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zhenek [66]
2 years ago
14

If a neutral object such as paper comes close to a positively charged plastic rod, what type of charge accumulates on the side o

f the paper closest to the positive rod?
Physics
2 answers:
Tju [1.3M]2 years ago
7 0

Explanation:

When a neutral object such as paper comes close to a positively charged plastic rod then an opposite charge, that is, negative charge will develop on the paper.

As a result, the side of paper with negative charge will get attracted towards the positive charge of plastic rod. As it is known that opposite charges attract each other and like charges repel each other.


Nataliya [291]2 years ago
3 0
The answer would be negative charge because +, and - dont like each other so they retract from each other.
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A 1500 kg car traveling at 20 m/s suddenly runs out of gas while approaching the valley shown in the figure. The alert driver im
geniusboy [140]

Answer:

v_f = 17.4 m / s

Explanation:

For this exercise we can use conservation of energy

starting point. On the hill when running out of gas

          Em₀ = K + U = ½ m v₀² + m g y₁

final point. Arriving at the gas station

         Em_f = K + U = ½ m v_f ² + m g y₂

energy is conserved

         Em₀ = Em_f

         ½ m v₀ ² + m g y₁ = ½ m v_f ² + m g y₂

        v_f ² = v₀² + 2g (y₁ -y₂)

         

we calculate

        v_f ² = 20² + 2 9.8  (10 -15)

        v_f = √302

         v_f = 17.4 m / s

8 0
2 years ago
3-m-high large tank is initially filled with water. The tank water surface is open to the atmosphere, and a sharp-edged 10-cm-di
irinina [24]

Answer:

The initial velocity of the water from the tank is 5.42 m/s

Explanation:

By applying Bernoulli equation between  point 1 and 2

\dfrac{P_1}{\rho g}+\dfrac{V_1^2}{2g}+Z_1=\dfrac{P_2}{\rho g}+\dfrac{V_2^2}{2g}+Z_2+h_L

At the point 1

P₁=0  ( Gauge pressure)

V₁= 0 m/s

Z₁=3 m

At point 2

P₂=0  ( Gauge pressure)

Z₂= 0 m/s

h_L=1.5\ m

Now by putting the values

\dfrac{P_1}{\rho g}+\dfrac{V_1^2}{2g}+Z_1=\dfrac{P_2}{\rho g}+\dfrac{V_2^2}{2g}+Z_2+h_L

Z_1-h_L=\dfrac{V_2^2}{2g}

3-1.5=\dfrac{V_2^2}{2\times 9.81}

V_2=\sqrt{2\times 1.5\times 9.81}\ m/s

V₂= 5.42 m/s

The initial velocity of the water from the tank is 5.42 m/s

3 0
3 years ago
There are many ways to lose weight, but a sustainable plan generally involves behavior modification related to diet and exercise
Anna007 [38]

The answer is adequate nutrition, regular physical activity, and practical ways to reduce calories while retaining important nutrients.

Explanation:

Despite that adequate nutrition, regular physical activity, and practical ways to reduce calories while retaining important nutrients is one of the best strategy to reducing weight, most time it is very difficult for those that want to reduce or control their weight to discipline themselves enough to follow these routine. But one an individual that want to loose weight or live a healthy lifestyle is able to follow these procedures he/she will surely loose weight.

5 0
2 years ago
A hockey stick of mass ms and length L is at rest on the ice (which is assumed to be frictionless). A puck with mass mp hits the
krek1111 [17]

Answer:

L = mp*v₀*(ms*D) / (ms + mp)

Explanation:

Given info

ms = mass of the hockey stick

uis = 0 (initial speed of the hockey stick before the collision)

xis = D (initial position of center of mass of the hockey stick before the collision)

mp = mass of the puck

uip = v₀ (initial speed of the puck before the collision)

xip = 0 (initial position of center of mass of the puck before the collision)

If we apply

Ycm = (ms*xis + mp*xip) / (ms + mp)

⇒  Ycm = (ms*D + mp*0) / (ms + mp)

⇒  Ycm = (ms*D) / (ms + mp)

Now, we can apply the equation

L = m*v*R

where m = mp

v = v₀

R = Ycm

then we have

L = mp*v₀*(ms*D) / (ms + mp)

5 0
2 years ago
Two vectors are presented as a=3.0i +5.0j and b=2.0i+4.0j find (a) a x b, ab (c) (a+b)b and (d) the component of a along the dir
Svet_ta [14]
Let's ask this question step by step:
 Part A) 
 a x b = (3.0i + 5.0j) x (2.0i + 4.0j) = (12-10) k = 2k
 ab = (3.0i + 5.0j). (2.0i + 4.0j) = 6 + 20 = 26
 Part (c)
 (a + b) b = [(3.0i + 5.0j) + (2.0i + 4.0j)]. (2.0i + 4.0j)
 (a + b) b = (5.0i + 9.0j). (2.0i + 4.0j)
 (a + b) b = 10 + 36
 (a + b) b = 46
 Part (d)
 comp (ba) = (a.b) / lbl
 a.b = (3.0i + 5.0j). (2.0i + 4.0j) = 6 + 20 = 26
 lbl = root ((2.0) ^ 2 + (4.0) ^ 2) = root (20)
 comp (ba) = 26 / root (20)
 answer
 2k
 26
 46
 26 / root (20)
3 0
2 years ago
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