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3 years ago

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A 47 kg mass is moving across a horizontal surface at 8 m/s. What is the force required to bring the mass to a stop in 4.1 secon

ds? (Hint: This is a multi-step calculation and your answer should come out to be negative). Write your answer to one decimal place.

1 answer:

crimeas [40]3 years ago

6 0

Answer:

Force = -91.7 Newton

Explanation:

Given the following data;

Mass = 47 kg

Time = 4.1 seconds

Initial velocity = 8 m/s

Since the object comes to a stop, its final velocity would be equal to zero.

To find the force required to bring it to stop;

First of all, we would determine the acceleration of the object;

Mathematically, acceleration is given by the equation;

$Acceleration (a) = \frac{final \; velocity - initial \; velocity}{time}$

Substituting into the equation;

$a = \frac{0 - 8}{4.1}$

$a = \frac{-8}{4.1}$

Acceleration, a = -1.95 m/s²

Next, we would determine the force required to bring the object to stop;

$Force = mass * acceleration$

$Force = 47 * -1.95$

Force = -91.65 ≈ 91.7 Newton

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What phase difference between two otherwise identical traveling waves, moving in the same direction along a stretched string, wi

Sergeu [11.5K]

Answer:

Ф = 142.674 degree

Ф = 2.490131057 rad

wavelength = 0.396316 λ

Explanation:

Given data

combined wave having an amplitude 0.64 times

to find out

answer in degrees, radians, and fraction of the wavelength

solution

let us consider these two wave equation

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and here Ф is phase difference

so we say

resultant wave is

Y = Asin(ωt) + Asin (ωt + Ф) = Asin(ωt + Ф/2) cos ( Ф/2)

so Y = A' sin(ωt + Ф/2)

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so 2Acos ( Ф/2) = 0.64 A

and Ф/2 = 71.3370

Ф = 142.674 degree

Ф = 2.490131057 rad

so wavelength is 2.490131057 / 2π

wavelength = 0.396316 λ

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Find the quantity of heat needed

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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

2 years ago