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3 years ago

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Homework help asap!!

1 answer:

ozzi3 years ago

7 0

Where is the picture cant6 see it

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In unit-vector notation, what is the torque about the origin on a particle located at coordinates (0 m, −3.0 m, 2.0 m) due to fo

irinina [24]

Answer:

The torque about the origin is $2.0Nm\hat{i}-8.0Nm\hat{j}-12.0Nm\hat{k}$

Explanation:

Torque $\overrightarrow{\tau}$ is the cross product between force $\overrightarrow{F}$ and vector position $\overrightarrow{r}$ respect a fixed point (in our case the origin):

$\overrightarrow{\tau}=\overrightarrow{r}\times\overrightarrow{F}$

There are multiple ways to calculate a cross product but we're going to use most common method, finding the determinant of the matrix:

$\overrightarrow{r}\times\overrightarrow{F} =-\left[\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k}\\ F1_{x} & F1_{y} & F1_{z}\\ r_{x} & r_{y} & r_{z}\end{array}\right]$

$\overrightarrow{r}\times\overrightarrow{F} =-((F1_{y}r_{z}-F1_{z}r_{y})\hat{i}-(F1_{x}r_{z}-F1_{z}r_{x})\hat{j}+(F1_{x}r_{y}-F1_{y}r_{x})\hat{k})$

$\overrightarrow{r}\times\overrightarrow{F} =-((0(2.0m)-0(-3.0m))\hat{i}-((4.0N)(2.0m)-(0)(0))\hat{j}+((4.0N)(-3.0m)-0(0))\hat{k})$

$\overrightarrow{r}\times\overrightarrow{F}=-2.0Nm\hat{i}+8.0Nm\hat{j}+12.0Nm\hat{k}=\overrightarrow{\tau}$

4 0

3 years ago

the electrical resistance of copper wire varies directly with its length and inversely with the square of the diameter of the wi

leonid [27]

Answer:24.47

Explanation:

Given

$L_1=30 m$

$d_1=3 mm$

$R_1=25 \Omega$

$L_2=40 m$

$d_2=3.5 mm$

we know Resistance $R=\frac{\rho L}{A}$

Where R=resistance

$\rho =resistivity$

L=Length

A=area of cross-section

$A=\frac{\pi d^2}{4}$

Thus $R\propto \frac{L}{d^2}$

therefore

$R_1\propto \frac{L_1}{d_1^2}$ --------1

$R_2\propto \frac{L_2}{d_2^2}$ --------2

divide 1 and 2 we get

$\frac{R_1}{R_2}=\frac{L_1}{L_2}\times \frac{d_2^2}{d_1^2}$

$\frac{R_1}{R_2}=\frac{30}{40}\times \frac{3.5^2}{3^2}$

$R_2=25\times 0.734\times 1.33$

$R_2=24.47 \Omega$

8 0

3 years ago

Chromosomes are a form of blood cells.

Inessa05 [86]

Answer:

false im just trying to get it if you'd like to give it to me

8 0

3 years ago

Read 2 more answers

Help plzzz!!! I only have 10 minutes to turn in

Lostsunrise [7]

The mechanic did 5406 Joules of work pushing the car.

That's the energy he put into the car. When he stops pushing, all the energy he put into the car is now the car's kinetic energy.

Kinetic energy = (1/2) (mass) (speed²)

And there we have it

The car's mass is 3,600 kg.
Its speed is 'v' m/s .
(1/2) (mass) (v²) = 5,406 Joules

(1/2) (3600 kg) (v²) = 5406 joules

1800 kg (v²) = 5406 joules

v² = (5406 joules) / (1800 kg)

v² = (5406/1800) (joules/kg)

= = = = = This section is just to work out the units of the answer:

v² = (5406/1800) (Newton-meter/kg)

v² = (5406/1800) (kg-m²/s² / kg)

v² = (5406/1800) (m²/s²)

= = = = =

v = √(5406/1800) m/s

<em>v = 1.733 m/s</em>

4 0

3 years ago

A car moving with an intial velocity of 60m/s is brought to rest in 30 seconds calculate the acceleration

gregori [183]

Answer:

a = 2 [m/s^2]

Explanation:

To solve this problem we must use the expressions of kinematics, we must bear in mind that when a body is at rest its velocity is zero.

$v_{f} = v_{i} - (a*t)$

where:

Vf = final velocity = 0

Vi = initial velocity = 60 [m/s]

a = desacceleration [m/s^2]

t = time = 30 [s]

Note: the negative sign of the above equation means that the car is slowing down, i.e. its speed decreases.

0 = 60 - (a*30)

a = 2 [m/s^2]

6 0

3 years ago