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Scorpion4ik [409]
4 years ago
6

for any object suspended by any number of ropes, wires, or chains, how is the total amount of tension (tension in each rope adde

d together) related to the weight of the object?
Physics
1 answer:
Sveta_85 [38]4 years ago
7 0

Answer:

To calculate the tension on a rope holding 1 object, multiply the mass and gravitational acceleration of the object. If the object is experiencing any other acceleration, multiply that acceleration by the mass and add it to your first total.

Explanation:

The tension in a given strand of string or rope is a result of the forces pulling on the rope from either end. As a reminder, force = mass × acceleration. Assuming the rope is stretched tightly, any change in acceleration or mass in objects the rope is supporting will cause a change in tension in the rope. Don't forget the constant acceleration due to gravity - even if a system is at rest, its components are subject to this force. We can think of a tension in a given rope as T = (m × g) + (m × a), where "g" is the acceleration due to gravity of any objects the rope is supporting and "a" is any other acceleration on any objects the rope is supporting.[2]

For the purposes of most physics problems, we assume ideal strings - in other words, that our rope, cable, etc. is thin, massless, and can't be stretched or broken.

As an example, let's consider a system where a weight hangs from a wooden beam via a single rope (see picture). Neither the weight nor the rope are moving - the entire system is at rest. Because of this, we know that, for the weight to be held in equilibrium, the tension force must equal the force of gravity on the weight. In other words, Tension (Ft) = Force of gravity (Fg) = m × g.

Assuming a 10 kg weight, then, the tension force is 10 kg × 9.8 m/s2 = 98 Newtons.

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Scientists plan to release a space probe that will enter the atmosphere of a gaseous planet. The temperature of the gaseous plan
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Answer:

Scientists plan to release a space probe that will enter the atmosphere of a gaseous planet. The temperature of the gaseous planet increases linearly with the height of the atmosphere as measured from the top of a visible boundary layer, defined as 0 kilometers in altitude. The instruments on board can withstand a temperature of 601 K. At what altitude will the probe's instruments fail? A. 50 kilometers B. 80 kilometers C. 83 kilometers D. 100 kilometers E. 111 kilometers

Explanation:

A. 50 kilometers

8 0
3 years ago
An open pipe of length 0.39m vibrates in the third harmonic with a frequency of 1400Hz. What is the distance from the center of
Gnoma [55]

Length of the pipe = 0.39 m

Third harmonic frequency = 1400 Hz

For the third harmonic:

Wavelength = \frac{2L}{3}

The center of the open pipe will host a node and the nearest anti - node from the center will be at the 0.25 × wavelength

Distance from center  = 0.25 × wavelength

Distance = 0.25 x \frac{2L}{3}

Plugging the value of the length of the pipe (L) = 0.39 m = 39 cm

Distance = 0.25 x \frac{2 \times 39}{3}

Distance from the center to the nearest anti - node = 6.5 cm

Hence, the nearest distance to the anti - node from the center = 6.5 cm

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7 0
3 years ago
A cat jumps off a piano that is 1.3m high. The initial velocity of the cat is 3m/s at an angle of 37degrees above the horizontal
SpyIntel [72]

Answer:

x=1.75m

Explanation:

From the exercise we have that

y_{o}=1.3m\\v_{o}=3m/s, \beta  =37\\

<em><u>To find how far from the edge of the piano does the cat strike the floor, we need to calculate its time first </u></em>

y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}

At the end of the motion y=0m

0=1.3+3sin(37)t-\frac{1}{2}(9.8)t^{2}

Solving for t

t=-0.36 s or t=0.73s

Since the <u>time</u> can't be negative the answer is t=0.73

Knowing that we can calculate how far does the cat strike the floor

x=v_{ox}t=3cos(37)(0.73)=1.75m

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4 years ago
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