Answer:
The triboelectic charging process (a.k.a., charging by friction) results in a transfer of electrons between the two objects that are rubbed together. Rubber has a much greater attraction for electrons than animal fur.
Explanation:
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Answer:
Kinematics
given,
time (t)=100 s, distance (s)=1 km=1000 m
V
b
=10m/s (relative speed r.p to bus)
Velocity (v)=
time
distance
=
100
1000
V
s
= velocity of scooter
V
b
→ Velocity of bus
V=V
s
−V
b
→As we know
10=V
s
−10
20=V
s
V
s
=20 m/s
Velocity with which scooterist
should chase the bus →20 m/s
Explanation:
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Answer:
<h3>C no.</h3>
Explanation:
<h2><em>M</em><em>a</em><em>r</em><em>k</em><em> </em><em>m</em><em>e</em><em> </em><em>m</em><em>e</em><em> </em><em>b</em><em>r</em><em>a</em><em>i</em><em>n</em><em>l</em><em>i</em><em>e</em><em>s</em><em>t</em><em> </em><em>p</em><em>l</em><em>z</em><em> </em><em>i</em><em> </em><em>r</em><em>e</em><em>a</em><em>l</em><em>l</em><em>y</em><em> </em><em>n</em><em>e</em><em>e</em><em>d</em><em> </em><em>i</em><em>t</em><em> </em><em>(⌒▽⌒)</em></h2>
Answer:
Power= 6.84×10⁸ W
Explanation:
Given Data
Niagara falls at rate of=1.4×10⁶ kg/s
falls=49.8 m
To find
Power Generated
Solution
Regarding this problem
GPE (gravitational potential energy) declines each second is given from that you will find much the kinetic energy of the falling water is increasing each second.
So power can be found by follow
Power= dE/dt = d/dt (mgh)
Power= gh dm/dt
Power= 1.4×10⁶ kg/s × 9.81 m/s² × 49.8 m
Power= 6.84×10⁸ W
Answer:
AFter 3.5 s, the wagon is moving at: 
Explanation:
Let's start by finding first the net force on the wagon, and from there the wagon's acceleration (using Newton's 2nd Law):
Net force = 250 N + 178 N = 428 N
Therefore, the acceleration from Newton's 2nd Law is:

So now we apply this acceleration to the kinematic expression for velocity in an object moving under constant acceleration:
