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Dima020 [189]
3 years ago
11

What units are assumed for speed limit signs in the United States

Physics
1 answer:
asambeis [7]3 years ago
6 0

Answer:

Miles

Explanation:

M.P.H = miles per hour

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. Determine if approximate cylindrical symmetry holds for the following situations. State why or why not. (a) A 300-cm long copp
MA_775_DIABLO [31]

Answer:

a) Yes

b) No

Explanation:

In the first case, part a, yes we can say for certainty that cylinderical symmetry holds. Why so? You may ask. This is because from the question, we are told that the length of the rod is 300 cm. And this said length is longer than the distance to the point from the center of the rod, which is 5 cm.

In the second half of the question, I beg to disagree that cylindrical symmetry holds. Again, you may ask why, this is because the length of the rod in this case, is having the same order of magnitude as the distance to the center of the rod. Thus, it is not symmetrical.

6 0
3 years ago
What is the largest ecosystem on the planet?
asambeis [7]
C. 

hope that helped you!!!


5 0
3 years ago
Read 2 more answers
If we have less power, we most likely have
katrin [286]
<span>the same amount of work being done over a longer period of time.</span>
3 0
3 years ago
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An ideal spring hangs from the ceiling. A 2.15 kg mass is hung from the spring, stretching the spring a distance d = 0.0895 m fr
Igoryamba

Answer:

The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

Explanation:

Given that,

Mass = 2.15 kg

Distance = 0.0895 m

Amplitude = 0.0235 m

We need to calculate the spring constant

Using newton's second law

F= mg

Where, f = restoring force

kx=mg

k=\dfrac{mg}{x}

Put the value into the formula

k=\dfrac{2.15\times9.8}{0.0895}

k=235.41\ N/m

We need to calculate the kinetic energy of the mass

Using formula of kinetic energy

K.E=\dfrac{1}{2}mv^2

Here, v = A\omega

K.E=\dfrac{1}{2}m\times(A\omega)^2

Here, \omega=\sqrt{\dfrac{k}{m}}^2

K.E=\dfrac{1}{2}m\times A^2\sqrt{\dfrac{k}{m}}^2

K.E=\dfrac{1}{2}kA^2

Put the value into the formula

K.E=\dfrac{1}{2}\times235.41\times(0.0235)^2

K.E=0.06500\ J

Hence, The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

8 0
3 years ago
A bird is flying north with a mass of 2.5kg has a momentum of 17.5kg m/s at what velocity is it flying
klasskru [66]
Momentum = mass • velocity
v= 17.5/2.5
= 7 m/s
8 0
3 years ago
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