⭐ 𝑬𝒍𝒆𝒄𝒕𝒓𝒊𝒄 𝒄𝒂𝒓 𝑫𝒖𝒃𝒊𝒏𝒂 𝑬𝒗𝒐 ⭐

Task 19.1

Determine the thickness of the ecosuede layer if 400 W was spent on changing the temperature of a surface area of ​​1 m^2 from 20 °C to 36 °C. The thermal conductivity coefficient of ecosuede is 0.25 W/m·K

Q = k· A· ΔT / d

d = k· A· ΔT / Q

d - thickness of the ecosuede layer, m;

k - thermal conductivity coefficient of ecosuede, W/m·K;

A - surface area, m^2;

ΔT - temperature difference, K;

Q - heating power, W

d = 0.25 · 1 · 16 / 400 = 0.01 m = 1 cm

Task 19.2

The driver exerts a pressure of 500 N on the back of a seat made of fiberglass with an area of ​​0.3 m^2 and a thickness of 5 mm. Determine how much the seat back will deflect if the Young's modulus of fiberglass is 70 GPa.

ΔL = F·d / A·E

ΔL - deflection of the seat back, m

F - force of pressure, F

d - thickness of fiberglass, m

A - surface area, m^2

E - Young's modulus, Pa

ΔL = 500·0.005 / 0.3·70·10^9 ≈ 0.119 nm

Task 19.3

A car seat is made of fiberglass with a density of 2 g/cm^3. The volume of the seat is 4 l. Determine the mass of the seat.

m = V · p

m = 0.004 m^3 · 2000 kg/m^3 = 8 kg

Task 20.1

An electric motor in the form of a cylinder made of aluminum weighing 50 kg cooled from 90 °C to 20 °C. What amount of heat did the motor give off if the specific heat capacity of aluminum is 900 J/kg·°C

Q = c · m · ΔT

Q = 900 · 50 · (90 - 20) = 3150000 J = 3150 kJ

Task 20.2

In a rectangular parallelepiped with two sides of 0.42 m and one side of 0.75 m, there is a cylinder with a base diameter of 0.24 m and a height of 0.7 m. Find the free volume in the parallelepiped.

Vf = Vp – Vc

Vp = a · b · c

Vc = π · r^2 · h

Vp = 0.42 · 0.42 · 0.75 = 0.1323 m^3

Vc = 3.14 · 0.0144 · 0.7 = 0.0316 m^3

Vf = 0.1323 – 0.0316 ≈ 0.1 m^3

Task 20.3

An electric motor with a surface area of ​​1.6 m^2 has a surface temperature of 90 °C. The motor is cooled by a turbulent air flow with a temperature of 20 °C at a speed of 9 m/s. Determine the heat flow from the motor to the air.

Q = h · A · ΔT

h = 10.45 + 10 · υ^0.8

Q – heat flux, W

h – heat transfer coefficient in turbulent flow W/(m^2 · °C)

A – surface area, m^2

ΔT – temperature difference, °C

υ – air flow velocity, m/s

h = 10.45 + 10 · 9^0.8 = 84.65 W/(m^2 · °C)

Q = 84.65 · 1.6 · 70 = 9471 W ≈ 9.5 kW

Task 21.1

The front trunk of a car has the shape of a regular truncated quadrangular pyramid. The sides of the lower base of the trunk are 80 cm and 45 cm. The sides of the upper base are 130 cm and 55 cm. The height of the trunk is 40 cm. Determine the volume of the trunk.

V = 1/3 · h · (S1 + S2 + (S1 · S2)^1/2)

S1 = a1 · b1

S2 = a2 · b2

S1 = 80 · 45 = 3600 cm^2

S2 = 130 · 55 = 7150 cm^2

V = 1/3 · 40 · (3600 + 7150 + 5073.5) = 210980 cm^3 = 211 l

Task 21.2

The rear trunk of a car has the shape of a regular truncated quadrangular pyramid. The sides of the lower base of the trunk are 48 cm and 75 cm. The sides of the upper base are 57 cm and 115 cm. The height of the trunk is 50 cm. Determine the volume of the trunk.

