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zubka84 [21]
4 years ago
10

A plastic rod that has been charged to -15.0nC touches a metal sphere. Afterward, the rod's charge is -10.0nC. How many charged

particles were transferred?
Physics
1 answer:
max2010maxim [7]4 years ago
7 0

Answer:

3.125 X 10^{10}

Explanation:

Net charge= -15.0 nC- -10.0 nC=-5nC

\frac {-5 X 10^{-9}}{ -1.6 X 10^{-19}} = 3.125 X 10^{10} electrons were transferred

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Ivanshal [37]
The mass of water moving around on Earth’s surface causes the amount of gravitational force.
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3 years ago
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Una furgoneta circula por una carretera a 55km/h. Diez km atrás , un coche circula en el mismo sentido a 85km/h ¿ En cuanto tiem
statuscvo [17]

Answer:

t = 0.33h = 1200s

x = 18.33 km

Explanation:

If the origin of coordinates is at the second car, you can write the following equations for both cars:

car 1:

x=x_o+v_1t    (1)

xo = 10 km

v1 = 55km/h

car 2:

x'=v_2t    (2)

v2 = 85km/h

For a specific value of time t the positions of both cars are equal, that is, x=x'. You equal equations (1) and (2) and solve for t:

x=x'\\\\x_o+v_1t=v_2t\\\\(v_2-v_1)t=x_o\\\\t=\frac{x_o}{v_2-v_1}

t=\frac{10km}{85km/h-55km/h}=0.33h*\frac{3600s}{1h}=1200s

The position in which both cars coincides is:

x=(55km/h)(0.33h)=18.33km

6 0
3 years ago
Jefferson refers to
Trava [24]

Answer: natural

Explanation:

7 0
3 years ago
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A block of mass m = 150 kg rests against a spring with a spring constant of k = 880 N/m on an inclined plane which makes an angl
weeeeeb [17]

Answer:

b)  k Δx - W cos θ - μ mg cos θ = m a ,  c)  θ = 86.6º, d)  Δx = 1.18 m

Explanation:

a) In the attachment we can see a diagram of the forces in this problem and the coordinate axes for its decomposition.

F is the force applied by the spring, while it is compressed, this force disappears when the block leaves the spring

b) Let's apply Newton's second law for when the spring is compressed

let's use trigonometry to break down the weight

            sin θ = Wₓ / W

            cos θ = W_y / W

             Wₓ = W sin θ

             W_y = W cos θ

Y axis  

               N - W_y = 0

               N = W_y

               N = W cos θ

X axis

           F -Wₓ -fr = ma

the force applied by the spring is given by hooke's law

           F = k Δx

friction force has the expression

           fr = μ N

           fr = μ W cos θ

we substitute

            k Δx - W cos θ - μ mg cos θ = m a           ( 1)

c) If the plane has no friction, what is the angle so that Δx = 0.1m

             

We write the equation 1, with fr = 0 and since the system is still a = 0

            k Δx - W cos θ -0 = 0

            cos θ = \frac{k \Delta x}{ m g}

            cos θ = \frac{880 \ 0.1}{ 150 \ 9.8}

            cos θ = 0.0598

            θ = cos⁻¹ 0.0598

            θ = 86.6º

d) In this part they give the angle θ = 45º and there is no friction, they ask the compression

the acceleration is zero, we substitute in 1

            k Δx - W cos θ - 0 = 0

            Δx = \frac{mg \ cos \  \theta}{k}

            Δx = \frac{ 150 \ 9.8 \ cos45}{880}

            Δx = 1.18 m

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