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2 years ago

How can I solve this?

Physics

1 answer:

Fofino [41]2 years ago

8 0

Explanation:

The equivalent capacitance of capacitors in parallel can be determined as

$C_{eq} = C_1 + C_2 + C_3$

$\:\:\:\:\:= 40\:\text{F} + 10\:\text{F} + 50\:\text{F} = 100\:\text{F}$

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A 1.0 kg red superball moving at 5.0 m/s collides head-on with stationary blue superball of mass 4.0 kg in an elastic collision.

STatiana [176]

Initial conditions:
m1 = 1.0 ; v1 = 5
m2 = 4.0 ; v2 = 0

In the case where the second object (sometimes called the target) is at rest the velocities after the condition are

v1' = v1* (m1-m2)/(m1+m2)
v2' = 2v1*m1/(m1+m2)

For this we get
v1' = 5*(-3)/5 = -3m/s (moving in the opposite direction as before at 3m/s
v2' = 2*5*(1)/5 = 2m/s in the same direction as the original ball was moving
you can see these directions by looking at the signs. The momenta also add to the initial momentum as required.

5 0

3 years ago

Read 2 more answers

A 205 kg log is pulled up a ramp by means of a rope that

Volgvan

Answer:2.737 kN

Explanation:

Given

mass of log(m)=205 kg

ramp inclination $=30^{\circ}$

coefficient of kinetic friction between log and ramp is ( $\mu _k$ )=0.9

log has an acceleration of 0.8 m/s^2

Let T be the tension in the rope

$T-mgsin\theta -f_r=ma$

Where $mgsin\theta$ =Sin component of weight

$f_r=friction\ Force=\mu _KN$ (Where N is Normal reaction)

$T-mgsin\theta -\mu _k\left ( mgcos\theta \right )=ma$

$T=m\left ( gsin\theta +\mu _kcos\theta +a\right )$

sin30=0.5

cos30=0.866

$T=205\times \left ( 9.81\times 0.5+0.9\times 9.81\times 0.866+0.8\right )$

$T=205\times 13.35=2.737 kN$

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2 years ago

which type of geoscientist interprets data from certain instruments to help detect earthquakes and identify fault lines

Volgvan

Seismologists would be your answer.

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3 years ago

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A 60-m-long chain hangs vertically from a cylinder attached to a winch. Assume there is no friction in the system and that the

Mariulka [41]

Answer:

part (a). 176580 J

part (b). 197381 J

Explanation:

Given,

Density of the chain = $\rho\ =\ 10\ kg/m.$
Length of the chain = L = 60 m
Acceleration due to gravity = g = 9.81 $m/s^2$

part (a)

Let dy be the small element of the chain at a distance of 'y' from the ground.

mass of the small element of the chain = $\rho dy$

Work done due to the small element,

$dw\ =\ \rho g (60\ -\ y)dy\\$

Total work done to wind the entire chain = w

$w\ =\ \displaystyle\int_{0}^{L} \rho g(60\ -\ y)dy\\\Rightarrow w\ =\ \rho g\left |(60y\ -\ \dfrac{y^2}{2})\ \right |_{0}^{60}\\\Rightarrow w\ =\ 10\times 9.81\times (60\times 60\ -\ \dfrac{60^2}{2})\\\Rightarrow w\ =\ 176580\ J$