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PolarNik [594]
3 years ago
15

The acceleration due to the earth's gravity, in si units, is 9.8 m/s2. in the absence of air friction, a ball is dropped from re

st. its speed on striking the ground is exactly 60 m/s. for what time interval was the ball falling?
Physics
1 answer:
gogolik [260]3 years ago
7 0
We don't know anything about the amount of distance it travels, but that's okay. The only equation we need here is 

velocity(final) = velocity(initial) + acceleration * time
vf = vi + (a * t)

The ball is dropped from rest, so vi = 0 m/s.
We want it so that the ball hits the ground with a final velocity of 60 m/s, so vf = 60 m/s. 
We are given the acceleration due to gravity, a = 9.8 m/s^2.
We are solving for the time, t = ?.

Now we just plug in the values.
vf = vi + (a * t)
60 m/s = 0 m/s + (9.8 m/s^2)*(t)

60 = 9.8t

60 / 9.8 = t

t = 6.122 s

Hopefully this is the right answer.

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one car accelerates at half the rate of another how much longer does it take the first car to travel a quarter mile
Orlov [11]

Answer:

t=1/4v1

Explanation:

Given data

Car one

Speed =v1

Time =t

Distance =1/4 mile

Given data

Car two

Speed =v1/2

Time =t

Distance =d

Speed =distance/time

v1=1/4/t

v1t=1/4

t=1/4*1/v1

t=1/4v1 seconds

7 0
2 years ago
How much heat is required to raise the temperature of 100 g of water from 5 degree celsius to 90 degree celsius?​
Scorpion4ik [409]

Explanation:

1-How many moles of NazCOs are in 10.0 ml of a 2.0 M solution?

2-How many moles of NaCl are contained in 100.0 ml of a 0.20 M solution?

3- What weight (in grams) of H2SO4 would be needed to make 750.0 ml of

2.00 M solution?

4-What volume (in ml) of 18.0 M H2SO4 is needed to contain 2.45 g H2S04?

8 0
2 years ago
A 45.0-kg girl is standing on a 168-kg plank. The plank, originally at rest, is free to slide on a frozen lake, which is a flat,
muminat

Answer:

The speed of the plank relative to the ice is:

v_{p}=-0.33\: m/s

Explanation:

Here we can use momentum conservation. Do not forget it is relative to the ice.

m_{g}v_{g}+m_{p}v_{p}=0 (1)

Where:

  • m(g) is the mass of the girl
  • m(p) is the mass of the plank
  • v(g) is the speed of the girl
  • v(p) is the speed of the plank

Now, as we have relative velocities, we have:

v_{g/b}=v_{g}-v_{p}=1.55 \: m/s (2)

v(g/b) is the speed of the girl relative to the plank

Solving the system of equations (1) and (2)

45v_{g}+168v_{p}=0

v_{g}-v_{p}=1.55

v_{p}=-0.33\: m/s

I hope it helps you!      

8 0
3 years ago
Which of the following is an example of homeostasis?
Bogdan [553]
The answer is : D

Reasoning:
Homeostasis is the body’s balance
3 0
3 years ago
Read 2 more answers
25 POINTS FOR ANSWER How are Newton’s Laws used to describe the motion of planets? Justify your response in two or more complete
Alexus [3.1K]

Pour la seule et simple raison qu'elle s'exerce entre tous les corps de l'univers ( objet, astres etc..

Si on tient compte des frottements liés aux chocs successifs des billes les une sur les autre, au bout d'un certain temps, le mouvement va cesser.

Si on dit que toute l'énergie potentielle de pesanteur est transformée en énergie cinétique, et réciproquement, donc que l'énergie mécanique est conservée au fil des chocs et des rebonds, alors, le mouvement est perpétuel. Le nombre de billes qui remontent est toujours égal au nombre de billes qu'on a lâchées.

La première loi concerne des systèmes immobiles, ou plutôt on considère des systèmes dit "isolé", c'est à dire qu'ils ne sont pas soumis à d'autre force que celle que l'on connait.

Ce qu'il faut retenir de celui ci c'est ça :

Si j'ai un système en mouvement rectiligne uniforme OU immobile, alors :

Avec F1 F2 F3, trois forces s'exercant sur mon système

Attention ! Ici je n'ai pas mit les flèches sur les différentes forces mais elles sont obligatoires ! On parle de vecteur force !

Pour la deuxième loi :

C'est le même principe, la différence c'est que l'on est en mouvement.

 

Avec a le vecteur accélération. Il y a beaucoup de ressource sur le net, n'hésite pas à regarder, la j'ai simplement pu te donner les expressions les plus connus. Mais il faudra les manipuler, et sans exercice sur lequel se baser, c'est plus difficile ! 

La troisième loi est bien moins importante que les deux autres, mais n'hésite pas à regarder sur le net, tu trouveras l'énoncé. C'est la même logique.

4 0
3 years ago
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