Why would an orange roll of your cafeteria tray if you stopped suddenly

1 answer:

6 0

Because the frictional force between the orange skin peel is great enough when you are walking for it to be carried on the tray, along with the gravitational force downwards onto the tray. When you stop, the force that you exerted moving forward it the same as on the tray and on the orange. So when you stop, the force is still on the orange as the same velocity as your we’re traveling, while the tray and you stop.

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If your speed changes from 10 km/h to 6 km/h, you have a(n) ______ acceleration.

Delicious77 [7]

If your speed changes from 10 km/h to 6 km/h then
you have an acceleration.

Whether it's a positive or negative one completely depends
on which direction you decided to call the positive direction,
when you started considering your speed and its changes.

If you decided to call the direction in which you're traveling
the positive direction, then a decrease in your speed is a
negative acceleration.

But you could just as easily have said that you're traveling
in the negative direction. If you did that, then a decrease in
your speed would be a positive acceleration.

It's completely up to you, and how you define things.

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3 years ago

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A person throws a tennis ball 6 m/s straight up. How long does it take for it to come back to their hand?

Alja [10]

(t) = 2t = 1.22 sec. I believe ...

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3 years ago

A 225-g object is attached to a spring that has a force constant of 74.5 N/m. The object is pulled 6.25 cm to the right of equil

snow_lady [41]

Answer:

1.137278672 m/s

+5.9 cm or -5.9 cm

Explanation:

A = Amplitude = 6.25 cm

m = Mass of object = 225 g

k = Spring constant = 74.5 N/m

Maximum speed is given by

$v_m=A\omega\\\Rightarrow v_m=A\sqrt{\dfrac{k}{m}}\\\Rightarrow v_m=6.25\times 10^{-2}\times \sqrt{\dfrac{74.5}{0.225}}\\\Rightarrow v_m=1.137278672\ m/s$

The maximum speed of the object is 1.137278672 m/s

Velocity is at any instant is given by

$\dfrac{v_m}{3}=\omega\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A\omega}{3}=\omega\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A}{3}=\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A^2}{9}=A^2-x^2\\\Rightarrow A^2-\dfrac{A^2}{9}=x^2\\\Rightarrow x^2=\dfrac{8}{9}A^2\\\Rightarrow x=A\sqrt{\dfrac{8}{9}}\\\Rightarrow x=6.25\times 10^{-2}\sqrt{\dfrac{8}{9}}\\\Rightarrow x=\pm 0.0589255650989\ m$

The locations are +5.9 cm or -5.9 cm

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3 years ago

A mechanic is trying to loosen a nut with a wrench, but it is stuck. What could the mechanic do to help loosen the nut? (Hint: T

SIZIF [17.4K]

C. Use a longer wrench, this is because it would create a greater turning moment as moment=force*distance

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3 years ago

What are the equipment requirements for windshields and side windows

8_murik_8 [283]

Answer: the glass are not to be covered or treated.

Explanation:

The windshield should be made with safety glass material and it must not be covered or treated with a material that will make it reflective or non-transparent.

Side windows located rear of the driver and rear windows having a material in together with glazing material that has a luminous reflectance of 35% plus or minus three per cent or less.

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