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2 years ago

Which of the following settings best create fear or anxiety?

Physics

2 answers:

MAVERICK [17]2 years ago

6 0

Answer:

option d) is correct

Explanation:

option d) is correct

castle ruins is the setting best create fear or anxiety because ruining castle means that there is war going on.

War is the condition where every one is tense and this condition creates fear and anxiety in the people facing the scenario of fear.

so, the best option is castle ruins.

Papessa [141]2 years ago

3 0

"Castle ruins" is the one settings among the following choices given in the question that <span>create fear or anxiety. The correct option among all the options that are given in the question is the last option or the fourth option. I hope that this is the answer that you were looking for and the answer has actually come to your help.</span>

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A straight wire 20 cm long, carrying a current of 4 A, is in a uniform magnetic field of 0.6 T. What is the force on the wire wh

zheka24 [161]

Answer:

Magnetic force, F = 0.24 N

Explanation:

It is given that,

Current flowing in the wire, I = 4 A

Length of the wire, L = 20 cm = 0.2 m

Magnetic field, B = 0.6 T

Angle between force and the magnetic field, θ = 30°. The magnetic force is given by :

$F=ILB\ sin\theta$

$F=4\ A\times 0.2\ m\times 0.6\ T\ sin(30)$

F = 0.24 N

So, the force on the wire at an angle of 30° with respect to the field is 0.24 N. Hence, this is the required solution.

7 0

2 years ago

Read 2 more answers

At time t, gives the position of a 3.0 kg particle relative to the origin of an xy coordinate system ( ModifyingAbove r With rig

Elena-2011 [213]

Complete Question

The complete Question is shown on the first uploaded image

Answer:

The torque acting on the particle is $\tau = 48t \r k$

The magnitude of the angular momentum increases relative to the origin

Explanation:

From the equation we are told that

The position of the particle is $\= r = 4.0 t^2 \r i - (2.0 t - 6.0 t^2 ) \r j$

The mass of the particle is $m = 3.0 kg$

The time is t

The torque acting on the particle is mathematically represented as

$\tau = \frac{ d \r l }{dt}$

where $\r l$ is change in angular momentum which is mathematically represented as

$\r l = m (\r r \ \ X \ \ \r v)$

Where X mean cross- product

$\r v$ is the velocity which is mathematically represented as

$\r v = \frac{d \r r }{dt}$

Substituting for $\r r$

$\r v = \frac{d }{dt} [ 4 t^2 \r i - (2t + 6t^2 ) \r j]$

$\r v = 8t \r i - (2 + 12 t) \r j$

Now the cross product of $\r r \ and \ \r v$ is mathematically evaluated as

$\r r \ \ X \ \ \r v = \left[\begin{array}{ccc}{\r i}&{\r j}&{\r k}\\{4t^2}&{-2t -6t^2}&0\\{8t}&{-2 -12t}&0\end{array}\right]$

$= 0 \r i + 0 \r j + (- 8t^2 -48t^3 + 16t^2 + 48t^3 ) \r k$

$\r r \ \ X \ \ \r v = 8t^2 \r k$

So the angular momentum becomes

$\r l = m (8t^2 \r k)$

Substituting for m

$\r l = 3 * (8t^2 \r k)$

$\r l =24t^2 \r k$

Substituting into equation for torque

$\tau = \frac{d}{dt} [24t^2 \r k]$

$\tau = 48t \r k$

The magnitude of the angular momentum can be evaluated mathematically as

$|\r l| = \sqrt{(24 t^2) ^2}$

$|\r l| = 24 t^2$

From the is equation we see that the magnitude of the angular momentum is varies directly with square of the time so it would relative to the origin because at the origin t= 0s and we move out from origin t increases hence angular momentum increases also