All categories

2 months ago

A cheetah can run at 30 m/s, but only for about 12s. How far will it run in that time

1 answer:

7 0

A cheetah is capable of running at 30 m/s, but only for a duration of about 12s. Therefore, it will cover a distance of 12x30=360 miles during that time.

You might be interested in

Rosa studies the position-time graph of two race cars. A graph titled Position versus Time shows time in hours on the x axis, nu

Sav [3153]

Answer:

B. Truck X was ahead, not truck Y.

Explanation:

Let's analyze the information provided.

Truck X moved from the point (0,20) to (2.8,50). This indicates that it began at the 20th kilometer and reached 50 km in 2.8 hours. Thus, its speed is v1 = (s2 - s1) / t

v1 = (50 - 20) / 2.8

v1 = 10.7 km/h

Given that it started from the 20th km, it indeed had a head start. Since the line on the graph is linear, this shows its speed was constant without any change in direction.

On the other hand, Truck Y's movement went from the origin (0,0) to (5,20), meaning it took 5 hours to travel 20 km, resulting in a speed of v2 = 20 / 5

v2 = 4 km/h

Again, the straightness of its graph line signifies it maintained a constant speed in a single direction.

Thus, it is evident that Rosa erred in her assumption that Truck Y had a head start.

5 0

3 months ago

Read 2 more answers

A positive point charge q is placed at the center of an uncharged metal sphere insulated from the ground. The outside of the sph

Softa [3030]

B. The charge on A is -q; B has no charge. Given that a positive charge is situated at the center of an uncharged metallic sphere which is insulated and disconnected from the ground, a negative charge (-q) will appear on the inner surface A of the sphere. Should the exterior surface B be grounded, it will become neutral, resulting in no charge remaining on surface B.

4 0

2 months ago

An infinite sheet of charge, oriented perpendicular to the x-axis, passes through x = 0. It has a surface charge density σ1 = -2

Maru [3345]

1) For x = 6.6 cm, $E_x=3.47\cdot 10^6 N/C$

2) For x = 6.6 cm, $E_y=0$

3) For x = 1.45 cm, $E_x=-3.76\cdot 10^6N/C$

4) For x = 1.45 cm, $E_y=0$

5) Surface charge density at b = 4 cm: $+62.75 \mu C/m^2$

6) At x = 3.34 cm, the x-component of the electric field equals zero

7) Surface charge density at a = 2.9 cm: $+65.25 \mu C/m^2$

8) None of these regions

Explanation:

The electric field from an infinite charge sheet is perpendicular to it:

$E=\frac{\sigma}{2\epsilon_0}$

where

$\sigma$ is the surface charge density

$\epsilon_0=8.85\cdot 10^{-12}F/m$ represents vacuum permittivity

Outside the slab, the electric field behaves like that of an infinite sheet.

Consequently, the electric field at x = 6.6 cm (situated to the right of both the slab and sheet) results from the combination of the fields from both:

$E=E_1+E_2=\frac{\sigma_1}{2\epsilon_0}+\frac{\sigma_2}{2\epsilon_0}$

where

$\sigma_1=-2.5\mu C/m^2 = -2.5\cdot 10^{-6}C/m^2\\\sigma_2=64 \muC/m^2 = 64\cdot 10^{-6}C/m^2$

The field from the sheet points left (negative, inward), and the slab’s field points right (positive, outward).

Thus,

$E=\frac{1}{2\epsilon_0}(\sigma_1+\sigma_2)=\frac{1}{2(8.85\cdot 10^{-12})}(-2.5\cdot 10^{-6}+64\cdot 10^{-6})=3.47\cdot 10^6 N/C$

and the negative sign indicates a rightward direction.

Both the sheet’s and slab’s fields are perpendicular to their surfaces, directing along the x-axis, hence there's no y-component for the total field.

<pThus, the y-component totals zero.

This happens because both the sheet and slab stretch infinitely along the y-axis. Choosing any x-axis point reveals that the y-component of the field, generated by a surface element dS of either the sheet or slab, $dE_y$ , will be equal and opposite to the corresponding component from the opposite side, $-dE_y$ . Thus, the combined y-direction field is always zero.

This scenario resembles part 1), but the point here is

x = 1.45 cm

which lies between the sheet and the slab. The fields from both contribute leftward as the slab has a negative charge (resulting in an outward field). Thus, the total field computes to

$E=E_1-E_2$

Replacing with expressions from part 1), we get

$E=\frac{1}{2\epsilon_0}(\sigma_1-\sigma_2)=\frac{1}{2(8.85\cdot 10^{-12})}(-2.5\cdot 10^{-6}-64\cdot 10^{-6})=-3.76\cdot 10^6N/C$

where the negative illustrates a leftward direction.

This portion parallels part 2). Since both fields remain perpendicular to the slab and sheet, no component exists along the y-axis, thus the electric field's y-component is zero.

Notably, the slab behaves as a conductor, signifying charge mobility within it.

The net charge on the slab is positive, indicating a surplus of positive charge. With the negatively charged sheet on the left of the slab, positive charges shift towards the left slab edge (at a = 2.9 cm), while negative charges move to the right edge (at b = 4 cm).

The surface charge density per unit area of the slab is

$\sigma=+64\mu C/m^2$

This average denotes the surface charge density on both slab sides at points a and b:

$\sigma=\frac{\sigma_a+\sigma_b}{2}$ (1)

Additionally, the infinite sheet at x = 0 negatively charged $\sigma_1=-2.5\mu C/m^2$ , induces an opposite net charge on the slab's left surface, thus

$\sigma_a-\sigma_b = +2.5 \mu C/m^2$ (2)

Having equations (1) and (2) allows for solving the surface charge densities at a and b, yielding:

$\sigma_a = +65.25 \mu C/m^2\\\sigma_b = +62.75 \mu C/m^2$

We aim to compute the x-component of the electric field at

x = 3.34 cm

This point lies inside the slab, bounded at

a = 2.9 cm

b = 4.0 cm

In a conducting slab, the electric field remains at zero owing to charge equilibrium; thus, the x-component thereof in the slab is zero

From part 5), we determined the surface charge density at x = a = 2.9 cm is $\sigma_a = +65.25 \mu C/m^2$

As mentioned in part 6), conductors have zero electric fields internally. Since the slab is conductive, the electric field inside remains zero; therefore, the regions where the electric field is null are

2.9 cm < x < 4 cm

Thus, the suitable answer is

"none of these regions"

Learn more about electric fields:

8 0

3 months ago