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r-ruslan [8.4K]

3 years ago

14

A 10 kg block slides on a frictionless surface at 10 m/s . It hits a rough patch 3 meters long with a coefficient of kinetic fri

ction of 0.4. When it reenters the frictionless surface , how fast is it going?

1 answer:

Valentin [98]3 years ago

8 0

12 MPH

I DIDNT do the math my brother did hes in colledge so good lucks guys

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Part AIf the potential of plate 1 is V, then, in equilibrium, what are the potentials of plates 3 and 6? Assume that the negativ

Nana76 [90]

Answer:

Part a: The potential at point 3 and point 6 are V and 0 respectively.

Part b: The charges Q1, Q2 and Q3 are CV, 2CV and 3CV respectively.

Part c: The net charge is 6CV.

Part d: The equivalent capacitance is 6C

Explanation:

As the question is not given here ,the complete question is found online and is attached herewith.

Part a:

V1 = V, V3 = V1 = V

and, V6 = V1-V = 0

The potential at point 3 and point 6 are V and 0 respectively

Part b

Q1 =CV = Q,

Q2 = 2C *V = 2Q

Q3 = 3 C*V = 3Q

So the charges Q1, Q2 and Q3 are CV, 2CV and 3CV respectively.

Part c

Total charge of the system,

Q_net =Q1+Q2+Q3= (1+2+3) CV = 6 CV

So the net charge is 6CV.

Part d:

Equivalent Capacitance = Net charge / Voltage

Eq. C = 6CV/V = 6C

So the equivalent capacitance is 6C

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A stockroom worker pushes a box with mass 11.2 kg on a horizontal surface with constant speed of 3.50m/s. The coefficient of kin

stellarik [79]

Answer:

3.125 m

Explanation:

We are given that

Mass of box=m=11.2 kg

Speed of box=u=3.5m/s

Coefficient of kinetic friction= $\mu_k=0.2$

Final velocity,v=0

a.We have to find the horizontal force applied by worker to maintain the motion.

According to question

Horizontal force=F= $f=\mu_kmg$

$g=9.8m/s^2$

Substitute the values

Horizontal force= $F=0.2\times 11.2\times 9.8=21.95 N$

b.According to work-energy theorem

$W=\frac{1}{2}mv^2-\frac{1}{2}mu^2$

$-\mu mg s=\frac{1}{2}(11.2)(0)^2-\frac{1}[2}(11.20)(3.5)^2$

$-\mu mg s=-\frac{1}{2}(11.2)(3.5)^2$

$0.2\times (11.2)\times 9.8\times s=\frac{1}{2}(11.2)(3.5)^2$

$s=\frac{11.2\times (3.5)^2}{2\times 0.2\times 11.2\times 9.8}$

$s=3.125 m$

Hence, the box slide before coming to rest=3.125 m

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