A study occasionally the effect of anxiety (low vs. high) and stress (low vs. moderate vs. high) on test.
Everyone experiences anxiety occasionally, but persistent anxiety can reduce your quality of life. Though likely best known for altering behavior, worry can have negative effects on our physical health. Anxiety speeds up our heartbeat and breathing, concentrating blood flow to the parts of our brains that need it. You are getting ready for a challenging situation by having this extremely bodily reaction. Test performance may be impacted by anxiety. According to studies, pupils with low levels of test anxiety perform better on multiple-choice question (MCQ) exams than pupils with high levels of anxiety. Studies have occasionally that female students have greater levels of test anxiety than male students.
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Answer:
y = 17 m
Explanation:
For this projectile launch exercise, let's write the equation of position
x = v₀ₓ t
y = 
t - ½ g t²
let's substitute
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
the maximum height the ball can reach where the vertical velocity is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
0 = v₀ sin θ - 9.8 t
Let's write our system of equations
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
0 = v₀ sin θ - 9.8 t
We have a system of three equations with three unknowns for which it can be solved.
Let's use the last two
v₀ sin θ = 9.8 t
we substitute
10 = (9.8 t) t - ½ 9.8 t2
10 = ½ 9.8 t2
10 = 4.9 t2
t = √ (10 / 4.9)
t = 1,429 s
Now let's use the first equation and the last one
45 = v₀ cos θ t
0 = v₀ sin θ - 9.8 t
9.8 t = v₀ sin θ
45 / t = v₀ cos θ
we divide
9.8t / (45 / t) = tan θ
tan θ = 9.8 t² / 45
θ = tan⁻¹ ( 9.8 t² / 45
)
θ = tan⁻¹ (0.4447)
θ = 24º
Now we can calculate the maximum height
v_y² = 
- 2 g y
vy = 0
y = v_{oy}^2 / 2g
y = (20 sin 24)²/2 9.8
y = 3,376 m
the other angle that gives the same result is
θ‘= 90 - θ
θ' = 90 -24
θ'= 66'
for this angle the maximum height is
y = v_{oy}^2 / 2g
y = (20 sin 66)²/2 9.8
y = 17 m
thisis the correct
Answer:
option C
Explanation:
given,
Force by the engine on plane in West direction = 350 N
Frictional force on the runway = 100 N in east
force exerted by the wind = 100 N in east
net force and direction = ?
consider west to be positive and east be negative.
when airplane will be moving there will be frictional as well as wind resistance will be acting in opposite direction of airplane
Net force = 350 N - 100 N - 100 N
= 150 N
as our answer comes out to be positive so the airplane will be moving in West
hence, the correct answer is option C
Answer:
Coefficient of friction = 0.5
Explanation:
Given:
Mass of box = 5 kg
Force applied = 20 N
Acceleration = 2 m/s²
Find:
Coefficient of friction
Computation:
Friction force = Mass x Acceleration.
Friction force = 5 x 2
Friction force = 10 N
Coefficient of friction = Friction force / Force applied
Coefficient of friction = 10 / 20
Coefficient of friction = 0.5
So the dimensions of acceleration is
Any answer that comes under that definition is correct.
