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diamong [38]
3 years ago
13

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:
Maru [420]3 years ago
8 0
The aid of a free-body diagram is essential for solving this mechanics problem. The summation of horizontal and vertical forces in the system must be equal to zero since the body is in equilibrium. 

Summation of x-forces:

T*cos(40)-T*cos(40) = 0

y forces:

2T*sin(40) - 150 = 0

T = 150/2*sin(40) = 100.66 N

Therefore, the tension in each wire would be 100.66 Newtons.

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When a fixed amount of ideal gas goes through an isobaric expansion A) its internal (thermal) energy does not change.B) the gas
Bingel [31]
<h2>Answer: its temperature must increase.</h2>

Explanation:

In an isobaric process the pressure remains constant, which means the initial pressure and the final pressure will be the same.

In addition, during this thermodynamic process, the volume of the ideal gas expands or contracts in such a way that the variation of pressure \Delta P is neutralized.

Now, according to the First law of Thermodynamics that establishes the conservation of energy:

\Delta U=\Delta Q-\Delta W   (1)

Where:

\Delta U is the internal energy

\Delta Q is the heat transferred

\Delta W is the work

Now, for an isobaric process:

\Delta W=P\Delta V    (2)

Where:

P is the pressure (<u>always positive</u>)

\Delta V is the volume variation of the gas

<u />

<u>Here we have two possible results:</u>

-If the gas expands (positive \Delta V), the work is positive.

-If the gas compresses (negative \Delta V), the work is negative.

In this case we are talking about the first result (work is positive).

Then, according to the above, equation (1) can be written as follows:

\Delta U=\Delta Q - P\Delta V   (3)

Clearing \Delta Q:

\Delta Q=\Delta U+P \Delta V    (4)

Then, for an ideal gas in an isobaric process, part of the heat (Q) added to the system will be used to do work (positive in this case) and the other part <u>will increase the internal energy</u>, hence <u>the temperature will increase as well.</u>

7 0
3 years ago
A truck skids for a distance of 25 m with the road pushing on its tires with force of 1500 N as its brakes
MakcuM [25]

Answer:-6800J

Explanation: 8.0m x 850N = 6800

Rewritten as a negative when brake/stop -6800

6 0
3 years ago
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What relationship exists betwen air resistance and acceleration of falling objects
ololo11 [35]
They both make a thing go faster and slower but the relationship is force.
5 0
4 years ago
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Mars has twice the mass of Mercury and is 4 times further away from the Sun. Calculate theratio of the gravitational force from
svetoff [14.1K]

Answer:

F(Mars) = 2 G m M / (4 R)^2   force of Sun on Mars

F(Merc) = G m M / R^2    force of force of Sun on Mercury

R = distance of Sun from Mercury, m = mass of Mercury

F(Merc) / F(Mars) = 4^2 / 2 = 8

6 0
2 years ago
a 1500 kg car traveling at 15 m/s to the south collides with a 4500 kg truck that is intially at rest at a spotlight. The car an
harkovskaia [24]

Answer:

3.75 m/s south

Explanation:

Momentum before collision = momentum after collision

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

Since the car and truck stick together, v₁ = v₂.

m₁ u₁ + m₂ u₂ = (m₁ + m₂) v

Given m₁ = 1500 kg, u₁ = -15 m/s, m₂ = 4500 kg, and u₂ = 0 m/s:

(1500 kg) (-15 m/s) + (4500 kg) (0 m/s) = (1500 kg + 4500 kg) v

-22500 kg m/s = 6000 kg v

v = -3.75 m/s

The final velocity is 3.75 m/s to the south.

4 0
3 years ago
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