V = 1/3 · h · (S1 + S2 + (S1 · S2) ^ 1/2)

S1 = a1 · b1

S2 = a2 · b2

S1 = 48 · 75 = 3600 cm ^ 2

S2 = 57 · 115 = 6555 cm ^ 2

V = 1/3 · 50 · (3600 + 6555 + 4857.7) = 250211.7 cm ^ 3 = 250 l

Task 22.1

An arc with a central angle of 230 °C is formed on a circle with a diameter of 31 inches. What is the length of the arc?

L = α · C / 360°

C = π · d

L - length of the arc, m

α - central angle, °

C - length of the circle, m

d - diameter of the circle, m

C = 3.14 · 0.786 = 2.468

L = 230 · 2.468 / 360 = 1.577 m ≈ 1.58 m

Task 22.2

What is the area of ​​a sector of a circle with a diameter of 31 inches and an angle of 230 °?

Ss = α · Sc / 360°

Sc = π · r^2

Ss – area of ​​the sector, m ^ 2;

Sc – area of ​​the circle, m ^ 2;

α – angle of the sector, °

r – radius of the circle, m

Sc = 3.14 · 0.7874 ^ 2 = 1.9468 m^2

Ss = 230° · 1.9468 / 360° = 1.2438 m^2 ≈ 1,24 m^2

Task 22.3

A car wheel has a diameter of 55 cm and a width of 30 cm. What is the minimum width of the wheel arch if the wheel turns 30°?

L = Lw + l

l = 30°/360° ⋅ π ⋅ d

L - wheel arch width, cm

Lw - wheel width, cm

l - arc length when turning, cm

l = 30°/360° ⋅ 3.14 ⋅ 55 = 14.4 cm

L = 30 + 14.4 = 44.4 cm

Task 23.1

The bottom of a car consists of 3 parts, two semicircles with a diameter of 1.9 m and a rectangle with sides of 1.9 m, 2 m. Find the area of ​​the bottom.

Sb = 2 · S1 + S2

S1 = ½ π · r^2

S2 = a · b

S1 = ½ · 3.14 · (0.95)^2 = 1.4184 m^2

S2 = 1.9 · 2 = 3.8 m^2

Sb = 2 · 1.4184 + 3.8 ≈ 6.64 m^2

Task 24.1

A car door weighing 15 kg opens vertically upwards forming an angle of 90° with the car body. Find the force required to hold the door open.

F = m · g

F – force, N

m – door mass, kg

g – acceleration of gravity, m/s^2

F = 15 · 9.81 ≈ 147 N

Task 24.2

A car door weighing 15 kg was lifted to a height of 1 m. What is the potential energy of the door?

Ep = m · g · h

Ep = 15 · 9.8 · 1 = 147 J

Task 25.1

A ray of light falls on a flat boundary of air and glass at an angle of 30°. The refractive index of air is 1, the refractive index of glass is 1.5. Determine the angle of the refracted ray.

n1 · sin(Ɵ1) = n2 · sin (Ɵ2)

1 · sin(30°) = 1.5 · sin(Ɵ2)

sin(30°) = 0.5

0.5 = 1.5 · sin(Ɵ2)

sin(Ɵ2) = 0.5/1.5 = 0.3333

Ɵ2 = arcsin (0.3333)

Ɵ2 ≈ 19.5°

Task 25.2

A car windshield has an area of ​​1.2 m^2. The oncoming air flow velocity is 30 m/s, and the air density is 1.2 kg/m^3 . Determine the air resistance force on the windshield if the aerodynamic drag coefficient of the glass is 0.3

F = Cd · P · A

P = ½ · p · v^2

F - air resistance force, N;

Cd - aerodynamic drag coefficient of the glass;

P - aerodynamic pressure, Pa;

A - glass area, m^2;

p - air density, kg/m^3;

v - air flow velocity, m/s

P = ½ · 1.2 · 30^2 = 540 Pa

F = 0.3 · 540 · 1.2 = 194.4 N

Task 25.3

What is the refractive index of a polymer film if the angle of incidence of a light beam is 50° and the angle of refraction is 30°

n = sin α/sin β ≈ 1.5

n = sin 50°/sin 30° = 0.766/0.5 ≈ 1.5

Task 25.4

In a rectangular trapezoid, the bases are 1 m and 1.2 m. The lateral side forms an angle of 70° with the base. Calculate the smaller lateral side.

h = Δ · tan(70°)

Δ = b – a

Δ = 1.2 – 1 = 0.2 m

h = 0.2 · 2.747 = 0.549 m

Task 26.1

A windshield wiper blade moves along the plane of the glass. The force of the blade on the glass is 5 N. The friction force is 1 N. Determine the coefficient of sliding friction.

