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3 years ago

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A 2 kg object is given a displacement ∆~s = (5 m)ˆı + (3 m) ˆ + (−4 m) ˆk along a straight line. During the displacement, a con

stant force of F~ = (3 N)ˆı + (−1 N) ˆ + (7 N) ˆk acts on the object. Find the work done by F~ for this displacement.

1 answer:

nadya68 [22]3 years ago

8 0

Answer: 15Nm

Explanation:

Work is said to be done when a force cause a body to move through a distance. Mathematically,

Work = Force (F) × distance (s)

If F = F~ = (3 N)ˆı + (−1 N) ˆj + (7 N) ˆk

S = (5 m)ˆı + (3 m) ˆ + (−4 m) ˆk

According to vector notation,

i.I = j.j =k.k = 1

Multiplication of different component will give 'zero'

F×s = (3 N)ˆi × (5 m)ˆı

Work done = 15Nm

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Joe is measuring the time it takes for a ball to roll down a ramp. In this experiment Joe takes the measurement 5 times and gets

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Step by step solution :

standard deviation is given by :

$\sigma = \sqrt\dfrac{{\sum (x-\bar{x})^2}}{n}$

where, $\sigma$ is standard deviation

$\bar{x}$ is mean of given data

n is number of observations

From the above data, $\bar{x}=24.88$

Now, if $x=24.8$ , then $(x-\bar{x})^2=0.0064$

If $x=23.9$ , then $(x-\bar{x})^2=0.9604$

if $x=26.1$ , then $(x-\bar{x})^2=1.4884$

If $x=25.1$ , then $(x-\bar{x})^2=0.0484$

If $x=24.5$ , then $(x-\bar{x})^2=0.1444$

so, $\sum (x-\bar{x})^2 =\frac{0.0064+0.9604+1.4884+0.0484+0.1444}{5}$

$\sum (x-\bar{x})^2 =2.648$

$\sqrt{\sum \frac{(x-\bar{x})^2}{n}}$

$\sigma =0.7277$

No, Joe's value does not agree with the accepted value of 25.9 seconds. This shows a lots of errors.

6 0

2 years ago

A racquetball strikes a wall with a speed of 30 m/s and rebounds in the opposite direction with a speed of 26 m/s. The collision

Fudgin [204]

Answer:

The average acceleration of the ball during the collision with the wall is $a=2,800m/s^{2}$

Explanation:

<u>Known Data</u>

We will asume initial speed has a negative direction, $v_{i}=-30m/s$ , final speed has a positive direction, $v_{f}=26m/s$ , $\Delta t=20ms=0.020s$ and mass $m_{b}$ .

<u>Initial momentum</u>

$p_{i}=mv_{i}=(-30m/s)(m_{b})=-30m_{b}\ m/s$

<u>final momentum</u>

$p_{f}=mv_{f}=(26m/s)(m_{b})=26m_{b}\ m/s$

<u>Impulse</u>

$I=\Delta p=p_{f}-p_{i}=26m_{b}\ m/s-(-30m_{b}\ m/s)=56m_{b}\ m/s$

<u>Average Force</u>

$F=\frac{\Delta p}{\Delta t} =\frac{56m_{b}\ m/s}{0.020s} =2800m_{b} \ m/s^{2}$

<u>Average acceleration</u>

$F=ma$ , so $a=\frac{F}{m_{b}}$ .

Therefore, $a=\frac{2800m_{b} \ m/s^{2}}{m_{b}} =2800m/s^{2}$

8 0

3 years ago

What are the units for volume and/or mass? Please answer and explain clearly thanks!

Paul [167]

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6 0

3 years ago

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nikdorinn [45]

Answer:

2.78 m

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Given:

a = -1.6 m/s²

v₀ = 2.98 m/s

v = 0 m/s

x₀ = 0 m

Find:

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x = 2.775 m

Rounded to 3 sig-figs, the astronaut halloweener reaches a maximum height of 2.78 meters.

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3 years ago

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9966 [12]

Answer:

$2.44156\times 10^{13}\ m^3$

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V = Volume

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r = Radius = $\dfrac{d}{2}=\dfrac{10}{2}=5\ km$

v = Velocity = 11 km/s

$H_v$ = Heat vaporization of water = $2.26\times 10^6\ J/kg$

$\Delta T$ = Change in temperature = 100-20

Mass is given by

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Heat is given by

$Q=mc\Delta T+mH_v\\\Rightarrow 6.33556\times 10^{22}=m\times (4186\times (100-20)+2.26\times 10^6)\\\Rightarrow m=\dfrac{ 6.33556\times 10^{22}}{4186\times (100-20)+2.26\times 10^6}\\\Rightarrow m=2.44156\times 10^{16}\ kg$

Mass of water is $2.44156\times 10^{16}\ kg$

Volume is $\dfrac{2.44156\times 10^{16}}{10^3}=2.44156\times 10^{13}\ m^3$

Amount of water is $2.44156\times 10^{13}\ m^3$

If it were a cube

$h=V^{\dfrac{1}{3}}\\\Rightarrow h=(2.44156\times 10^{13})^{\dfrac{1}{3}}\\\Rightarrow h=29010.53917\ m$

The height of the water would be 29010.53917 m

4 0

2 years ago