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2 years ago

What type of energy slows down

Physics

1 answer:

Tema [17]2 years ago

4 0

It is Potential energy it's at rest

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Two electrons are passing 20.0 mm apart. What is the electric repulsive force that they exert on each other

vladimir1956 [14]

15.0 I’m pretty sure that’s the answer to your question

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2 years ago

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A certain alarm clock ticks four times each second, with each tick representing half a period. The balance wheel consists of a t

Semenov [28]

Answer:

a. $I=2.77x10^{-8} kg*m^2$

b. $K=4.37 x10^{-6} N*m$

Explanation:

The inertia can be find using

$I = m*r^2$

$m = 0.95 g * \frac{1 kg}{1000g}=9.5x10^{-4} kg$

$r=0.54 cm * \frac{1m}{100cm} =5.4x10^{-3}m$

$I = 9.5x10^{-4}kg*(5.4x10^{-3}m)^2$

$I=2.77x10^{-8} kg*m^2$

now to find the torsion constant can use knowing the period of the balance

T=0.5 s

$T=2\pi *\sqrt{\frac{I}{K}}$

Solve to K'

$K = \frac{4\pi^2* I}{T^2}=\frac{4\pi^2*2.7702 kg*m^2}{(0.5s)^2}$

$K=4.37 x10^{-6} N*m$

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2 years ago

Determine the velocity that a car should have while traveling around a frictionless curve of radius 100m and that is banked 20 d

alex41 [277]

Answer:

$v=18.89\frac{m}{s}$

Explanation:

From the free-body diagram for the car, we have that the normal force has a vertical component and a horizontal component, and this component act as the centripetal force on the car:

$\sum F_x:Nsin(20^\circ)=ma_c(1)\\\sum F_y:Ncos(20^\circ)=mg(2)$

Solving N from (2) and replacing in (1):

$N=\frac{mg}{cos(20^\circ)}\\(\frac{mg}{cos(20^\circ)})sin(20^\circ)=ma_c\\gtan(20^\circ)=a_c$

The centripetal acceleration is given by:

$a_c=\frac{v^2}{r}$

Replacing and solving for v:

$gtan(20^\circ)=\frac{v^2}{r}\\v=\sqrt{grtan(20^\circ)}\\v=\sqrt{9.8\frac{m}{s^2}(100m)tan(20^\circ)}\\v=18.89\frac{m}{s}$

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3 years ago

In a clock, the average speed is maximum for the tip of<br />a) Seconds hand b) Hours hand c) Minutes hand

Simora [160]

A, hope this helped! I didn’t really get it but I think it’s correct?

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3 years ago

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When t= 1 a stone B is thrown vertically upwards from ground level with speed 5ms (ii) find t when a and b collide

Ksju [112]

Answer:

i don't know if this is good for you but

Explanation:

ignoring frictional air resistance (drag) the speed on return is the same as when it left the ground (5 m/s but in the opposite direction).

Note: this points out a good reason for not firing live bullets into the air..they will return somewhere and at the same speed.

However, if you take into account the atmospheric drag the reurn speed will be somewhat smaller (but in the case of a bullet, probably still lethal.) Drag depends on many factors and is difficult to calculate.

5 0

2 years ago