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pav-90 [236]
3 years ago
8

? A body is suspended from the ceiling with two wires that make an angle of 40° with the ceiling. The weight of the body is 150N

. What is the tension in each wire?
Physics
1 answer:
andriy [413]3 years ago
4 0
So, the weight is supported by the tension in the two wires, but only part of the tension goes in holding the body: the vertical component.

If the angle between the ceiling and the wire is 40 deg, then the vertical component is

T_y = T * sin(40),

There are two wires, so together they do

2*T*sin(40)

And all this holds the weight of the body, so:

2*T*sin(40)=150 ----> T = 75/sin(40)~116.68 N,

Notice how it is larger than 150N (both wires is ~ 233 N), as the angle becomes smaller, there is less tension that is vertical, so one needs more tension to make the vertical component as large as (half) the weight of the body

Hope it helps!

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At a sudden contraction in a pipe the diameter changes fromD1toD2. The pressure drop,Δp, which develops across the contraction i
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The Great Red Spot, a hurricane-like storm with twice the diameter of earth, is found on (a.)Jupiter (b.)Saturn (c.)Titan (d.)Th
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3 0
3 years ago
Read 2 more answers
A curve that has a radius of 90 m is banked at an angle of =10.8∘. If a 1100 kg car navigates the curve at 75 km/h without skidd
PilotLPTM [1.2K]

The minimum coefficient of static friction  between the pavement and the tires is 0.69.

The given parameters;

  • <em>radius of the curve, r = 90 m</em>
  • <em>angle of inclination, θ = 10.8⁰</em>
  • <em>speed of the car, v = 75 km/h = 20.83 m/s</em>
  • <em>mass of the car, m = 1100 kg</em>

The normal force on the car is calculated as follows;

F_n = mgcos(\theta)

The frictional force between the car and the road is calculated as;

F_k = \mu_k F_n\\\\F_k = \mu_k mgcos(\theta)

The net force on the car is calculated as follows;

mgsin(\theta) +  \mu_s mgcos(\theta) = \frac{mv^2}{r} \\\\mg(sin\theta \ + \ \mu_s cos\theta)= \frac{mv^2}{r} \\\\g(sin\theta \ + \ \mu_s cos\theta)= \frac{v^2}{r}\\\\sin\theta \ + \ \mu_s cos\theta = \frac{v^2}{rg}\\\\\mu_s cos\theta = sin\theta \  + \ \frac{v^2}{rg}\\\\\mu_s = \frac{sin\theta}{cos \theta} + \frac{v^2}{cos (\theta)rg}\\\\\mu_s = tan(\theta) +   \frac{v^2}{cos (\theta)rg}\\\\\mu_s = tan(10.8) +  \frac{(20.83)^2}{cos(10.8) \times 90 \times 9.8} \\\\\mu_s = 0.19 + 0.5\\\\

\mu_s = 0.69

Thus, the minimum coefficient of static friction  between the pavement and the tires is 0.69.

Learn more here:brainly.com/question/15415163

8 0
3 years ago
If a ball dropped from a tower reaches the ground after 3.5 seconds, what is the height of the tower?
solmaris [256]

Answer:

60.025m.

Explanation:

S= ut + at^2/2 (2nd equation of motion)

S= 0 + (9.8)(3.5)^2 /2 (free fall case, initial velocity = 0)

S = 4.9 x 12.25

S= 60.025 m.

Disclaimer: did math in my head, so you better double check with a calculator.

6 0
3 years ago
In an engine, an almost ideal gas is compressed adiabatically to half its volume. In doing so, 2820 J of work is done on the gas
Dmitry_Shevchenko [17]

Answer:

  1. Q=0
  2. U=2820
  3. Energy increases

Explanation:

From the question we are told that

Work done W=2820

a)Generally the heat flow for an adiabatic process is 0 (zero)

Q= U + W =>0

Q=0

b)Generally Change in internal energy of gas is mathematically given by

Since W=-2820J

Therefore

U=2820

Giving

Q= 2820 -2820

Q=O

c)With increases in internal energy brings increase in temperature

Therefore

Energy increases

3 0
3 years ago
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