Answer:

Angle θ = 30.82°

Explanation:

From Malus’s law, since the intensity of a wave is proportional to its amplitude squared, the intensity I of the transmitted wave is related to the incident wave by; I = I_o cos²θ

where;

I_o is the intensity of the polarized wave before passing through the filter.

In this question,

I is 0.708 W/m²

While I_o is 0.960 W/m²

Thus, plugging in these values into the equation, we have;

0.708 W/m² = 0.960 W/m² •cos²θ

Thus, cos²θ = 0.708 W/m²/0.960 W/m²

cos²θ = 0.7375

Cos θ = √0.7375

Cos θ = 0.8588

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3 years ago

The slope of a linear velocity-time graph tells us the _____ of the object.

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Speed of the object !!!!

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3 years ago

Draw a stationary wave and show the position of node and antinode

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I hope that the attachment helps you..

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2 years ago

2 Ten identical lengths of wire are laid closely side-by-

Mice21 [21]

Answer:

(a) 0.71 mm

(b) 0.158 cubic cm

Explanation:

The width of one wire is the diameter of the wire.

(a) Let the diameter of each wire is d.

So, 10 d = 14.2 mm

d = 1.42 mm

radius of each wire, r = d/2 = 1.42/2 = 0.71 mm

(b) Length, L = 10 cm

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2 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:

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

3 0

3 years ago