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Sauron [17]
3 years ago
15

How do i figure out the net force of an object? explain please :)

Physics
1 answer:
Savatey [412]3 years ago
3 0
Draw a free body diagram to help you visualize. draw the object and use arrows to label and determine where all the forces acting upon it will be. then u add up the sum of all forces. if two forces of equal magnitude and opposite directions will cancel each other out.
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8.
svetoff [14.1K]

Answer:

9.5 seconds

Explanation:

You need to make the ending speed into meters per sec.

So it would be 120km/h = 33.33m/s

If the acceleration rate is 3.5m/s, then the final answer would equal to 9.5 seconds

Also btw rounding it up is a good idea too so 10 seconds if you round

8 0
3 years ago
An astrophysicist mounts two thin lenses along a single optical axis (the lenses are at right angles to the line connecting them
Alexandra [31]

Answer:

1 / i + 1 / o = 1 / f     thin lens equations

i = o f / (o - f)   rearranging

Lens 1:   object = 30 cm    f = 15.2 cm

i1 = 30 * 15.2 / (30 - 15.2) = 30.8 cm

o2 = 40.2 - 30/8 = 9.4 cm    distance of image 1 from lens 2

i2 = 9.4 * 15.2 / (9.4 - 15.2) = - 24.6 cm

The final image is 24.6 cm to the left of lens 2

The first image is inverted

The second image is erect (as seen from the first image)

So the final image is inverted

M = m1 * m2 = (-30.8 / 30) * (24.6 / 9.4) = -2.69

7 0
3 years ago
An Alpha particle moving in north direction give reasons​
Mazyrski [523]

Your question has been heard loud and clear.

An alpha particle , can move in any direction randomly. But with a magnetic field , we can deflect the alpha particle in any direction we want.

So , the magnetic field must be placed to the west of the alpha particle , so that the particle gets deflected and moves towards the north direction.

Thank you.

7 0
3 years ago
Read 2 more answers
A mass is attached to the end of a spring and set into oscillation on a horizontal frictionless surface by releasing it from a c
Aneli [31]

Answer

given,

x = (3.9 cm)sin[(9.3 rad/s)πt]

general equation of displacement

x = A sin ω t

A is amplitude

now on comparing

c) Amplitude  =3.9 cm

a) frequency =

     f = \dfrac{\omega}{2\pi}

     f = \dfrac{9.3\pi}{2\pi}

           f = 4.65 Hz

b) period of motion

        T= \dfrac{1}{f}

        T= \dfrac{1}{4.65}

        T = 0.215 s

d) time when displacement is equal to x= 2.6 cm

x = (3.9 cm)sin[(9.3 rad/s)πt]

2.6 = (3.9 cm)sin[(9.3 rad/s)πt]

sin[(9.3 rad/s)πt] = 0.667

9.3 π t = 0.73

t = 0.025 s

4 0
3 years ago
Read 2 more answers
PLEASE HELP !! ILL GIVE BRAINLIEST A projectile is launched from the top of a 15 m tall building at a speed of 25ms at an angle
mihalych1998 [28]

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5 0
3 years ago
Read 2 more answers
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