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sesenic [268]
2 years ago
14

A PERSON GETTING OUT OF MOVING BUS FALLS IN THE DIRECTION OF MOTION OF THE BUS. WHY? ​

Physics
2 answers:
Usimov [2.4K]2 years ago
6 0

Answer:

When the person steps on the ground, his feet do not move but his upper body moves in the direction of the bus due to inertia of its motion. Since his upper body moves in the forward direction and his lower body does not move, the person falls in the forward direction.

djverab [1.8K]2 years ago
3 0

Answer:

PLEASE MARK AS BRAINLIEST!!

Explanation:

A getting passenger getting down from a moving bus, falls in the direction of the motion of the bus. This is because his feet come to rest on touching the ground and the remaining body continues to move due to inertia of motion.

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What distance will a car traveling at a speed of 15 m/s cover in 300 s?
AnnZ [28]

Distance formula: d = st

d = 15 * 300

d = 4,500m

Best of Luck!

3 0
3 years ago
A cannonball is launched from the ground at an angle of 30 degrees above the horizontal and a speed of 30 m/s. Ideally (no air r
marusya05 [52]

As we know that here no air resistance while ball is moving in air

So here we will say that

initial total energy = final total energy

KE_i + U_i = KE_f + U_f

here we know that

Ui = U_f = 0 (as it will be on ground at initial and final position)

so we will say

KE_i = KE_f

since mass is always conserved

so we will say that final speed of the ball must be equal to the initial speed of the ball

so we have

v_f = v_i = 30 m/s

6 0
3 years ago
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Red blood cells can be modeled as spheres of 6.53 μm diameter with −2.55×10−12 C excess charge uniformly distributed over the su
yKpoI14uk [10]

Complete Question

Red blood cells can be modeled as spheres of 6.53 μm diameter with −2.55×10−12 C excess charge uniformly distributed over the surface. Find the electric field at the following locations, with radially outward defined as the positive direction and radially inward defined as the negative direction. The permittivity of free space ????0 is 8.85×10−12 C/(V⋅m). What is the electric field

E⃗ 1 inside the cell at a distance of 3.05 μm from the center?

E⃗ 2 Just inside the surface of the cell

E⃗ 3 Just outside the surface of the cell

E⃗ 4 At a point outside the cell 3.05 μm from the surface

Answer:

E⃗ 1

      0 V/m

E⃗ 2

      0 V/m

E⃗ 3

         E_3 =  2.153 *10^{9} \  V/m

E⃗ 4

E_4 =  5.754 *10^ {8} \  V/m

Explanation:

From the question we are told that

The diameter is d =  6.53 \mu m  = 6.53*10^{-6}\  m

The charge is Q =  -.2.55 *10^{-12} \  C

The permittivity of free space is \epsilon_o  =  8.85* 10^{-12}\  C / V.m

The distance considered is d =  3.05 \mu m  =  3.05 *10^{-6} \ m

Generally the electric field inside the cell at a distance of 3.05 μm from the center is

0 V/m

This because there is no electric field felt inside the cell according Gauss the cell is taken as a point charge

Generally the electric field just inside the surface of the cell is 0 V/m

This because there is no electric field felt inside the cell according Gauss the cell is taken as a point charge

Generally the electric field just outside the cell is mathematically represented as

E_3 =  \frac{ k  *  |Q|}{ r^2 }

Here k is the coulomb constant with value

k  =   9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}

r is the radius of the sphere which is mathematically as

r =  \frac{d}{2} =   \frac{6.53*10^{-6}}{2}  = 3.265 *10^{-6} \  m

E_3 =  \frac{ 9*10^{9}  *  |-2.55 *10^{-12} |}{ [3.265 *10^{-6} ]^2 }

E_3 =  2.153 *10^{9} \  V/m

Generally the electric field at a point outside the cell 3.05 μm from the surface is mathematically represented as

E_4 =  \frac{ k  *  |Q|}{ R^2 }

Here R is mathematically represented as

R  =  3.265 *10^{-6} +  3.05 *10^{-6}

=>       R  =  6.315 *10^{-6}

So

E_4 =  \frac{ 9*10^{9}  *  |-2.55 *10^{-12} |}{ [ 6.315 *10^{-6} ]^2 }

E_4 =  5.754 *10^ {8} \  V/m

3 0
2 years ago
20 POINTSSS!!!!!!!!
Soloha48 [4]
Answer: A
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KatRina [158]

The animal that is hunted and consumed is considered the prey

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