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ioda
4 years ago
13

Why might it be necessary to ignore some of the data points just before and just after the collision?

Physics
1 answer:
krek1111 [17]4 years ago
7 0
We have no idea. We need to examine the experimental set-up. You've given us no information, except that there may have been some sort of collision.
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When calculating acceleration, what about the answer would tell you that the object is slowing down?
Soloha48 [4]

If the object's velocity and acceleration have <em>opposite signs</em>, then the object is slowing down.

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4 years ago
A 1.8 kg book has been dropped from the top of the football stadium. Its speed is 4.8 m/s when it is 2.9 meters above the ground
Alexus [3.1K]

Answer:

Answer:

72.936 Joule

Explanation:

Mechanical Energy = Potential Energy + Kinetic Energy

Kinetic Energy = (1/2) x m x V² = (1/2) x 1.8 x 4.8² = 20.736 J

Potential Energy = m x g x h = 1.8 x 10 x 2.9 = 52.2 J

Total Mechanical Energy = 20.736 + 52.2 = 72.936 Joule

Explanation:

8 0
3 years ago
Natalie and Will are discussing socialization. Natalie says that socialization occurs when an animal becomes accustomed to the p
skad [1K]
The correct answer for this question is this one: "C. Neither Natalie nor Will." Natalie and Will are discussing socialization. Natalie says that socialization occurs when an animal becomes accustomed to the people in the household. <span>Will says that socialization is easily attained if the animal is first exposed to humans after 12 weeks of age.</span>
7 0
3 years ago
Read 2 more answers
The electric current leaves the battery through the --- complete this blank
aev [14]
The bulb

Hope this helps!
5 0
4 years ago
An undiscovered planet, many lightyears from Earth, has one moon in a periodic orbit. This moon takes 2060 × 103 seconds (about
Fittoniya [83]

Answer:

Acceleration, a=2.22\times 10^{-3}\ m/s^2

Explanation:

It is given that,

Time period of revolution of the moon, T=2060\times 10^3\ s

If the distance from the center of the moon to the surface of the planet is, h=235\times 10^6\ m

The radius of the planet, r=3.9\times 10^6\ m

Let a is the moon's radial acceleration. Mathematically, it is given by :

a=R\times \omega^2, R is the radius of orbit

Since, \omega=\dfrac{2\pi}{T}

The radius of orbit is,

R=r+h

R=3.9\times 10^6\ m+235\times 10^6\ m=238900000\ m

So, a=\dfrac{4\pi^2 R}{T^2}

a=\dfrac{4\pi^2 \times 238900000}{(2060\times 10^3)^2}

a=2.22\times 10^{-3}\ m/s^2

Hence, this is the required solution for the radial acceleration of the moon.

5 0
3 years ago
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