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Fantom [35]
3 years ago
11

The 8.00-cm long second hand on a watch rotates smoothly.

Physics
1 answer:
dolphi86 [110]3 years ago
4 0
The watch hand covers an angular displacement of 2π radians in 60 seconds.

ω = 2π/60
ω = 0.1 rad/s

v = ωr
v = 0.1 x 0.08
v = 8 x 10⁻³ m/s
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Kay [80]

Answer:

one that acts in the direction of the acceleration is the static friction force. The ... Express fs,max in terms of Fn in the x ... ma. FF = − g n. Fn − w = may = 0 or, because ay = 0 and Fg = mg, mg. F = n.

5 0
3 years ago
A mine car, whose mass is 440kg, rolls at a speed of 0.50m/s on ahorizontal track, as the drawing shows. A 150kg chunk of coalha
ella [17]

Answer: 0.56 m/s

Explanation:

hello, there is 25° inclination angle for the chute in the drawing. Thankfully, I know this problem. The conservation of momentum.

so there are X and Y components for the momentum in this problem. The Y component is not conserved as when the coal gets in the cart, the normal force exerted by the surface reduces it to 0.

Now, the X component is definitely conserved here.

so you have the momentum of the cart which is 440*0.5 added to the momentum of the chunk which is 150*0.8*cos(25°), that is the momentum before the coupling between the objects. Afterwards both objects will have the same velocity, so we write the equation like this:

440*0.5 + 150*0.8*cos(25) = 440*v_{final} + 150*v_{final} \\ => 220+120*cos(25) = (440+150)v_{final} => v_{final} = \frac{220+120*cos(25)}{590}  = 0.56 m/s

3 0
3 years ago
Why are force fields used to describe magnetic force?
Illusion [34]

Answer:

If I'm not working I think the answer is C.

7 0
4 years ago
A solid with flat sides that meet at sharp edges and corners is called
Illusion [34]
1.) Product is a solid
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3 0
3 years ago
According to coulombs law, what will happen to the force between two charged particles if the magnitude are increased by 6 times
seropon [69]

Answer: The electrostatic force will be the same

Explanation:

According to Coulomb's Law, when two electrically charged bodies come closer, appears a force that attracts or repels them, depending on the sign of the charges of this two bodies or particles.  

In this sense, this law states the following:

"The electrostatic force F_{E} between two point charges q_{1} and q_{2} is proportional to the product of the charges and inversely proportional to the square of the distance d that separates them, and has the direction of the line that joins them"  

 

F_{E}= K\frac{q_{1}.q_{2}}{d^{2}}  (1)

Being K is a proportionality constant.  

Now, if each q_{1} and q_{2} are increased by 6, and the distance between them as well, we will have the following:

F_{E}= K\frac{6 q_{1}. 6 q_{2}}{(6d)^{2}}  (2)

F_{E}= 36 K\frac{q_{1}. q_{2}}{36d^{2}}  (3)

Simplifying:

F_{E}= K\frac{q_{1}.q_{2}}{d^{2}}  (4)

Comparing (1) with (4) we can see the electrostatic force is the same.

4 0
4 years ago
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