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Whitepunk [10]
3 years ago
12

Which has the greater acceleration, a person going from 0 m/s to 10 m/s in 10 seconds or an ant going from 0 m/s to 0.25 m/s in

2.0 seconds?​
Physics
1 answer:
monitta3 years ago
4 0

<u>P</u><u>e</u><u>r</u><u>s</u><u>o</u><u>n</u><u>-</u><u>1</u>

  • Initial velocity=u=0m/s
  • Final velocity=v=10m/s
  • Time=10s=t

\\ \sf\longmapsto Acceleration=\dfrac{v-u}{t}

\\ \sf\longmapsto Acceleration=\dfrac{10-0}{10}

\\ \sf\longmapsto Acceleration=\dfrac{10}{10}

\\ \sf\longmapsto Acceleration=1m/s^2

<u>P</u><u>e</u><u>r</u><u>s</u><u>o</u><u>n</u><u>-</u><u>2</u>

  • initial velocity=0m/s=u
  • Final velocity=v=0.25m/s
  • Time=t=2s

\\ \sf\longmapsto Acceleration=\dfrac{0.25-0}{2}

\\ \sf\longmapsto Acceleration=\dfrac{0.25}{2}

\\ \sf\longmapsto Acceleration=0.125m/s^2

Person-1 is accelerating faster.

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Find the following answer based on the image.
saul85 [17]

<u>We are given:</u>

Mass of Neptune = 1.03 * 10²⁶ kg

Distance from the center of Neptune (r) = 2.27 * 10⁷

now, computing the value of the acceleration due to gravity (g)

<u>Finding g:</u>

We know the formula:

g = G(mass of planet) / (r)²

g = [6.67 * 10⁻¹¹ * 1.03*10²⁶] / (2.27*10⁷)                      [since G is 6.67*10⁻¹¹]

g = (6.87 * 10¹⁵) / (5.15 * 10¹⁴)

which can be rewritten as:

g = (6.87 * 10¹⁵ * 10⁻¹⁴) / 5.15

g = (6.87 * 10¹⁵⁻¹⁴) / 5.15

g = (6.87/5.15) * 10

g = 1.34 * 10

g = 13.4 m/s² <em>(approx)</em>

5 0
3 years ago
A 2,000 kg car is moving at 15m/s when it collides with a 1200kg car sitting still. Briefly compare the impulse imparted on the
I am Lyosha [343]

impulse = F × t

The greater the impulse exerted on something, the greater will be the change in momentum.

impulse = change in momentum

Ft = ∆(mv)

4 0
3 years ago
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3 years ago
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Problem 4: A uniform flat disk of radius R and mass 2M is pivoted at point P A point mass of 1/2 M is attached to the edge of th
brilliants [131]

From the case we know that:

  1. The moment of inertia Icm of the uniform flat disk witout the point mass is Icm = MR².
  2. The moment of inerta with respect to point P on the disk without the point mass is Ip = 3MR².
  3. The total moment of inertia (of the disk with the point mass with respect to point P) is I total = 5MR².

Please refer to the image below.

We know from the case, that:

m = 2M

r = R

m2 = 1/2M

distance between the center of mass to point P = p = R

Distance of the point mass to point P = d = 2R

We know that the moment of inertia for an uniform flat disk is 1/2mr². Then the moment of inertia for the uniform flat disk is:

Icm = 1/2mr²

Icm = 1/2(2M)(R²)

Icm = MR² ... (i)

Next, we will find the moment of inertia of the disk with respect to point P. We know that point P is positioned at the arc of the disk. Hence:

Ip = Icm + mp²

Ip = MR² + (2M)R²

Ip = 3MR² ... (ii)

Then, the total moment of inertia of the disk with the point mass is:

I total = Ip + I mass

I total = 3MR² + (1/2M)(2R)²

I total = 3MR² + 2MR²

I total = 5MR² ... (iii)

Learn more about Uniform Flat Disk here: brainly.com/question/14595971

#SPJ4

8 0
1 year ago
If you speed up from rest to 12m/s in 3 seconds, what is your acceleration?
boyakko [2]


b) 4m/s/s

This is because you divide the speed you reach, by the time it takes to get to that speed:

12m/s ÷ 3s = 4m/s/s

The units come from what you divide, meters per second ÷ seconds this can be written as m/s/s or ms-² 

6 0
3 years ago
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