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3 years ago

Why did friction make it difficult to discover newtons first law of motion

2 answers:

7 0

Friction made it difficult because an object in motion may not always stay in motion due to the friction caused by its movement. For instance, if you roll a ball down a straight platform, it will not go on forever. The friction of the platform and the ball with eventually slow it down and stop it.

Hope this helps!!

V125BC [204]3 years ago

6 0

Because Static Friction and Inertia can be confused, because mass defines how much inertia a body has, that is, the difficulty of putting some acceleration in it. And it's basically the same with friction. Static friction says that, for example, for you to pull a table with your hands, you need to put force (obviously), as measured by Newtons. From the Newton that the table start to accelerate, then there is the static friction.

For example:

To speed up the table, you put five Newtons, and the table does not move. Put ten, still not moving. It puts 11, and it begins to move (accelerate). So the static friction force is 10 Newtons.

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Fill in the correct answers. A big wheel with twice the circumference of a small wheel will rotate with ____ the force and ___ t

Lina20 [59]

Answer:

Explanation:

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3 years ago

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What is the freezing point of radiator fluid that is 50% antifreeze by mass? k f for water is 1.86 ∘ c/m?

densk [106]

Ethylene glycol is termed as the primary ingredients in antifreeze.
The ethylene glycol molecular formula is C₂H₆O₂.
Molar mass of C₂H₆O₂ is = (2×12) +(6×1) + (216) = 62g/mol
Now that antifreeze by mass is 50%, then there is 1kg of ethylene glycol which is present in 1kg of water.
ΔTf = Kf×m
ΔTf = depression in the freezing point.
= freezing point of water freezing point of the solution
= O°c - Tf
= -Tf
Kf = depression in freezing constant of water = 1.86°C/m
M is the molarity of the solution.
=(mass/molar mass) mass of solvent in kg
=1000g/62 (g/mol) /1kg
=16.13m
If we plug the value we get
-Tf = 1.86 × 16.13 = 30
Tf = -30°c

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3 years ago

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An Olympic track runner starts from rest and has an acceleration of 2.4 m/s2 for 3.6 s, then has zero acceleration for the remai

rjkz [21]

Answer:

The runner's speed at the following times would remain 8.64 m/s.

Explanation:

Acceleration definition: Acceleration is rate of change in velocity of an object with respect to time.

In this case, after 3.6 seconds the acceleration is zero, it means that the velocity of the runner after 3.6 seconds is not changing and it will remain constant for the remainder of the race. Now, we have to find the velocity of the runner that he had after 3.6 seconds and that would be the runner's speed for the remainder of the race. For this we use first equation of motion.

First equation of motion: Vf = Vi + a×t

Vf stands for final velocity

Vi stands for initial velocity

a stands for acceleration

t stands for time

In the question, it is mentioned that the runner starts from rest so its initial velocity (Vi) will be 0 m/s.

The acceleration (a) is given as 2.4 m/s²

The time (t) is given as 3.6 s

Now put the values of Vi, a and t in first equation of motion

Vf = Vi + a×t

Vf = 0 + 2.4×3.6

Vf = 2.4×3.6

Vf = 8.64 m/s

So,the runner's speed at the following times would remain 8.64 m/s.

5 0

2 years ago

An unusual fog that is denser than air begins to fill a room. Should the people in the room walk out or crwl out? The people sho

horsena [70]

Walk out. If it's denser than air, it'll settle to the bottom

8 0

2 years ago

a desktop computer and monitor together draw about 2 A of current they plug into a wall outlet that is 120 V what is the Resista

Delvig [45]

Answer:

$60 \Omega$

Explanation:

the relation between current, voltage and resistance in an electrical circuit is given by Ohm's law:

$V=IR$

where V is the voltage, I is the current and R is the resistance. In this problem, the current is I=2 A, the voltage is V=120 V, therefore we can arrange the previous equation and find the resistance:

$R=\frac{V}{I}=\frac{120 V}{2 A}=60 \Omega$

7 0

2 years ago