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stellarik [79]
3 years ago
8

An object acted on by three forces moves with constant velocity. one force acting on the object is in the positive x direction a

nd has a magnitude of 6.3 n; a second force has a magnitude of 4.3 n and points in the negative y direction. find the direction and magnitude of the third force acting on the object.
Physics
1 answer:
Afina-wow [57]3 years ago
4 0
10.6 is the answer i think im not the best at math
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A child is sliding down a slide at the playgound. is mechanicalenergy conserved
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No. Mechanical energy is not conserved.  There's quite a bit of friction on the slide.  So some of the potential energy is lost to heat on the way down, and the child arrives at the bottom with hot pants and less kinetic energy than you might expect.

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Step 2: Apply NEwton's second law Apply ∑Fy = may , what should ay be equal to, since the block doesn't move in the y direction
andrey2020 [161]

Answer:

∑Fy = 0, because there is no movement, N = m*g*cos (omega)

Explanation:

We can solve this problem with the help of a free body diagram where we show the respective forces in each one of the axes, y & x. The free-body diagram and the equations are in the image attached.

If the product of mass by acceleration is zero, we must clear the normal force of the equation obtained. The acceleration is equal to zero because there is no movement on the Y-axis.

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3 years ago
The acceleration of a cart rolling down a ramp depends on __________.
zmey [24]

The angle that the cart rolls with the horizontal. The closer the ramp gets to 90 degrees the faster the cart will accelerate.

8 0
3 years ago
If a hypothesis is falsifiable, _____.
san4es73 [151]

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Explanation:

4 0
3 years ago
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A subway train starts from rest at a station and accelerates at a rate of 1.68 m/s2 for 14.2 s. It runs at constant speed for 68
liubo4ka [24]

Answer:

total distance = 1868.478 m

Explanation:

given data

accelerate = 1.68 m/s²

time = 14.2 s

constant time = 68 s

speed = 3.70 m/s²

to find out

total distance

solution

we know train start at rest so final velocity will be after 14 .2 s is

velocity final = acceleration × time      ..............1

final velocity = 1.68 × 14.2

final velocity = 23.856 m/s²

and for stop train we need time that is

final velocity = u + at

23.856 = 0 + 3.70(t)

t = 6.44 s

and

distance = ut + 1/2 × at²     ...........2

here u is initial velocity and t is time for 14.2 sec

distance 1 = 0 + 1/2 × 1.68 (14.2)²

distance 1 = 169.37 m

and

distance for 68 sec

distance 2= final velocity × time

distance 2= 23.856 × 68

distance 2 = 1622.208 m

and

distance for 6.44 sec

distance 3 = ut + 1/2 × at²

distance 3 = 23.856(6.44) - 0.5 (3.70) (6.44)²

distance 3 = 76.90 m

so

total distance = distance 1 + distance 2 + distance 3

total distance = 169.37 + 1622.208 + 76.90

total distance = 1868.478 m

7 0
3 years ago
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