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3 years ago

Using the midpoint and the distance formulas, calculate he coordinate of the midpoint and the length of the segment.

Physics

1 answer:

arsen [322]3 years ago

7 0

Answer:

Explanation:

Formula to calculate the distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ is,

d = $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Therefore, length of CD having coordinates C(3,1) and D(3, 3),

CD = $\sqrt{(3-3)^2+(1-3)^2}$

= 2 units

Formula to find the midpoint of CD is,

M = $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

= $(\frac{3+3}{2},\frac{1+3}{2})$

= (3, 2)

Therefore, length of CD = 2 units and (3, 2) are the coordinates of the midpoint of CD.

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Answer:

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3 years ago

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A bicyclist travels 4.5 km west, then travels 6.7 km at an angle 27.0 degrees South of West. What is the magnitude of the bicycl

Dimas [21]

Answer:

10.90km

Explanation:

Magnitude of the total displacement is expressed using the equation

d = √dx²+dy²

dx is the horizontal component of the displacement

dy is the vertical component of the displacement

dy = -6.7sin27°

dy = -6.7(0.4539)

dy = -3.042

For the horizontal component of the displacement

dx = -4.5 - 6.7cos27

dx = -4.5 -5.9697

dx = -10.4697

Get the magnitude of the bicyclist's total displacement

Recall that: d = √dx²+dy²

d = √(-3.042)²+(-10.4697)²

d = √9.2538+109.6146

d = √118.8684

d = 10.90km

Hence the magnitude of the bicyclist's total displacement is 10.90km

6 0

3 years ago

It is measured that 3/4 of a body's volume is submerged in oil of density 800kg/m³

Evgesh-ka [11]

Complete question:

It is measured that 3/4 of a body's volume is submerged in oil of density 800kg/m³. What is the specific gravity of oil?

Answer:

The specific gravity of the oil is 0.8.

Explanation:

Given;

density of the oil, $\rho_o$ = 800 kg/m³

density of water, $\rho_w$ = 1000 kg/m³

The specific gravity of any substance is the ratio of the substance density to the density of water.

Specific gravity of the oil = density of the oil / density of water

Specific gravity of the oil = 800/1000

Specific gravity of the oil = 0.8

Therefore, the specific gravity of the oil is 0.8.

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3 years ago

A compact disc (CD) stores music in a coded pattern of tiny pits 10−7m deep. The pits are arranged in a track that spirals outwa

andreev551 [17]

(a) 50 rad/s

The angular speed of the CD is related to the linear speed by:

$\omega=\frac{v}{r}$

where

$\omega$ is the angular speed

v is the linear speed

r is the distance from the centre of the CD

When scanning the innermost part of the track, we have

v = 1.25 m/s

r = 25.0 mm = 0.025 m

Therefore, the angular speed is

$\omega=\frac{1.25 m/s}{0.025 m}=50 rad/s$

(b) 21.6 rad/s

As in part a, the angular speed of the CD is given by

$\omega=\frac{v}{r}$

When scanning the outermost part of the track, we have

v = 1.25 m/s

r = 58.0 mm = 0.058 m

Therefore, the angular speed is

$\omega=\frac{1.25 m/s}{0.058 m}=21.6 rad/s$

(c) 5550 m

The maximum playing time of the CD is

$t =74.0 min \cdot 60 s/min = 4,440 s$

And we know that the linear speed of the track is

v = 1.25 m/s

If the track were stretched out in a straight line, then we would have a uniform motion, therefore the total length of the track would be:

$d=vt=(1.25 m/s)(4,440 s)=5,550 m$

(d) $-6.4\cdot 10^{-3} rad/s^2$

The angular acceleration of the CD is given by

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_f = 21.6 rad/s$ is the final angular speed (when the CD is scanned at the outermost part)

$\omega_i = 50.0 rad/s$ is the initial angular speed (when the CD is scanned at the innermost part)

$t=4440 s$ is the time elapsed

Substituting into the equation, we find

$\alpha=\frac{21.6 rad/s-50.0 rad/s}{4440 s}=-6.4\cdot 10^{-3} rad/s^2$

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3 years ago

Read 2 more answers

A student places an object with a mass of m on a disk at a position r from the center of the disk. The student starts rotating t

koban [17]

Answer:

The coefficient of static friction between the object and the disk is 0.087.

Explanation:

According to the statement, the object on the disk experiments a centrifugal force due to static friction. From 2nd Newton's Law, we can represent the object by the following formula:

$\Sigma F_{r} = \mu_{s}\cdot N = m\cdot \frac{v^{2}}{R}$ (1)

$\Sigma F_{y} = N - m\cdot g = 0$ (2)

Where:

$N$ - Normal force from the ground on the object, measured in newtons.

$m$ - Mass of the object, measured in newtons.

$g$ - Gravitational acceleration, measured in meters per square second.

$v$ - Linear speed of rotation of the disk, measured in meters per second.

$R$ - Distance of the object from the center of the disk, measured in meters.

By applying (2) on (1), we obtain the following formula:

$\mu_{s}\cdot m\cdot g = m\cdot \frac{v^{2}}{R}$

$\mu_{s} = \frac{v^{2}}{g\cdot R}$

If we know that $v = 0.8\,\frac{m}{s}$ , $g = 9.807\,\frac{m}{s^{2}}$ and $R = 0.75\,m$ , then the coefficient of static friction between the object and the disk is:

$\mu_{s} = \frac{\left(0.8\,\frac{m}{s} \right)^{2}}{\left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.75\,m)}$

$\mu_{s} = 0.087$

The coefficient of static friction between the object and the disk is 0.087.

5 0

3 years ago