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Tcecarenko [31]
3 years ago
11

A mason stretches a string between two points 70 ft apart on the same level with a tension of 20 lb at each end. If the string w

eighs 0.18 lb, determine the sag h at the middle of the string.
Physics
1 answer:
sergejj [24]3 years ago
4 0

Answer:

Explanation:

Let tension be T in each string . Let angle with horizontal be θ in the middle

The sum of vertical components of tension of two string will balance the weight

2Tsinθ = mg

2 x 20 sinθ = .18

sinθ = .18 / 40 = .0045

θ = .25783 degree

If sag be y

y / 70 = tan .25783

y = 70 x tan.25783

= .3150 m

= 31.50 cm .

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In an engine, an almost ideal gas is compressed adiabatically to half its volume. In doing so, 1850 J of work is done on the gas
Oliga [24]

Answer:

The value of change in internal l energy of the gas = 1850 J

Explanation:

Work done on the gas (W) =  - 1850 J

Negative sign is due to work done on the system.

From the first law  we know that Q = Δ U + W ------------- (1)

Where Q = Heat transfer to the gas

Δ U = Change in internal energy of the gas

W = work done on the gas

Since it is adiabatic compression of the gas so heat transfer to the gas is zero.

⇒ Q = 0

So from equation (1)

⇒ Δ U = - W ----------------- (2)

⇒ W = - 1850 J (Given)

⇒ Δ U = - (- 1850)

⇒ Δ U = + 1850 J

This is the value of change in internal energy of the gas.

6 0
3 years ago
A 3.0kg weight W is initially at rest on incline AB, which is raised 40° above the horizontal. The effective coefficient is
lakkis [162]

(a) The acceleration of the system is determined as 1.58 m/s².

(b) The relative weight of P is pounds is determined as 0.14 lb.

<h3>Acceleration of the system</h3>

The acceleration of the system is calculated as follows;

W - T = m₂a --- (1)

T = m₁a ----(2)

μmgsinθ - m₁a = m₂a

(0.3 x 3 x 9.8 x sin40) - (0.4 + 0.2)a = 3a

5.67 - 0.6a = 3a

5.67 = 3.6a

a = 5.67/3.6

a = 1.58 m/s²

<h3> Relative Weight of P</h3>

W = ma

W = 0.4 x 1.58

W = 0.632 N = 0.14 lb

Learn more about weight here: brainly.com/question/2337612

#SPJ1

3 0
2 years ago
Marvin Martian is standing on planet Potatoine.
prisoha [69]

Answer:

W = 113.98 N

Explanation:

Given that,

Radius of PotatoineR=6.4\times 10^6\ m

Mass of Potatoine, M=2\times 10^{24}\ kg

Mass of Marvin, m = 35 kg

We need to find his weight on Potatoine. Weight of an object is given by :

W = mg

g is acceleration due to gravity, g=\dfrac{GM}{R^2}

So,

W=\dfrac{GMm}{R^2}\\\\W=\dfrac{6.67\times 10^{-11}\times 35\times 2\times 10^{24}}{(6.4\times 10^6)^2}\\\\W=113.98\ N

So, his weight on Potatoine is 113.98 N.

4 0
3 years ago
Am I right? Please help!
JulsSmile [24]

I believe you are right.

3 0
3 years ago
You throw a small rock straight up from the edge of a highway bridge that crosses a river. The rock passes you on its way down,
Agata [3.3K]

Answer:

Explanation:

for vertical movement , time to reach the top = time to reach the hand = 2.5 s

v = u - gt

At the top , v = 0 , time t = 2.5 s

0 = u - g x 2.5

u = 2.5 x 9.8 = 24.5 m /s

velocity of throw = 24.5 m /s

So , when it passes the hand on its way down , it will have velocity equal to 24.5 m /s and it will accelerate downwards . Let its velocity down by 22 m be v

v² = u² + 2 g s

= 24.5² + 2 x 9.8 x 22

= 600.25 + 431.2

= 1031.45

v = 32.11 m /s .

7 0
3 years ago
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