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elena55
3 months ago
11

Zamir and Talia raced through a maze. Zamir walked 2 m north, 2 m east, 4 m south, 2 m east, 4 m north, 2 m east, 3 m south, 4 m

east and 4 m north. Talia walked 2 m north, 6 m east, 3 m south, 4 m east, and 4 m north. Compare their displacements:
Physics
2 answers:
Maru [3.3K]3 months ago
6 0

The displacement of Zamir is the same as Talia’s displacement.

Keith_Richards [3.2K]3 months ago
6 0

Response: The displacements of Zamir and Talia are the same.

Clarification:

Displacement refers to the shift in an object's position.

Zamir travels a total of 27 m while Talia covers 19 m; however, both their initial and final locations are identical.

Thus, we can conclude that Zamir's displacement matches Talia's displacement.

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The cluster of decisions that managers make to assist the organization to achieve its goals is known as:
Softa [3030]
The answer is option C. SWOT analysis. The compilation of decisions made by managers to aid the organization in reaching its objectives is referred to as SWOT Analysis.
Here are the options. A. Strategy B. Scenario planning C. SWOT analysis D. Diversification E. Related diversification
8 0
3 months ago
The drawing shows an adiabatically isolated cylinder that is divided initially into two identical parts by an adiabatic partitio
Yuliya22 [3333]

Answer:

the temperature on the left side is 1.48 times greater than that on the right

Explanation:

GIVEN DATA:

\gamma = 5/3

T1 = 525 K

T2 = 275 K

It is known that

P_1 = \frac{nRT_1}{v}

P_2 = \frac{nrT_2}{v}

n and v are constant on both sides. Therefore we have

\frac{P_1}{P_2} = \frac{T_1}{T_2} = \frac{525}{275} = \frac{21}{11}

P_1 = \frac{21}{11} P_2..............1

let the final pressure be P and the temperature T_1 {f} and T_2 {f}

P_1^{1-\gamma} T_1^{\gamma} = P^{1 - \gamma}T_1 {f}^{\gamma}

P_1^{-2/3} T_1^{5/3} = P^{-2/3} T_1 {f}^{5/3}..................2

similarly

P_2^{-2/3} T_2^{5/3} = P^{-2/3} T_2 {f}^{5/3}.............3

divide equation (2) by equation (3)

\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}

T_1 {f} = 1.48 T_2 {f}

thus, the left side temperature equals 1.48 times the right side temperature

6 0
3 months ago
The electric field must be zero inside a conductor in electrostatic equilibrium, but not inside an insulator. It turns out that
serg [3582]

Response:

Reasoning:

We will utilize a Gaussian surface that resembles the curved wall of a cylinder, with a radius of 3mm and a length of 1 unit directed parallel to the wire axis.

The charge within this cylinder amounts to 250 x 10⁻⁹ C.

Let E denote the electric field at the curved surface, perpendicular to it.

The total electric flux leaving the curved surface

is calculated as 2π r x 1 x E

or 2 x 3.14 x 3 x 10⁻³ E

According to Gauss's law, the total flux is given by the charge within divided by ε (the charge inside the cylinder being 250 x 10⁻⁹C)

equals 250 x 10⁻⁹ / 2.5 x 8.85 x 10⁻¹²   (where ε = 2.5 ε₀ = 2.5 x 8.85 x 10⁻¹²)

resulting in 11.3 x 10³ weber.

Thus,

2 x 3.14 x 3 x 10⁻³ E = 11.3 x 10³

E =  11.3 x 10³ /  2 x 3.14 x 3 x 10⁻³

=.599 x 10⁶ N /C.

4 0
2 months ago
Suppose an electrical wire is replaced with one having every linear dimension doubled (i.e., the length and radius have twice th
Softa [3030]

Response:

The new resistance is half of the original resistance.

Explanation:

Resistance in a wire is represented by:

R=\dfrac{\rho L}{A}

\rho = resistivity of the material

L and A are the physical dimensions

If a wire is exchanged for one where all linear dimensions are doubled, i.e. l' = 2l and r' = 2r

The updated resistance of the wire can be calculated as follows:

R'=\dfrac{\rho L'}{A'}

R'=\dfrac{\rho (2L)}{\pi (2r)^2}

R'=\dfrac{1}{2}\dfrac{\rho L}{A}

R'=\dfrac{1}{2}R

The new resistance equals half of the original resistance. Thus, this provides the solution needed.

4 0
3 months ago
For a projectile, which of the following quantities are constant during the flight: x, y, vx, vy, v, ax, ay? Check all that appl
Yuliya22 [3333]

Response:

C. vx

F. ax

G. ay

Clarification:

The projectile follows a curved trajectory toward the ground, causing changes in x and y positions.

Since there is no external force acting in the x-direction, the acceleration in x remains at zero. Consequently, ax and vx remain unchanged.

The projectile is subject to the force of gravity, directed downwards, leading to an increase in its velocity due to the rise in its y-component.

Meanwhile, the y-component of acceleration remains constant due to gravitational acceleration.

5 0
3 months ago
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