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kakasveta [241]
3 years ago
15

Si un vector tiene una dirección de 2300 a partir del eje x positivo, ¿Qué signos tendrán sus componentes x y y? Si la la razón

de R/R, es negativa. ¿cuáles son los ángulos posibles de R, medidos a partir del eje x positivo?
Physics
1 answer:
Masteriza [31]3 years ago
8 0
If you can write this in English I can help
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An old clock has a spring that must be wound to make the clock hands move. Which statement describes the energy of the spring an
kondaur [170]

Answer:

D

Explanation:

When the spring is wound, then it gathers potential energy in the form of tension energy. As it slowly unwinds, the potential energy is converted to kinetic energy of the hands' movements of the clock. This energy is channeled through the use of cogs/gears in the clock.

7 0
3 years ago
Read 2 more answers
A horizontal clothesline is tied between 2 poles, 16 meters apart. When a mass of 3 kilograms is tied to the middle of the cloth
Alla [95]

Answer:

The magnitude of the tension on the ends of the clothesline is 41.85 N.

Explanation:

Given that,

Poles = 2

Distance = 16 m

Mass = 3 kg

Sags distance = 3 m

We need to calculate the angle made with vertical by mass

Using formula of angle

\tan\theta=\dfrac{8}{3}

\thta=\tan^{-1}\dfrac{8}{3}

\theta=69.44^{\circ}

We need to calculate the magnitude of the tension on the ends of the clothesline

Using formula of tension

mg=2T\cos\theta

Put the value into the formula

3\times9.8=2T\times\cos69.44

T=\dfrac{3\times9.8}{2\times\cos69.44}

T=41.85\ N  

Hence, The magnitude of the tension on the ends of the clothesline is 41.85 N.

4 0
4 years ago
Read 2 more answers
An aluminum bar 600mm long, with diameter 40mm long has a hole drilled in the center of the bar.The hole is 30mm in diameter and
Svetradugi [14.3K]

Answer:

Total contraction on the Bar  = 1.22786 mm

Explanation:

Given that:

Total Length for aluminum bar = 600 mm  

Diameter for aluminum bar  = 40 mm

Hole diameter  = 30 mm

Hole length = 100 mm

elasticity for the aluminum is 85GN/m² = 85 × 10³ N/mm²

compressive load P = 180 KN = 180  × 10³ N

Calculate the total contraction on the bar = ???

The relation used in  calculating the contraction on the bar is:

\delta L = \dfrac{P *L }{A*E}

The relation used in  calculating the total contraction on the bar can be expressed as :

Total contraction in the Bar = (contraction in part of bar without hole + contraction in part of bar with hole)

i.e

Total contraction on the Bar = \dfrac{P *L_1 }{A_1*E} +  \dfrac{P *L_2 }{A_2 *E}

Let's find the area of cross section without the hole and with the hole

Area of cross section without the hole is :

Using A = πd²/4

A = π (40)²/4

A = 1256.64 mm²

Area of cross section with the hole is :

A = π (40²-30²)/4

A = 549.78 mm²

Total contraction on the Bar = \dfrac{P *L_1 }{A_1*E} +  \dfrac{P *L_2 }{A_2 *E}

Total contraction on the Bar  = \dfrac{180 *10^3 \N  }{85*10^3 \ N/mm^2} [\dfrac{500}{1256.64}+ \dfrac{100}{549.78}]

Total contraction on the Bar  = 2.117( 0.398 + 0.182)

Total contraction on the Bar  = 2.117*(0.58)

Total contraction on the Bar  = 1.22786 mm

5 0
4 years ago
A circle has an initial radius of 50ft when the radius begins decreasing at a rate of 2ft/s. what is the rate of change of the a
valkas [14]
The area of the circle with radius r is
A = πr²

The rate of change of area with respect to time is
\frac{dA}{dt} = \frac{dA}{dr} . \frac{dr}{dt} =2 \pi r. \frac{dr}{dt}

The rate of change of the radius is given as
\frac{dr}{dt} =-2 \,  \frac{ft}{s}
Therefore
\frac{dA}{dt} =-4 \pi r \,  \frac{ft^{2}}{s}

When r = 10 ft, obtain
\frac{dA}{dt}|_{r=10 \, ft} = -40 \pi  \,  \frac{ft^{2}}{s}

Answer: - 40π ft²/s (or - 127.5 ft²/s)
7 0
4 years ago
Based on the graph, describe how momentum changes with time for an object in free fall. If you can help, I would be so grateful.
pochemuha
With time, momentum increases as it builds speed assuming their is nothing in the way to stop it. Based on the graph, you can see that example being displayed as the line on the graph gets higher
4 0
3 years ago
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