µ = F/N

µ = 1/5 = 0.2

Task 26.2

A 0.6 m long windshield wiper blade moves along an arc with a rotation angle of 120°. Find the area of ​​the glass being cleaned by the wiper.

S = π · R^2 · 120°/360°

S – sector area, m^2

R – blade length, m

Ɵ – blade rotation angle, rad

S = 3.14 · 0.6^2 · 120°/360° = 0.376 m^2

Task 26.3

When moving a windshield wiper 1 m away, it does 5 J of work. What is the friction force?

F = A/S

F = 5/1 = 5 N

Task 27.1

The surface area of ​​the side panel of the car body is 1.4 m^2. What volume of paint is needed for painting if the consumption is 0.25 l/m^2

V = S · R

V = 1.4 · 0.25 = 0.35 l

Task 27.2

A segment DE is drawn in triangle ABC parallel to side AC (end D of the segment lies on side AB, and E on side BC). Find AD if AB = 160 cm; AC = 200 cm; DE = 150 cm

Segment DE parallel to side AC divides the triangle into two similar triangles ΔADE and ΔABC. According to the property of similar triangles, the ratio of the lengths of the corresponding sides is the same.

AD/AB = DE/AC

AD/160 = 150/200

AD = 160 · ¾ = 120 cm

Task 28.1

What is the area of ​​parallelogram ABCD if side AB = 0.2 m; BC = 1 m; and angle A = 70°

S = AB · BC · sin(180°-α)

S = 0.2 · 1 · 0.9397 = 0.188 m^2

Task 28.2

What volume of paint is required to paint a fender with an area of ​​0.2 m^2 if the paint consumption is 0.25 l/m^2

V = S · R

V = 0.2 · 0.25 = 0.05 l

Task 28.3

In parallelogram ABCD perimeter ABCD = 2.4 m, perimeter ABD = 2.1 m. Calculate BD.

P ABD = AB + BD + AD

BD = P ABD - (AB + AD)

AB + AD = P ABCD/2

BD = 2.1 – 1.2 = 0.9 m

Task 29.1

An air flow with a speed of 25 m/s acts on a car bumper with an area of ​​1.5 m^2. The aerodynamic drag coefficient of the bumper is Cd = 0.3, air density is p = 1.2 kg/m^3. Determine the air resistance force acting on the bumper.

F = ½ · Cd · p · A · V^2

F = ½ · 0.3 · 1.2 · 1.5 · 25^2 ≈ 169 N

Task 29.2

In an isosceles trapezoid, the bases are 2 m and 0.8 m, the lateral side is 0.7 m. What is the area of ​​the trapezoid?

AF = (AD – BC)/2

BF = (AB^2 – AF^2)^0,5

S = ½ (AD + BC) · BF

BF = (0,7^2 – 0,6^2)^0,5 = (0,13)^0,5 = 0,36

S = 0,5 · (2 + 0,8) · 0,36 ≈ 0,5 m

Task 29.3

A car weighing 1200 kg moves at a speed of 5 m/s and collides with a stationary obstacle. The car's bumper deforms and absorbs the impact energy. Calculate the work done by the bumper.

W = ½ · m · υ^2

W = ½ · 1200 · 5^2 = 15000 J = 15 kJ

Task 30.1

A LED lamp with a diameter of 5 cm creates a light spot through a diverging lens with a focal length of 8 cm. Calculate the diameter of the light spot if the distance from the lamp to the lens is 8 cm, and the distance from the lens to the spot is 10 m.

Di = M · D0

M = di/d0

Di - diameter of the light spot, m

M - magnification

M = 100/8 = 125

Di = 125 · 5 = 625 cm