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3 years ago

Neuroticism, psychoticism, and extraversion are the three dominant personality traits according to

Physics

2 answers:

Anna71 [15]3 years ago

6 0

The answer is b I just took the test

Alex787 [66]3 years ago

3 0

Answer:

B. Hans Eysenck

Explanation:

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A car is on a circular off ramp of an interstate and is traveling at exactly 25 mph around the curve. Does the car have velocity

netineya [11]

Answer:

The car has velocity and acceleration but is not decelerating

Explanation:

Since the car is traveling at 25 mph around the curve, it has a tangential velocity. This tangential velocity is constantly changing in direction (so the car could adapt to the curve and not moving forward in a straight line), there should be a centripetal acceleration in play here. This acceleration does not slow down the car so it's not decelerating.

6 0

3 years ago

In which type of wave do air particles move together or apart parallel to the direction of the wave?

Kitty [74]

C. Sound waves is the answer =P

7 0

4 years ago

Consider a spring mass system (mass m1, spring constant k) with period T1. Now consider a spring mass system with the same sprin

tatuchka [14]

Answer:

Assuming that both mass here move horizontally on a frictionless surface, and that this spring follows Hooke's Law, then the mass of $m_2$ would be four times that of $m_1$ .

Explanation:

In general, if the mass in a spring-mass system moves horizontally on a frictionless surface, and that the spring follows Hooke's Law, then

$\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2$ .

Here's how this statement can be concluded from the equations for a simple harmonic motion (SHM.)

In an SHM, if the period is $T$ , then the angular velocity of the SHM would be

$\displaystyle \omega = \frac{2\pi}{T}$ .

Assume that the mass starts with a zero displacement and a positive velocity. If $A$ represent the amplitude of the SHM, then the displacement of the mass at time $t$ would be:

$\mathbf{x}(t) = A\sin(\omega\cdot t)$ .

The velocity of the mass at time $t$ would be:

$\mathbf{v}(t) = A\,\omega \, \cos(\omega\, t)$ .

The acceleration of the mass at time $t$ would be:

$\mathbf{a}(t) = -A\,\omega^2\, \sin(\omega \, t)$ .

Let $m$ represent the size of the mass attached to the spring. By Newton's Second Law, the net force on the mass at time $t$ would be:

$\mathbf{F}(t) = m\, \mathbf{a}(t) = -m\, A\, \omega^2 \, \cos(\omega\cdot t)$ ,

Since it is assumed that the mass here moves on a horizontal frictionless surface, only the spring could supply the net force on the mass. Therefore, the force that the spring exerts on the mass will be equal to the net force on the mass. If the spring satisfies Hooke's Law, then the spring constant $k$ will be equal to:

$\begin{aligned} k &= -\frac{\mathbf{F}(t)}{\mathbf{x}(t)} \\ &= \frac{m\, A\, \omega^2\, \cos(\omega\cdot t)}{A \cos(\omega \cdot t)} \\ &= m \, \omega^2\end{aligned}$ .

Since $\displaystyle \omega = \frac{2\pi}{T}$ , it can be concluded that:

$\begin{aligned} k &= m \, \omega^2 = m \left(\frac{2\pi}{T}\right)^2\end{aligned}$ .

For the first mass $m_1$ , if the time period is $T_1$ , then the spring constant would be:

$\displaystyle k = m_1\, \left(\frac{2\pi}{T_1}\right)^2$ .

Similarly, for the second mass $m_2$ , if the time period is $T_2$ , then the spring constant would be:

$\displaystyle k = m_2\, \left(\frac{2\pi}{T_2}\right)^2$ .

Since the two springs are the same, the two spring constants should be equal to each other. That is:

$\displaystyle m_1\, \left(\frac{2\pi}{T_1}\right)^2 = k = m_2\, \left(\frac{2\pi}{T_2}\right)^2$ .

Simplify to obtain:

$\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2$ .

6 0

3 years ago

Determine the number of ways to throw two die and get the number 11, as well as the probability of getting 11.

Ivenika [448]

Answer:

1/18

Explanation:

The number of ways in which two die can be thrown is the sample space of the experiment

(1,1), (1,2) (1,3), (1,4) (1,5) (1,6)

(2,1), (2,2) (2,3), (2,4) (2,5) (2,6)

(3,1), (3,2) (3,3), (3,4) (3,5) (3,6)

(4,1), (4,2) (4,3), (4,4) (4,5) (4,6)

(5,1), (5,2) (5,3), (5,4) (5,5) (5,6)

(6,1), (6,2) (6,3), (6,4) (6,5) (6,6)

From here it can be seen that there are only two cases where the sum on the two die is 11 i.e., (6,5) and (5,6)

$\text{Probability of getting 11}=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\\\Rightarrow \text{Probability of getting 11}=\frac{2}{36}\\\Rightarrow \text{Probability of getting 11}=\frac{1}{18}=0.056$

∴ Probability of getting 11 is 1/18

8 0

3 years ago

The same book is placed on a table .9m high. What type of energy does it have and how much?

Lina20 [59]

mass (m)= 5 kg

height (h)= 9m

gravity (g)= 9.8 m/s^2

It has Potential energy. (PE)

• PE = mgh

Replacing:

PE = 5 * 9.8 * 9 = 441 Joules

6 0

1 year ago