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just olya [345]
2 years ago
12

I AM TIMED !! 30 POINTS Which is the smallest possible particle of an element?

Physics
2 answers:
zimovet [89]2 years ago
4 0

Answer:

atoms

Explanation:

they are the smallest

melamori03 [73]2 years ago
3 0
The correct answer is Atoms!
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WHAT SHOULD I NAME MY DEAD RAT?
olga_2 [115]

Answer:

pablito

Explanation:

6 0
3 years ago
A ramp is 1.0 m high and 3.0 m long. What is the IMA of the ramp?
oksano4ka [1.4K]
To calculate the ideal mechanical advantage for an inclined plane, divide th length of the incline by the height of the incline. 
Therefore; IMA = L/h
L= 3.0 m, while h =1.0 m
 IMA = 3/1
         = 3
Therefore the IMA of the ramp is 3 
This means the ramp increases the force that is being exerted by 3 times.
4 0
3 years ago
Hi, I need your help with this Physics exercise, I hope you can help me A pulse moving to the right along the x axis is represen
igomit [66]

Answer:

Velocity = 0.309 m/s

Along negative x axis

Explanation:

A pulse moving to the right along the x axis is represented by the wave function

y(x,t) = 2/ (x - 3t)² + 1

At t =0

y(x,0) = 2/ ((x - 3(0))² + 1)

        =2 / (x² + 1)

At t = 1

y(x,t) = 2/ ((x - 3(1))² + 1)

= 2 /(( x - 3)² + 1)

At t = 2

y(x,t) = 2/ ((x - 3(2))² + 1)

= 2 /(( x - 6)² + 1)

For the pulse with expression y(x,t) = 4.5e^{-(8.73x + 2.70t)}²

The Velocity is

V = 2.7 / 8.73

= 0.309 m/s

3 0
3 years ago
Assuming that the hill on the left will provide all of the potential energy for the ride, would this roller coaster actually mak
trapecia [35]
Yes I believe it would I think the answer is D
3 0
3 years ago
Two equal forces are applied to a door. The first force is applied at the midpoint of the door, the second force is applied at t
Orlov [11]

Answer:

D) the second at the doorknob

Explanation:

The torque exerted by a force is given by:

\tau = Fdsin \theta

where

F is the magnitude of the force

d is the distance between the point of application of the force and the centre of rotation

\theta is the angle between the direction of the force and d

In this problem, we have:

- Two forces of equal magnitude F

- Both forces are perpendicular to the door, so \theta=90^{\circ}, sin \theta=1

- The first force is exerted at the midpoint of the door, while the 2nd force is applied at the doorknob. This means that d is the larger for the 2nd force

--> therefore, the 2nd force exerts a greater torque

4 0
3 years ago
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