Applied sciences refers to the study of scientific principles. True False

What force must be exerted on the master cylinder of a hydraulic lift to support the weight of a 2000-kg car (a large car) resti

Nataliya [291]

Archimedes principle states that

F1 / A1 = F2 / A2

F2 = (A2 / A1) * F1

Also, formula for the force is F = mg. Formula for the area of the cylinder is A = πr^2, therefore we get

F2 = (πr2^2 / πr1^2) * mg

Since the diameter of the cylinders are 2 cm and 24 cm, r1 = 12 and r2 = 1.

Substituting the values to the derived equation, we get

F2 = (π 1^2 / π 12^2) * 2400 * 9.8

F2 = 163.3333 N

6 0

1 year ago

A glider is gliding through the air at a height of 416 meters with a speed of 45.2 m/s. The glider dives to a height of 278 mete

Verdich [7]

Answer:

The glider's new speed is 68.90 m/s

Explanation:

Principle Of Conservation Of Mechanical Energy

The mechanical energy of a system is the sum of its kinetic and potential energy. When the only potential energy considered in the system is related to the height of an object, then it's called the gravitational potential energy. The kinetic energy of an object of mass m and speed v is

$\displaystyle K=\frac{1}{2}mv^2$

The gravitational potential energy when it's at a height h from the zero reference is

$U=mgh$

The total mechanical energy is

$M=K+U$

$\displaystyle M=\frac{1}{2}mv^2+mgh$

The principle of conservation of mechanical energy states the total energy is constant while no external force is applied to the system. One example of a non-conservative system happens when friction is considered since part of the energy is lost in its thermal manifestation.

The initial conditions of the problem state that our glider is glides at 416 meters with a speed of 45.2 m/s. The initial mechanical energy is

$\displaystyle M_1=\frac{1}{2}m(45.2)v_o^2+m(9.8)(416)$

Operating in terms of m

$\displaystyle M_1=1021.52m+4076.8m$

$\displaystyle M_1=5098.32m$

Then we know the glider dives to 278 meters and we need to know their final speed, let's call it $v_f$ . The final mechanical energy is

$\displaystyle M_2=\frac{1}{2}mv_f^2+m(9.8)(278)$

Operating and factoring

$\displaystyle M_2=m(\frac{1}{2}v_f^2+2724.4)$

Both mechanical energies must be the same, so

$\displaystyle m(\frac{1}{2}v_f^2+2724.4)=5098.32m$

Simplifying by m and rearranging

$\displaystyle \frac{v_f^2}{2}=5098.32-2724.4$

Computing

$v_f=\sqrt{4747.84}=68.90\ m/s$

The glider's new speed is 68.90 m/s

8 0

2 years ago

The air around a pool and the water in the pool receive equal amounts of energy from the sun. Why does the air experience a grea

labwork [276]

Answer:

A

Explanation:

4 0

1 year ago

Which statements accurately describe Dmitri Mendeleev’s contributions to the development of the periodic table? Check all that a

Mama L [17]

Mendeleev produced the first orderly arrangement of known elements. Mendeleev used patterns to predict undiscovered elements.

6 0

1 year ago

Read 2 more answers

Calculate the final temperature of a mixture of 0.350 kg of ice initially at 218°C and 237 g of water initially at 100.0°C.

kramer

Answer:

115 ⁰C

Explanation:

Step 1: The heat needed to melt the solid at its melting point will come from the warmer water sample. This implies

$q_{1} +q_{2} =-q_{3}$ -----eqution 1

where,

$q_{1}$ is the heat absorbed by the solid at 0⁰C

$q_{2}$ is the heat absorbed by the liquid at 0⁰C

$q_{3}$ the heat lost by the warmer water sample

Important equations to be used in solving this problem

$q=m *c*\delta {T}$ , where -----equation 2

q is heat absorbed/lost

m is mass of the sample

c is specific heat of water, = 4.18 J/0⁰C

$\delta {T}$ is change in temperature

Again,

$q=n*\delta {_f_u_s}$ -------equation 3

where,

q is heat absorbed

n is the number of moles of water

tex]\delta {_f_u_s}[/tex] is the molar heat of fusion of water, = 6.01 kJ/mol

Step 2: calculate how many moles of water you have in the 100.0-g sample

$=237g *\frac{1 mole H_{2} O}{18g} = 13.167 moles of H_{2}O$

Step 3: calculate how much heat is needed to allow the sample to go from solid at 218⁰C to liquid at 0⁰C

$q_{1} = 13.167 moles *6.01\frac{KJ}{mole} = 79.13KJ$

This means that equation (1) becomes

79.13 KJ + $q_{2} = -q_{3}$

Step 4: calculate the final temperature of the water

$79.13KJ+M_{sample} *C*\delta {T_{sample}} =-M_{water} *C*\delta {T_{water}$

Substitute in the values; we will have,

$79.13KJ + 237*4.18\frac{J}{g^{o}C}*(T_{f}-218}) = -350*4.18\frac{J}{g^{o}C}*(T_{f}-100})$

79.13 kJ + 990.66J* $(T_{f}-218})$ = -1463J* $(T_{f}-100})$

Convert the joules to kilo-joules to get

79.13 kJ + 0.99066KJ* $(T_{f}-218})$ = -1.463KJ* $(T_{f}-100})$

$79.13 + 0.99066T_{f} -215.96388= -1.463T_{f}+146.3$

collect like terms,

2.45366 $T_{f}$ = 283.133

∴ $T_{f} =$ = 115.4 ⁰C

Approximately the final temperature of the mixture is 115 ⁰C

6 0

2 years ago