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3 years ago

How do psychologist avoid false conclusion

1 answer:

7 0

A false positive<span> is a conclusion that some effect occurred when in fact it did not. hope it helped but i am not 100%

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A velocity vs time graph what is the acceleration of the object

tigry1 [53]

Answer:

If the object is moving with an acceleration of +4 m/s/s (i.e., changing its velocity by 4 m/s per second), then the slope of the line will be +4 m/s/s.

Explanation:

7 0

3 years ago

Read 2 more answers

This graph shows the energy of a reaction over time. Which statement is

kirill115 [55]

Answer: D.

Explanation: Just took the quiz.

3 0

3 years ago

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Consider two particles A and B. The angular position of particle A, with constant angular acceleration, depends on time accordin

Vera_Pavlovna [14]

Answer:

θ = θ₀ + ½ w₀ (t -t_1) + α (t -t_1)²

Explanation:

This is an angular kinematic exercise the equation for the angular position

the particle A

θ = θ₀ + ω₀ t + ½ α t²

They say for the particle B

w₀B = ½ w₀

αB = 2 α

In addition, the particle begins at a time t_1 after particle A, in order to use the same timer, we must subtract this time from the initial

t´ = t - t_1

et's write the equation of particle B

θ = θ₀ + w₀B t´ + ½ αB t´2

replace

θ = θ₀ + ½ w₀ (t -t_1) + ½ 2α (t -t_1)²

θ = θ₀ + ½ w₀ (t -t_1) + α (t -t_1)²

4 0

3 years ago

Two satellites, A and B are in different circular orbits

jek_recluse [69]

Answer:

The ratio of the orbital time periods of A and B is $\frac{1}{2}$

Solution:

As per the question:

The orbit of the two satellites is circular

Also,

Orbital speed of A is 2 times the orbital speed of B

$v_{oA} = 2v_{oB}$ (1)

Now, we know that the orbital speed of a satellite for circular orbits is given by:

$v_{o} = \farc{2\piR}{T}$

where

R = Radius of the orbit

Now,

For satellite A:

$v_{oA} = \farc{2\piR}{T_{a}}$

Using eqn (1):

$2v_{oB} = \farc{2\piR}{T_{a}}$ (2)

For satellite B:

$v_{oB} = \farc{2\piR}{T_{b}}$ (3)

Now, comparing eqn (2) and eqn (3):

$\frac{T_{a}}{T_{b}} = \farc{1}{2}$

6 0

3 years ago

A small ball with mass 1.20 kg is mounted on one end of a rod 0.860 m long and of negligible mass. The system rotates in a horiz

Rufina [12.5K]

Answer:

0.88752 kgm²

0.02236 Nm

Explanation:

m = Mass of ball = 1.2 kg

L = Length of rod = 0.86 m

$\theta$ = Angle = 90°

Rotational inertia is given by

$I=mL^2\\\Rightarrow I=1.2\times 0.86^2\\\Rightarrow I=0.88752\ kgm^2$

The rotational inertia is 0.88752 kgm²

Torque is given by

$\tau=FLsin\theta\\\Rightarrow \tau=2.6\times 10^{-2}\times 0.86sin90\\\Rightarrow \tau=0.02236\ Nm$

The torque is 0.02236 Nm

5 0

3 years ago