Which liquid is the most viscous?<br><br> syrup<br> water<br> milk<br> apple juice

If the torque required to loosen a nut that is holding a flat tire in place on a car has a magnitude of 41 N · m, what minimum f

Marina86 [1]

We have that the Force is mathematically given as

F=170.833N

<h3></h3><h3>Force </h3>

Question Parameters:

The torque required to loosen a nut that is holding a flat tire in place on a car has a magnitude of 41 N · m,
the end of a 24 cm-long wrench to <em>loosen </em>the nut

Generally the equation for the Force is mathematically given as

$F=\frac{T}{d}$

Therefore

F=41/0.24

F=170.833N

For more information on Force visit

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3 0

2 years ago

23

Svetllana [295]

C

Vegetable oil’s heat capacity is more as it can absorb more heat than water.

6 0

2 years ago

i need the answer asap plz A child rides her bike at a rate of 12.0 km/hr down the street. A squirrel suddenly runs in front of

Kruka [31]

-16 km/hr/sec because she slows down 4 km/hr (12 - 8) in 0.25 seconds. in one second (0.25 sec * 4) she would slow down by 16 km/hr (4 * 4). it’s negative acceleration because she’s slowing down

5 0

3 years ago

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A pair of eyeglass frames is made of epoxy plastic. At room temperature (20.0°C), the frames have circular lens holes 2.50 cm in

nignag [31]

Answer:

T₂ = 114 °C

Explanation:

Area or superficial expansion: can be defined as an increase in area, per unit area per degree rise in temperature. It unit is (1/K) or (1/°C).

β = ΔA/(A₁ΔT) ............................................. equation 1

β = 2α ......................................................... equation 2

Area of circle = πr².................................... equation 3

Where β = Area expansivity, α = linear expansivity, ΔA = increase in area = (A₂ - A₁), ΔT = change in temperature, A₁ = initial area, A₂ = Final area r = radius,

from the question, The coefficient of linear expansion for epoxy = 1.3 × 10⁻⁴ °C⁻¹,

∴ Coefficient of area expansion for epoxy = 2 × 1.3 × 10⁻⁴ =

β = 2.6 × 10⁻⁴ °C⁻¹,

Using equation 3 to calculate for area, and taking (π = 3.143)

r₁ = 2.5 cm ∴ A₁ = πr₁² = 3.143 × 2.5² =19.64 cm².

r₂ = 2.53 cm ∴ A₂ = πr₂² = 3.143 × 2.53² =20.12 cm².

ΔA = A₂ - A₁ = 20.12 - 19.64 = 0.48 cm².

Making ΔT the subject of the relation in equation 1.

ΔT = ΔA/(βA₁) ...................................... equation 4

Substituting the values above into equation 4,

ΔT = 0.48/(2.6 × 10⁻⁴ × 19.64)

ΔT = 0.48/(51.064 × 10⁻⁴)

ΔT =( 0.48/51.064) × 10000

ΔT = 0.0094 × 10000 = 94 °C

But, ΔT = T₂ -T₁,

Then, T₂ = ΔT + T₁

Where T₁ = 20 °C, ΔT =94 °C

∴ T₂ = 94 + 20 = 114 °C.

Therefore, temperature at which the frame must be heated if the the lenses 2.53 cm are to be inserted in them = 114 °C

3 0

3 years ago

Automobile A and B are initially 30 m apart travelling in adjacent highway lanes at speeds VA = 14.4 km/hr., VB 23.4 km/hr. at t

marshall27 [118]

Answer:

x = 240 m

Explanation:

This is a kinematics exercise

Let's fix our frame of reference on car A

x = x₀ₐ+ v₀ₐ t + ½ aₐ t²

the initial position of car a is zero

x = 0 + v₀ₐ t + ½ 0.8 t²

for car B

x = x_{ob} + v_{ob} t - ½ a_b t²

car B's starting position is 30 m

x = 30 + v_{ob} t - ½ 0.4 t²

at the point where they meet, the position of the two vehicles is the same

0 + v₀ₐ t + ½ 0.8 t² = 30 + v_{ob} t - ½ 0.4 t²

let's reduce the speeds to the SI system

v₀ₐ = 14.4 km / h (1000 m / 1 km) (1h / 3600s) = 4 m / s

v_{ob} = 23.4 km / h = 6.5 m / s

4 t + 0.4 t² = 30 + 6.5 t - 0.2 t²

0.2 t² - 2.5 t - 30 = 0

t² - 12.5 t - 150 = 0

we solve the quadratic equation

t = $\frac{12.5 \pm \sqrt{12.5^2 + 4 \ 150} }{2}$

t = $\frac{12.5 \ \pm 27.5}{2}$

t₁ = 20 s

t₂ = -7.5 s

time must be a positive quantity so the correct result is t = 20 s

let's look for the distance

x = 4 t + ½ 0.8 t²

x = 4 20 + ½ 0.8 20²

x = 240 m

8 0

3 years ago

Which liquid is the most viscous? syrup water milk apple juice