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alexira [117]
2 years ago
6

The pulley system has a mechanical advantage of 5. Determine how much input force is required to achieve 1000 N of lifting force

.
Physics
1 answer:
Katen [24]2 years ago
8 0
I don’t think I need anything from my dad to get my hair fixed so I’m just bored I don’t want you guys
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3 years ago
I'm not sure if the answer is A or B... someone help
Talja [164]
Its b i literally have had this exact question
8 0
3 years ago
A car going at v = 29.7 m/s (67 mph) rounds a curve of radius R = 50.0 m, where the road is banked at an angle of θ = 30.0°. Wha
Nikolay [14]

Answer:

μ = 0.6

Explanation:

given,

speed of car = 29.7 m/s

Radius of curve = 50 m

θ = 30.0°

minimum static friction = ?

now,

writing all the forces acting along y-direction

N cos θ - f sinθ = mg

N cos θ -μN sinθ = mg

N = \dfrac{m g}{cos\theta-\mu sin \theta}

now, writing the forces acting along x- direction

N sin θ + f cos θ = F_{net}

N cos θ + μN sinθ = F_{net}

\dfrac{m g}{cos\theta-\mu sin \theta}(cos \theta + \mu sin\theta)=F_{net}

taking cos θ from nominator and denominator

F_{net} =\dfrac{tan\theta + \mu}{1-\mutan\theta}. mg

\dfrac{mv^2}{r}=\dfrac{tan\theta + \mu}{1-\mutan\theta}. mg

\dfrac{v^2}{r}=\dfrac{tan\theta + \mu}{1-\mutan\theta}g

\mu=\dfrac{v^2 -r g tan\theta}{v^2tan\theta + r g}

now, inserting all the given values

\mu=\dfrac{29.7^2 -50 \times 9.8tan 30^0}{29.7^2\times tan 30^0 +50 \times 9.8}

μ = 0.6

7 0
3 years ago
the area of 1 end of U-tube is 0.0 metre square and that of the other end is 1 metre square when you force was applied on the li
Lapatulllka [165]

Answer:

The question is wrong Since if you apply Force on 0.0m²It would mean That the pressure exerted=F/A=F/0

An since we can't divide a number by 0, the question is wrong

6 0
3 years ago
A sample of a diatomic ideal gas has pressure P and volume V. When the gas is warmed, its pressure triples and its volume double
Andreyy89

Answer:

Amount of  Energy transferred =8.5PV

Explanation:

Given:

Initial volume=V

Initial pressure=P

Final volume=2V

Final pressure=3P

Now w know that the Energy transferred in constant pressure pressure is given by

E_1=nc_pdT\\\\E_1=n\times\dfrac{7R}{2}dT\\E_1=3.5(nRdT)\\E_1=3.5(V(2P-P))\\E_1=3.5PV

Now the Energy transferred in constant volume process is given by

E_2=nc_vdT\\\\E_2=n\times\dfrac{5R}{2}dT\\E_1=2.5(nRdT)\\E_1=2.5(V(3V-V))\\E_1=5PV

The total Energy transferred is given by

E_{total}=E_1+E_2\\E{total}=3.5PV+5PV\\=8.5PV

3 0
3 years ago
Read 2 more answers
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