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3 years ago

Which is NOT a " big idea " involving force and motion ?

1 answer:

3 0

A: In theory, no. According to Newton's first law, a body in motion will remain in motion. ... Objects do not stop because of a lack of force; they stop because a different force is being applied. For example, friction is a force that acts against an object's motion.

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Which one of the following parenting factors is associated with improved IQ scores in children?

lapo4ka [179]

I would say B. A stable home and varied activities

5 0

3 years ago

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

4 years ago

A mass suspended from a spring is oscillating up and down as indicated. Consider the following possibilities. A At some point du

dangina [55]

A mass suspended from a spring is oscillating up and down, (as stated but not indicated).

A). At some point during the oscillation the mass has zero velocity but its acceleration is non-zero (can be either positive or negative). <em>Yes. </em> This statement is true at the top and bottom ends of the motion.

B). At some point during the oscillation the mass has zero velocity and zero acceleration. No. If the mass is bouncing, this is never true. It only happens if the mass is hanging motionless on the spring.

C). At some point during the oscillation the mass has non-zero velocity (can be either positive or negative) but has zero acceleration. <em>Yes.</em> This is true as the bouncing mass passes through the "zero point" ... the point where the upward force of the stretched spring is equal to the weight of the mass. At that instant, the vertical forces on the mass are balanced, and the net vertical force is zero ... so there's no acceleration at that instant, because (as Newton informed us), A = F/m .

D). At all points during the oscillation the mass has non-zero velocity and has nonzero acceleration (either can be positive or negative). No. This can only happen if the mass is hanging lifeless from the spring. If it's bouncing, then It has zero velocity at the top and bottom extremes ... where acceleration is maximum ... and maximum velocity at the center of the swing ... where acceleration is zero.

7 0

3 years ago

Calculate the activation energy, E a Ea , in kilojoules per mole for a reaction at 57.0 ∘ C 57.0 ∘C that has a rate constant of

Alecsey [184]

Explanation:

Below is an attachment containing the solution

3 0

3 years ago

Use Stefan's law to find the intensity of the cosmic background radiation emitted by the fireball of the Big Bang at a temperatu

Rus_ich [418]

Complete Question

Use Stefan's law to find the intensity of the cosmic background radiation emitted by the fireball of the Big Bang at a temperature of 2.81 K. Remember that Stefan's Law gives the Power (Watts) and Intensity is Power per unit Area (W/m2).

Answer:

The intensity is $I = 3.535 *10^{-6} \ W/m^2$