Read the claim about caffeine. Caffeine improves mental alertness and motor coordination. A university research study was conduc

ted to examine how caffeine affects fine motor skills. Ten teen girls, ages 14 to 18, were given 200 mg of caffeine per day for 30 days. Thirty minutes after taking the caffeine, each subject took a timed fine motor skills test in which they fitted 20 small beads onto pegs. The graph compares each subject’s average results on the same test done 30 minutes before and after taking the caffeine. Which group would most benefit from the claim? people who sell coffee with caffeine people who sell decaffeinated tea diners that serve only drinks without caffeine manufacturers of caffeine-free bottled fruit juice

Physics

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A point charge with charge q1 is held stationary at the origin. A second point charge with charge q2 moves from the point (x1, 0

kicyunya [3294]

Answer:

$W=kq_1q_2(\dfrac{1}{x_1}-\dfrac{1}{\sqrt{x_2^2+y_2^2}})$

Explanation:

The position of the charge q₁ is established at (0,0)

Meanwhile, the charge q₂ is located at (x₁,0)

Thus, the electric potential energy between these two charges is determined by:

$U_1=k\dfrac{q_1q_2}{x_1}$

Now, the location of charge q₂ shifts from (x₁,0) to (x₂,y₂). The updated electric potential energy between the charges can be represented as:

$U_2=k\dfrac{q_1q_2}{\sqrt{x_2^2+y_2^2}}$

According to the work-energy theorem, the alteration in potential energy corresponds to the work performed. This is expressed mathematically as:

$W=-\Delta U$

$W=-(U_2-U_1)$

$W=(U_1-U_2)$

$W=(k\dfrac{q_1q_2}{x_1}-k\dfrac{q_1q_2}{\sqrt{x_2^2+y_2^2}})$

$W=kq_1q_2(\dfrac{1}{x_1}-\dfrac{1}{\sqrt{x_2^2+y_2^2}})$

Consequently, the work done by the electrostatic force on the moving charge is $kq_1q_2(\dfrac{1}{x_1}-\dfrac{1}{\sqrt{x_2^2+y_2^2}})$ . Therefore, this concludes the solution.

3 0

2 months ago

An ideal spring is mounted horizontally, with its left end fixed. The force constant of the spring is 170 N/m. A glider of mass

kicyunya [3294]

Answer:

A) The updated amplitude = 0.048 m

B) Period T = 0.6 seconds

Explanation: Please refer to the attached documents for the solution.

4 0

2 months ago

A Federation starship (8.5 ✕ 106 kg) uses its tractor beam to pull a shuttlecraft (1.0 ✕ 104 kg) aboard from a distance of 14 km

Keith_Richards [3271]

Consider the diagram below.

m₁ = 8.5 x 10⁶ kg, the starship's mass
m₂ = 10⁴ kg, the shuttlecraft's mass
a₁ = acceleration of the starship
a₂ = the acceleration of the shuttle
F = 4 x 10⁴ N, the force exerted

Let y represent the distance covered by the starship
Let x denote the distance covered by the shuttlecraft
If t indicates the travel time, then
y = 0.5a₁t² (1)
x = 0.5a₂t² (2)

F = m₁a₁ = m₂a₂ (3)
Additionally,
x + y = 14000 m (4)

From (2), we derive
a₁ = (4 x 10⁴ N)/(8.5 x 10⁶ kg) = 4.706 x 10⁻³ m/s²
a₂ = (4 x 10⁴ N)/(10⁴ kg) = 4 m/s²

From (1), (2) and (4), we find
0.5*(t s)²*(4 + 4.706 x 10⁻³ m/s²) = 14000 m
2.002353t² = 14000
t² = 6991.774 s²
t = 83.617 s

Thus
x = 0.5*4*6991.774 = 13984 m = 13.984 km
y = 0.5*4.706 x 10⁻³*6991.774 = 16.452 m

The starship moves roughly 16.5 meters while towing the shuttlecraft by 13.98 kilometers.

Result: The starship shifts by 16.5 m (to the nearest tenth)

3 0

2 months ago