3)What are the "bright stars" that seem to move against the sky? Are they really stars? What are they?

A force of 100 N acts on a body and moves at a distance of 2 m in the direction of the force. How much work has been done?

Flauer [41]

Answer:

200 joules

Explanation:

work=force×distance

5 0

3 years ago

1. A block is pulled to the right at constant velocity by a 20N force acting at 30o above the horizontal. If the coefficient of

DiKsa [7]

Answer:

44.6 N

Explanation:

Draw a free body diagram of the block. There are four forces on the block:

Weight force mg pulling down,

Normal force N pushing up,

Friction force Nμ pushing left,

and applied force F pulling right 30° above horizontal.

Sum of forces in the y direction:

∑F = ma

N + F sin 30° − mg = 0

N = mg − F sin 30°

Sum of forces in the x direction:

∑F = ma

F cos 30° − Nμ = 0

F cos 30° = Nμ

N = F cos 30° / μ

Substitute:

mg − F sin 30° = F cos 30° / μ

mg = F sin 30° + (F cos 30° / μ)

Plug in values:

mg = 20 N sin 30° + (20 N cos 30° / 0.5)

mg = 44.6 N

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3 years ago

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A disk 7.90 cm in radius rotates at a constant rate of 1 190 rev/min about its central axis. (a) Determine its angular speed. 12

Tanya [424]

Answer:

$124.62\ \text{rad/s}$

$3.71\ \text{m/s}$

$1.23\ \text{km/s}^2$

$20.28\ \text{m}$

Explanation:

r = Radius of disk = 7.9 cm

N = Number of revolution per minute = 1190 rev/minute

Angular speed is given by

$\omega=N\dfrac{2\pi}{60}\\\Rightarrow \omega=1190\times \dfrac{2\pi}{60}\\\Rightarrow \omega=124.62\ \text{rad/s}$

The angular speed is $124.62\ \text{rad/s}$

r = 2.98 cm

Tangential speed is given by

$v=r\omega\\\Rightarrow v=2.98\times 10^{-2}\times 124.62\\\Rightarrow v=3.71\ \text{m/s}$

Tangential speed at the required point is $3.71\ \text{m/s}$

Radial acceleration is given by

$a=\omega^2r\\\Rightarrow a=124.62^2\times 7.9\times 10^{-2}\\\Rightarrow a=1226.88\approx 1.23\ \text{km/s}^2$

The radial acceleration is $1.23\ \text{km/s}^2$ .

t = Time = 2.06 s

Distance traveled is given by

$d=vt\\\Rightarrow d=\omega rt\\\Rightarrow d=124.62\times 7.9\times 10^{-2}\times 2.06\\\Rightarrow d=20.28\ \text{m}$

The total distance a point on the rim moves in the required time is $20.28\ \text{m}$ .

8 0

3 years ago

Un alambre de teléfono de 120m de largo y de 2.2mm de diámetro se estira debido a una fuerza de 380 N cual es el esfuerzo longit

sergij07 [2.7K]

Respuesta: verifique amablemente la explicación

Explicación:

Dado lo siguiente:

Longitud (L) del cable = 120 m

Diámetro (d) = 2,2 mm (2,2 / 1000) = 2,2 * 10 ^ -3 m

Fuerza (F) = 380 N

Esfuerzo longitudinal = Fuerza / Área

Área = πd² / 4 = (π * (2.2 * 10 ^ -3) ^ 2) / 4

Área = (3.142 * 4.84 * 10 ^ -6)

Área = 0.00000380132 m²

Estrés = Fuerza / Área

Estrés = 380 / 0.00000380132

Esfuerzo longitudinal = 99952128.12 = 9.9952128 * 10^7 Nm^-2

Deformación longitudinal: extensión / longitud

Extensión = 0.10 m

Longitud = 120 m

Deformación longitudinal = 0,1 m / 120 m

Deformación longitudinal = 0.0008333 = 8.33 × 10 ^ -4

6 0

3 years ago

Where do you think the government should put the greatest support: solar energy, wind energy, clean coal, oil exploration, gas e

True [87]

The government should put it support in a combination of sources, as no source in the present can fully provide all energy requirements.

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3 years ago

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