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hodyreva [135]
3 years ago
7

Need help science the top and bottom! Please

Physics
1 answer:
Oksi-84 [34.3K]3 years ago
7 0

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A 12,000-N car is raised using a hydraulic lift, which consists of a U-tube with arms of unequal areas, filled with incompressib
Natasha_Volkova [10]

Answer:

F = 1958.4 N

Explanation:

By volume conservation of the fluid on both sides we can say that volume of fluid displaced on the side of the car must be equal to the volume of fluid on the other side

so we have

L_1A_1 = L_2A_2

1.20(\pi 18^2) = L_2(\pi 5^2)

L_2 = 15.55 m

so the car will lift upwards by distance 1.2 m and the other side will go down by distance 15.55 m

So here the net pressure on the smaller area is given as

P = P_{atm} + \frac{12,000}{\pi (0.18)^2} + \rho g (1.2 + 15.55)

excess pressure exerted on the smaller area is given as

P_{ex} = \frac{12000}{\pi (0.18)^2} + 800(9.81)(16.75)

P_{ex} = 2.49\times 10^5 Pascal

now the force required on the other side is given as

F = P_ex (area)

F = (2.49 \times 10^5)(\pi (0.05)^2)

F = 1958.4 N

3 0
4 years ago
What is the difference between pitch and volume?
Fantom [35]

Answer:

Pitch is a measure of how high or low something sounds and is related to the speed of the vibrations that produce the sound. Volume is a measure of how loud or soft something sounds and is related to the strength of the vibrations.

Explanation:

4 0
3 years ago
A 560-g squirrel with a surface area of 930cm2 falls from a 5.0-m tree to the ground. Estimate its terminal velocity. (Use a dra
bija089 [108]

Explanation:

Terminal velocity is given by:

v_t=\sqrt{\frac{2mg}{\rho C_dA}}

Here, m is the mass of the falling object, g is the gravitational acceleration,  C_d is the drag coefficient, \rho is the fluid density through which the object is falling, and A is the projected area of the object. in this case the projected area is given by:

A=\frac{A_s}{2}=\frac{930cm^2}{2}=465cm^2\\465cm^2*\frac{1m^2}{10^4cm^2}=0.0465m^2\\560g*\frac{1kg}{10^3g}=0.56kg

Recall that drag coefficient for a horizontal skydiver is equal to 1 and air density is 1.28\frac{kg}{m^3}.

v_t=\sqrt{\frac{2(0.56kg)(9.8\frac{m}{s^2})}{(1.28\frac{kg}{m^3}(1)(0.0465m^2)}}\\v_t=13.58\frac{m}{s}

Without drag contribution the motion of the person is an uniformly accelerated motion, thus:

v_f^2=v_o^2+2gh\\v_f=\sqrt{2gh}\\v_f=\sqrt{2(9.8\frac{m}{s^2})(5m)}\\v_f=9.9\frac{m}{s}

7 0
3 years ago
Two parallel-plate capacitors have the same plate area. Capacitor 1 has a plate separation twice that of capacitor 2, and the qu
Luba_88 [7]

Answer:

V_1=8 V_2

Explanation:

Given that:

  • Area of the plate of capacitor 1= Area of the plate of capacitor 2=A
  • separation distance of capacitor 2, d_2=d
  • separation distance of capacitor 1, d_1=2d
  • quantity of charge on capacitor 2, Q_2=Q
  • quantity of charge on capacitor 1, Q_1=4Q

We know that the Capacitance of a parallel plate capacitor is directly proportional to the area and inversely proportional to the distance of separation.

Mathematically given as:

C=\frac{k.\epsilon_0.A}{d}.....................................(1)

where:

k = relative permittivity of the dielectric material between the plates= 1 for air

\epsilon_0 = 8.85\times 10^{-12}\,F.m^{-1}

From eq. (1)

For capacitor 2:

C_2=\frac{k.\epsilon_0.A}{d}

For capacitor 1:

C_1=\frac{k.\epsilon_0.A}{2d}

C_1=\frac{1}{2} [ \frac{k.\epsilon_0.A}{d}]

We know, potential differences across a capacitor is given by:

V=\frac{Q}{C}..........................................(2)

where, Q = charge on the capacitor plates.

for capacitor 2:

V_2=\frac{Q}{\frac{k.\epsilon_0.A}{d}}

V_2=\frac{Q.d}{k.\epsilon_0.A}

& for capacitor 1:

V_1=\frac{4Q}{\frac{k.\epsilon_0.A}{2d}}

V_1=\frac{4Q\times 2d}{k.\epsilon_0.A}

V_1=8\times [\frac{Q.d}{k.\epsilon_0.A}]

V_1=8 V_2

6 0
3 years ago
A 1800-kg Jeep travels along a straight 500-m portion of highway (from A to B) at a constant speed of 10 m/s. At B, the Jeep enc
vladimir1956 [14]

Answer:

3600N

Explanation:

F=m\frac{v^2}{r}

4 0
3 years ago
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