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3 years ago

Pls need an answer asap

Physics

1 answer:

Andrei [34K]3 years ago

3 0

c] Electrons

Electron gets transferred

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6. If you lift an object, to find the work you did on the object, you must find

OleMash [197]

I think D the weight but I’m not sure sorry if I’m wrong

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3 years ago

The automobile has a weight of 2700 lb and is traveling forward at 4 ft>s when it crashes into the wall. If the impact occurs

alex41 [277]

Answer:

$F_b=153918\, lb.ft.s^{-2}$

Explanation:

Given:

mass, $m=2700 \,lb$

time, $t=0.06\,s$

velocity, $v=4\,ft.s^{-1}$

coefficient of kinetic friction between wheels & pavement, $\mu=0.3$

According to first condition,

$F\times \Delta t=m\times v$

$F\times 0.06= 2700\times 4$

$F=180000\,lb.ft.s^{-2}$

According to second condition,

<u>Magnitude of frictional force (which acts opposite to the direction of motion):</u>

$f=\mu.N$

where N is the normal reaction.

$f=0.3\times 2700\times 32.2$

$f=26082 \,lb.ft.s^{-2}$

Now, the impulsive force on the wall if the brakes were applied during the crash:

$F_b= F-f$

$F_b=180000-26082$

$F_b=153918\, lb.ft.s^{-2}$

3 0

3 years ago

Two devices with capacitances of 25 μf and 5.0 μf are each charged with separate 120 v power supplies. calculate the total energ

mash [69]

The energy stored in a capacitor is given by
$U= \frac{1}{2} CV^2$
where C is the value of the capacitance while V is the voltage difference applied to the capacitor.

Let's calculate the energy of the first capacitor:
$U_1 = \frac{1}{2} (25\cdot 10^{-6}F)(120 V)=1.5 \cdot 10^{-3}J$

And now the energy of the second capacitor:
$U_2 = \frac{1}{2} (5 \cdot 10^{-6}F)(120 V)=3 \cdot 10^{-4}J$

So, the total energy stored in the two capacitors is
$U=U_1 +U_2 = 1.8 \cdot 10^{-3}J$

3 0

3 years ago

Read 2 more answers

A 5 kg rock is raised 28 m above the ground level. What is the change in its potential energy?

marin [14]

Let's assume that ground level is the height 0 meters. The change in potential energy is going to be gravitational potential energy, which is given by PE=mgh.
ΔPE=mgh-mgy
=mg(h-y)
=50(28-0)
=1400 J

3 0

4 years ago

A child slides down the water slide at a swimming pool and enters the water at a final speed of 4.22 m/s. At what final speed wo

zheka24 [161]

Answer:

v'=5.97 m/s

Explanation:

If the friction and fluid friction are ignored, then by law of conservation of mechanical energy, potential energy at the top of the slide must be equal to the kinetic energy at the bottom of the slide. Thus, the height of slide and final speed at the bottom of the slide are related as:

$mgh = \frac {1}{2} mv^2\\v =\sqrt{2gh}$

let the final speed be v' when h' = 2 h

$\frac{v'}{v}=\frac{\sqrt h'}{\sqrt h}\\v'=\sqrt{\frac{2h}{h}}\times v\\v'=\sqrt2 v\\v'=5.97 m/s$

8 0

3 years ago