Charging phenomenon is basically transfer of electrons from one system to other. When a system loses the electrons the it becomes positively charged and when a system gains electrons then it will become negatively charged
so here if slab is rub against fur then one of them will lose the electrons while other will gain the electrons and hence one will get negatively charged and other will get positive charge
Here it is given that slab get negatively charged so correct options are
<em>3. Electrons move from the fur to the slab.
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<em>6. The fur becomes positive after rubbing the slab.
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The so-called "velocity-time" graph is actually a "speed-time" graph. At any point
on it, the 'x'-coordinate is a time, and the 'y'-coordinate is the speed at that time.
'Velocity' is a speed AND a direction. Without a direction, you do not have a velocity,
and these graphs never show the direction of the motion. It seems to me that it would be
pretty tough to draw a graph that shows the direction of motion at every instant of time,
so my take is that you'll never see a true "velocity-time" graph.
At best, it would need a second line on it, whose 'y'-coordinate referred to a second
axis, calibrated in angle and representing the 'bearing' or 'heading' of the motion at
each instant. The graph of uniform circular motion, for example, would have a straight
horizontal line for speed, and a 'sawtooth' wave for direction.
Answer:
B. A system cannot take in additional matter.
Explanation: The total amount of energy in the universe remains constant, it can merely change from one form to another.
Hope it helps you:)))
have a good day
Answer:
x = 240 m
Explanation:
This is a kinematics exercise
Let's fix our frame of reference on car A
x = x₀ₐ+ v₀ₐ t + ½ aₐ t²
the initial position of car a is zero
x = 0 + v₀ₐ t + ½ 0.8 t²
for car B
x = x_{ob} + v_{ob} t - ½ a_b t²
car B's starting position is 30 m
x = 30 + v_{ob} t - ½ 0.4 t²
at the point where they meet, the position of the two vehicles is the same
0 + v₀ₐ t + ½ 0.8 t² = 30 + v_{ob} t - ½ 0.4 t²
let's reduce the speeds to the SI system
v₀ₐ = 14.4 km / h (1000 m / 1 km) (1h / 3600s) = 4 m / s
v_{ob} = 23.4 km / h = 6.5 m / s
4 t + 0.4 t² = 30 + 6.5 t - 0.2 t²
0.2 t² - 2.5 t - 30 = 0
t² - 12.5 t - 150 = 0
we solve the quadratic equation
t =
t =
t₁ = 20 s
t₂ = -7.5 s
time must be a positive quantity so the correct result is t = 20 s
let's look for the distance
x = 4 t + ½ 0.8 t²
x = 4 20 + ½ 0.8 20²
x = 240 m