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inessss [21]
3 years ago
14

Kathryn was in a minor car accident while driving her sister to school six months ago. Since then, she experiences anxiety whene

ver she has to drive. Her mother has encouraged Kathryn to drive even though she finds it stressful. In this scenario, the most likely stressor causing Kathryn to experience anxiety is . Kathryn's stressor is most likely related to her .
Physics
2 answers:
Westkost [7]3 years ago
8 0

Answer:

Kathryn's stressor is most likely related to her car accident.

Explanation:

Stressor: In psychology, the term "stressor" is described as any specific environmental stimulus, event, or experience that tends to cause stress in a particular individual or person. Therefore, these experiences or events are being perceived as challenges or threats to the person and it can be either psychological or physical.

Example: Divorce, getting married, death of a loved one, etc.

In the question above, "the most likely stressor causing Kathryn to experience anxiety is car accident".

valina [46]3 years ago
5 0

Answer:

driving a car. and  past experience

Explanation:

You might be interested in
Lucinda frequently finds herself getting upset and yelling at her husband, co-workers, and friends. What therapeutic technique w
IrinaK [193]

The therapeutic technique Lucinda would start is behavioral self-monitoring.

<u>Explanation:</u>

Behavioral self-monitoring, introduced by Mark Snyder in the 1970s, is an ability of an individual to regulate the behavior to accommodate in various social situations.

Here, Lucinda frequently gets upset and yells at her husband, co-workers, and friends. She wants to keep a record of where, when, or with whom she might lose her temper in order to evaluate her behavior.

The individuals with high levels of self-monitoring possess greater skill at bridging social situations and in relationships they seem to be more flexible and adaptable.

6 0
4 years ago
Andy is waiting at the signal. As soon as the light turns green, he accelerates his car at a uniform rate of 8.00 meters/second2
Tatiana [17]
You can reason it out like this:

-- The car starts from rest, and goes 8 m/s faster every second.

-- After 30 seconds, it's going (30 x 8) = 240 m/s.

-- Its average speed during that 30 sec is  (1/2) (0 + 240) = 120 m/s

-- Distance covered in 30 sec at an average speed of 120 m/s

                                                                           =  3,600 meters .
___________________________________

The formula that has all of this in it is the formula for
distance covered when accelerating from rest:

       Distance = (1/2) · (acceleration) · (time)²

                       = (1/2) ·      (8 m/s²)     · (30 sec)²

                       =      (4 m/s²)          ·      (900 sec²)

                       =            3600 meters.

_________________________________

When you translate these numbers into units for which
we have an intuitive feeling, you find that this problem is
quite bogus, but entertaining nonetheless.

When the light turns green, Andy mashes the pedal to the metal
and covers almost 2.25 miles in 30 seconds.

How does he do that ?

By accelerating at 8 m/s².  That's about 0.82 G  !

He does zero to 60 mph in 3.4 seconds, and at the end
of the 30 seconds, he's moving at 534 mph ! 

He doesn't need to worry about getting a speeding ticket.
Police cars and helicopters can't go that fast, and his local
police department doesn't have a jet fighter plane to chase
cars with.
3 0
3 years ago
Adam drops a ball from rest from the top floor of a building at the same time Bob throws a ball horizontally from the same locat
guapka [62]

Answer:

Both balls hit the ground at the same time

Explanation:

Adam drops the ball from rest, so the ball just "<em>falls</em>" in vertical direction, being gravity its only acceleration, for cinematic movements we use that:

y(t)=y_{0}+v_{0y}t+\frac{1}{2}gt^{2}

In this case we have that gravity is negative, and as Adam drops the ball, v_{0y}=0

Bob throws the ball horizontally, so the movement will be a <em>parabola</em>, we can divide into horizontal direction, and vertical direction.

But we only need to analize the vertical movement, in wich again the only acceleration is gravity, and compare it with Adam's ball. Again we have that gravity is negative, and as the initial throw is horizontal, v_{0y}=0

Finally, we have that

y(t)=h-\frac{1}{2}gt^{2}

where

h=y_{0}

both for Adam's vertical drop, and for Bob's vertical component of the parabolic throw.

Now, if we put y(t)=0 (the origin of the vertical coordinate), we get for both cases that

h=\frac{1}{2}gt^{2}

where we can clear the value for the time t, of the fall, wich will be the same in both cases.

Hence, both balls hit the ground at the same time.

3 0
3 years ago
4. Consider a 1 kg block is on a 45° slope of ice. It is connected to a 0.4 kg block by a cable
icang [17]

If an icy surface means no friction, then Newton's second law tells us the net forces on either block are

• <em>m</em> = 1 kg:

∑ <em>F</em> (parallel) = <em>mg</em> sin(45°) - <em>T</em> = <em>ma</em> … … … [1]

∑ <em>F</em> (perpendicular) = <em>n</em> - <em>mg</em> cos(45°) = 0

Notice that we're taking down-the-slope to be positive direction parallel to the surface.

• <em>m</em> = 0.4 kg:

∑ <em>F</em> (vertical) = <em>T</em> - <em>mg</em> = <em>ma</em> … … … [2]

<em />

Adding equations [1] and [2] eliminates <em>T</em>, so that

((1 kg) <em>g</em> sin(45°) - <em>T </em>) + (<em>T</em> - (0.4 kg) <em>g</em>) = (1 kg + 0.4 kg) <em>a</em>

(1 kg) <em>g</em> sin(45°) - (0.4 kg) <em>g</em> = (1.4 kg) <em>a</em>

==>   <em>a</em> ≈ 2.15 m/s²

The fact that <em>a</em> is positive indicates that the 1-kg block is moving down the slope. We already found the acceleration is <em>a</em> ≈ 2.15 m/s², which means the net force on the block would be ∑ <em>F</em> = <em>ma</em> ≈ (1 kg) (2.15 m/s²) = 2.15 N directed down the slope.

8 0
3 years ago
Two gliders move toward each other on a linear air track, which we assume is frictionless. Glider A has a mass of 0.50 kg, and g
Nataly_w [17]

Answer:

-0.4 m/s

-3.552 m/s

Explanation:

m_1 = Mass of first glider = 0.5 kg

m_2 = Mass of second glider = 0.3 kg

u_1 = Initial Velocity of first glider = 2 m/s

u_2 = Initial Velocity of second glider = -2 m/s

v_1 = Final Velocity of first glider

v_2 = Final Velocity of second glider = 2 m/s

As the linear momentum of the system is conserved we have

m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\Rightarrow v_1=\dfrac{m_1u_1+m_2u_2-m_2v_2}{m_1}\\\Rightarrow v_1=\dfrac{0.5\times 2+0.3\times (-2)-0.3\times 2}{0.5}\\\Rightarrow v_1=-0.4\ m/s

The velocity of glider A is -0.4 m/s

u_1 = 0

u_2 = -5 m/s

v_2 = 0.92 m/s

m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\Rightarrow v_1=\dfrac{m_1u_1+m_2u_2-m_2v_2}{m_1}\\\Rightarrow v_1=\dfrac{0.5\times 0+0.3\times (-5)-0.3\times 0.92}{0.5}\\\Rightarrow v_1=-3.552\ m/s

The velocity of glider A is -3.552 m/s

8 0
3 years ago
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