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vovikov84 [41]
3 years ago
10

Which sequence of events in emotional responses is characteristic of the james-lange theory of emotion?

Physics
1 answer:
AlladinOne [14]3 years ago
4 0
According to James-Lange theory of emotion, a stimulus first leads to bodily arousal, and this is followed by our interpretation of it as an emotion.
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What happens when air from the balloon has exited the balloon? <br><br><br> How will motion change?​
tiny-mole [99]

Answer:

When you release the opening of the balloon, gas quickly escapes to equalize the pressure inside with the air pressure outside of the balloon. The escaping air exerts a force on the balloon itself. ... That opposing force—called thrust, in this case—propels the rocket forward.

6 0
3 years ago
What year did Badminton become a full-medal Olympic sport?
Basile [38]

Answer:

1992

Explanation:

Badminton made its debut as a demonstration sport at the 1972 Olympic Games in Munich. It was not until the 1992 Games in Barcelona that it was officially included on the Olympic programme, with men's and women's singles and doubles events.

6 0
3 years ago
Consider two identical objects of mass m = 0.250 kg and charge q = 4.00 μC. The first charge is held in place at the origin of a
Gnom [1K]

Answer:

a = 640 m/s²

Explanation:

From work-kinetic energy principles,

The net force acting on the second object is the gravitational force and the electric force due to the first object.

So, the gravitational force on the mass is F₁ = Gm₁m₂/r² since m₁ = m₂ = m, U = -Gm²/r²

Also, the electric force on the charge is F₂ = kq₁q₂/r² since q₁ = q₂ = q, U = kq²/r²

The net Force F = ma

So, -F₁ + F₂ = F     (F₁ is negative since it is an attractive force in the negative x -direction and F₂ is positive since it is a repulsive force in the positive x- direction)

-Gm²/r² + kq²/r² = ma

ma = -Gm²/r² + kq²/r²

a = (-Gm²/r² + kq²/r²)/m

a = (-G + kq²/m²)m/r²

Since m = 0.250 kg, q = 4.00 μC = 4.00 × 10⁻⁶ C, r = 3.00 cm = 3.00 × 10⁻² m, G = 6.67 × 10⁻¹¹ Nm²/kg², k = 9 × 10⁹ Nm²/C² and a = acceleration of second mass.

Substituting the variables into the equation, we have

a = (m/r²)(-G + k(q/m)²)]

a = (0.250 kg/{3.00 × 10⁻² m}²)(-6.67 × 10⁻¹¹ Nm²/kg² + 9 × 10⁹ Nm²/C²(4.00 × 10⁻⁶ C/0.250 kg)²)

a = (0.250 kg/9.00 × 10⁻⁴ m)(-6.67 × 10⁻¹¹ Nm²/kg² + 9 × 10⁹ Nm²/C²(16 × 10⁻⁶ C/kg)²)]

a = (0.250 kg/9.00 × 10⁻⁴ m)(-6.67 × 10⁻¹¹ Nm²/kg² + 9 × 10⁹ Nm²/C²(256 × 10⁻¹² C²/kg²)]

a = (0.250 kg/9.00 × 10⁻⁴ m)(-6.67 × 10⁻¹¹ Nm²/kg² + 2304 × 10⁻³ Nm²/kg²  ]

a = (0.250 kg/9.00 × 10⁻⁴ m)(2.304 Nm²/kg²)

a = 0.576 Nm²/kg /9.00 × 10⁻⁴ m²

a = 0.064 × 10⁴N/kg

a = 64 × 10 N/kg)

a = 640 m/s²

8 0
3 years ago
A stunt driver rounds a banked, circular curve. The driver rounds the curve at a high, constant speed, such that the car is just
bagirrra123 [75]

Answer: C

Frictional force

Explanation:

The description of the question above is an example of a circular motion.

For a car travelling in a curved path, the frictional force between the tyres and the road surface will provide the centripetal force.

Since the road is banked, and the cross section of the banked road is constructed like a ramp. The car drives transversely to the slope of the ramp, so that the wheels of one side of the car are lower than the wheels on the other side of the car, for cornering the banked road, the car will not rely only on the frictional force.

Therefore, the correct answer is option C - the frictional force.

5 0
3 years ago
Which statement supports the giant-impact hypothesis of the moon’s formation
V125BC [204]

Answer:

the moon lacks a sizeable iron core

Explanation:

7 0
3 years ago
Read 2 more answers
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