Calculate the current flowing if a charge of 36 kilocoulombs flows in 1 hour.

A paintball’s mass is 0.0032kg. A typical paintball strikes a target moving at 85.3 m/s.

vekshin1

Answer:

A) If the paintball stops completely the magnitude of the change in the paintball’s momentum is $p=0.273kg*m/s$

B) If the paintball bounces off its target and afterward moves in the opposite direction with the same speed, the change in the paintball’s momentum is $p=0.546kg*m/s$

C) A paintball bouncing off your skin in the opposite direction with the same speed hurts more than a paintball exploding upon your skin because of the strength exerted is twice than if it explodes.

Explanation:

Hi

A) We use the formula of momentum $p=mv$ , so we have $p=0.0032kg*85.3m/s=0.273kg*m/s$

B) We use the same formula above, then due we have a change of direction at the same speed, therefore the change in the momentum is the double so

$p=2*0.0032kg*85.3m/s=0.546kg*m/s$ .

C) The average strength of the force an object exerts during impact is determined by the amount the object’s momentum changes. therefore

$F=\frac{\Delta p}{\Delta t}$ , as we don't have any data about the impact time but we know momentum is twice, time does no matter and strength is twice too.

4 0

3 years ago

An electron with speed 2.45 x 10^7 m/s is traveling parallel to a uniform electric field of magnitude 1.18 x 10^4N/C . How much

cupoosta [38]

Answer:

time will elapse before it return to its staring point is 23.6 ns

Explanation:

given data

speed u = 2.45 × $10^{7}$ m/s

uniform electric field E = 1.18 × $10^{4}$ N/C

to find out

How much time will elapse before it returns to its starting point

solution

we find acceleration first by electrostatic force that is

F = Eq

here

F = ma by newton law

so

ma = Eq

here m is mass , a is acceleration and E is uniform electric field and q is charge of electron

so

put here all value

9.11 × $10^{-31}$ kg ×a = 1.18 × $10^{4}$ × 1.602 × $10^{-19}$

a = 20.75 × $10^{14}$ m/s²

so acceleration is 20.75 × $10^{14}$ m/s²

and

time required by electron before come rest is

use equation of motion

v = u + at

here v is zero and u is speed given and t is time so put all value

2.45 × $10^{7}$ = 0 + 20.75 × $10^{14}$ (t)

t = 11.80 × $10^{-9}$ s

so time will elapse before it return to its staring point is

time = 2t

time = 2 ×11.80 × $10^{-9}$

time is 23.6 × $10^{-9}$ s

time will elapse before it return to its staring point is 23.6 ns

7 0

2 years ago

You wad up a piece of paper and throw it into the wastebasket. How far will

vitfil [10]

The range of the piece of paper is C) 1.4 m

Explanation:

The motion of the piece of paper is the motion of a projectile, which consists of two separate motions:

- A uniform motion along the horizontal direction, with constant velocity

- A uniformly accelerated motion along the vertical direction, with constant acceleration (the acceleration of gravity, $g=9.8 m/s^2$ )

From the equation of motion, it is possible to find an expression for the range (the total horizontal distance covered) of a projectile, which is given by:

$d=\frac{u^2 sin 2\theta}{g}$