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vampirchik [111]

3 years ago

10

A ski hill is angled 20° above the horizontal. If a skier starts from rest at the top of the hill, how fast will she be travelin

g after she has traveled 62 m 62 ⁢ m down the slope? Assume that friction is negligible.

1 answer:

MaRussiya [10]3 years ago

6 0

Answer:

20 m/s

Explanation:

First, we calculate for the acceleration down the incline using $a = gsin\theta$ .

a = (9.80)sin(20°)

a = 3.35

Then, we use one of the kinematics equations, ${v_{f}}^{2} = {v_{i}}^{2} + 2 a\Delta s$ .

$v_{i}$ = 0 m/s

a = 3.35

$\Delta s$ = 62 m

$v_{f} = \sqrt{2(3.35)(62)}$

$v_{f} =20.381 = 20 m/s$

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The area of the piston to the master cylinder in a hydraulic braking system of a car is 0.4 square inches. If a force of 6.4 lb

Anit [1.1K]

Answer:

The force applied on one wheel during braking = 6.8 lb

Explanation:

Area of the piston (A) = 0.4 $in^{2}$

Force applied on the piston(F) = 6.4 lb

Pressure on the piston (P) = $\frac{F}{A}$

⇒ P = $\frac{6.4}{0.4}$

⇒ P = 16 $\frac{lb}{in^{2} }$

This is the pressure inside the cylinder.

Let force applied on the brake pad = $F_{1}$

Area of the brake pad ( $A_{1}$ )= 1.7 $in^{2}$

Thus the pressure on the brake pad ( $P_{1}$ ) = $\frac{F_{1} }{A_{1} }$

When brake is applied on the vehicle the pressure on the piston is equal to pressure on the brake pad.

⇒ P = $P_{1}$

⇒ 16 = $\frac{F_{1} }{A_{1} }$

⇒ $F_{1}$ = 16 × $A_{1}$

Put the value of $A_{1}$ we get

⇒ $F_{1}$ = 16 × 1.7

⇒ $F_{1}$ = 27.2 lb

This the total force applied during braking.

The force applied on one wheel = $\frac{F_{1} }{4}$ = $\frac{27.2}{4}$ = 6.8 lb

⇒ The force applied on one wheel during braking.

7 0

3 years ago

Qual o valor a ser pago, no final de 5 meses e 18 dias, correspondente a um empréstimo de $125.000,00, sabendo-se que a taxa de

liberstina [14]

25% ao semestre = 25%/6 ao mes.
Um mes tem em media 365/12 dias, ou 30,42 dias/mes.
18 dias = 18/30,42 mes

5 meses + 18 dias = 5 meses + 18/30,42 mes = 5,592 mes

a taxa de juros por 5 meses e 18 dias e 25%/6 * 5.592 = 23,3%

123,3% * $125.000 = $154.125,00

6 0

4 years ago

Read 2 more answers

(a) How many kilograms of water must evaporate from a 60.0-kg woman to lower their body temperature by 0.750ºC?

Mrrafil [7]

Answer:

69.69 g

Explanation:

Evaporation of water will take out latent heat of vaporization. Let the mass of water be m and latent heat of vaporization of water be 2260000 J per kg

Heat taken up by evaporating water

= 2260000 x m J

Heat lost by body

= mass x specific heat of body x drop in temperature

60 x 3500 x .750 ( specific heat of human body is 3.5 kJ/kg.k)

= 157500 J

Heat loss = heat gain

2260000 m= 157500

m = .06969 kg

= 69.69 g

5 0

4 years ago

What is the total entropy change if a 0.280 efficiency engine takes 3.78E3 J of heat from a 3.50E2 degC reservoir and exhausts i

babymother [125]

Answer:

$\Delta S=1.69J/K$

Explanation:

We know,

$\eta=1-\frac{T_2}{T_1}=1-\frac{Q_2}{Q_1}$ ..............(1)

where,

η = Efficiency of the engine

T₁ = Initial Temperature

T₂ = Final Temperature

Q₁ = Heat available initially

Q₂ = Heat after reaching the temperature T₂

Given:

η =0.280

T₁ = 3.50×10² °C = 350°C = 350+273 = 623K

Q₁ = 3.78 × 10³ J

Substituting the values in the equation (1) we get

$0.28=1-\frac{Q_2}{3.78\times 10^{3}}$

or

$\frac{Q_2}{3.78\times 10^3}=0.72$

or

$Q_2=3.78\times 10^3\times0.72$

⇒ $Q_2 =2.721\times 10^3 J$

Now,

The entropy change ( $\Delta S$ ) is given as:

$\Delta S=\frac{\Delta Q}{T_1}$

or

$\Delta S=\frac{Q_1-Q_2}{T_1}$

substituting the values in the above equation we get

$\Delta S=\frac{3.78\times 10^{3}-2.721\times 10^3 J}{623K}$

$\Delta S=1.69J/K$

7 0

3 years ago

Energy can do blank or bring about change

Mrac [35]

Work

Explanation:

Energy can do work or bring about change.

Energy is simply the ability to do work. When work is done, changes occur in a body.

There are different forms of energy and they are all related to work done.
When work is done on a body, a force causes a change in that body.
Both potential and kinetic energy are the same as work done.
Energy is needed to do work.
Work is done when a body changes energy.

Learn more:

Energy brainly.com/question/5416146

#learnwithBrainly

4 0

3 years ago