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Elza [17]
3 years ago
14

Some make-up removal products suggest shaking the solution before using it. Why would the manufacturers provide this suggestion?

Physics
2 answers:
maks197457 [2]3 years ago
6 0
B. The solution separates naturally.
SSSSS [86.1K]3 years ago
6 0
B.The solution separates naturally. 
If the ingredients are non-polar to water (which is usually the main ingredient in removal products) then the bottle would have to be physically shaken up to distribute both ingredients and get the desired result.  Over time the non-polar ingredient will float on top of the water.
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Often times we think about insulation, we think about keeping cold items cold, such as drinks in an ice chest or the cool temper
Burka [1]

Answer:

So, insulation essentially works by creating a sort of barrier between the hot and the cold object. This barrier helps to reduce heat transfer by either reflecting the thermal radiation or by decreasing thermal conduction and convection from one object to the other.

4 0
2 years ago
A spring is used to stop a 50-kg package which is moving down a 20º incline. The spring has a constant k = 30 kN/m and is held b
Elina [12.6K]

Answer:

0.3 m

Explanation:

Initially, the package has both gravitational potential energy and kinetic energy.  The spring has elastic energy.  After the package is brought to rest, all the energy is stored in the spring.

Initial energy = final energy

mgh + ½ mv² + ½ kx₁² = ½ kx₂²

Given:

m = 50 kg

g = 9.8 m/s²

h = 8 sin 20º m

v = 2 m/s

k = 30000 N/m

x₁ = 0.05 m

(50)(9.8)(8 sin 20) + ½ (50)(2)² + ½ (30000)(0.05)² = ½ (30000)x₂²

x₂ ≈ 0.314 m

So the spring is compressed 0.314 m from it's natural length.  However, we're asked to find the additional deformation from the original 50mm.

x₂ − x₁

0.314 m − 0.05 m

0.264 m

Rounding to 1 sig-fig, the spring is compressed an additional 0.3 meters.

8 0
3 years ago
On a straight road, a car speeds up at a constant rate from rest to 20 m/s over a 5 second interval and a truck slows at a const
IceJOKER [234]

Answer:

a)

Explanation:

  • Since the car speeds up at a constant rate, we can use the kinematic equation for distance (assuming that the initial position is x=0, and choosing t₀ =0), as follows:

        x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2}   (1)

  • Since the car starts from rest, v₀ =0.
  • We know the value of t = 5 sec., but we need to find the value of a.
  • Applying the definition of acceleration, as the rate of change of velocity with respect to time, and remembering that v₀ = 0 and t₀ =0, we can solve for a, as follows:

       a_{c} =\frac{v_{fc}}{t} = \frac{20m/s}{5s} = 4 m/s2  (2)

  • Replacing a and t in (1):

       x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2}  = \frac{1}{2}*a*t^{2} = \frac{1}{2}* 4 m/s2*(5s)^{2} = 50.0 m.  (3)

  • Now, if the truck slows down at a constant rate also, we can use (1) again, noting that v₀ is not equal to zero anymore.
  • Since we have the values of vf (it's zero because the truck stops), v₀, and t, we can find the new value of a, as follows:

       a_{t} =\frac{-v_{to}}{t} = \frac{-20m/s}{10s} = -2 m/s2  (4)

  • Replacing v₀, at and t in (1), we have:

       x_{ft} = 20m/s*10.0s + \frac{1}{2}*(-2 m/s2)*(10.0s)^{2} = 200m -100m = 100.0m   (5)

  • Therefore, as the truck travels twice as far as the car, the right answer is a).
7 0
3 years ago
A stone is thrown horizontally at 8.0 m/s from a cliff 78.4 m high. How far from the base of the cliff does the stone strike the
Ivenika [448]
64 meters from the base of the cliff.
4 0
3 years ago
Read 2 more answers
The velocity of the transverse waves produced by an earthquake is 5.05 km/s, while that of the longitudinal waves is 8.585 km/s.
sattari [20]

Answer:

d=691.71km

Explanation:

The time lag between the arrival of transverse waves and the arrival of the longitudinal waves is defined as:

t=\frac{d}{v_t}-\frac{d}{v_l}

Here d is the distance at which the earthquake take place and v_t, v_l is the velocity of the transverse waves and longitudinal waves respectively. Solving for d:

t=d(\frac{1}{v_t}-\frac{1}{v_l})\\d=\frac{t}{\frac{1}{v_t}-\frac{1}{v_l}}\\d=\frac{56.4s}{\frac{1}{5.05\frac{km}{s}}-\frac{1}{8.585\frac{km}{s}}}\\d=691.71km

8 0
3 years ago
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