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Neporo4naja [7]
3 years ago
8

Which describes a force acting on an object? a car moving in a straight line a ball that moves when kicked a billiard ball that

rolls at a constant speed
Physics
2 answers:
Amiraneli [1.4K]3 years ago
7 0
<span>Force is needed to change an objects state of motion and it is always a part of a mutual action that involves another force. The force can be push or pull.</span>
A ball that moves when kicked <span>describes a force acting on an object.
</span><span>The force is a vector quantity, because it that needs both magnitude and direction in order to be defined.</span>
AleksandrR [38]3 years ago
3 0

Answer:

  • a ball that moves when kicked
  • a ball that rolls at a constant speed

Explanation:

<h3>When a car moves in a straight line</h3>

For this statement we don’t know HOW the car is moving, it could be at constant speed, it could be changing its speed, we don’t know. So, this statement is not a very good description of how a force acts.

<h3>a ball that moves when kicked </h3>

The ball starts at rest, and, after being kicked, is moving. When a force acts in an object, the object is<u> accelerated</u>. This means, the force changes the velocity at which the object moves. That means that the force made by the kick changed its speed. This is the perfect description of how a force works.

<h3>a billiard ball that rolls at a constant speed</h3>

This one is a little tricky. One could say that this is just like the car, moving at constant speed, but that is not true, cause the car its modeled as an particle. The mechanics of rolling are more complicated. You see, if a ball spins, and there is no friction, the ball spins in its place. For the ball to start rolling, we need to consider the static friction that enables to rolls. So, there is a force acting on the object, the static friction force.

Of course, one could argue that there are friction forces acting on the car, and there is pressure, and so on, and so on, but in physics we model the systems as simple as necessary, and we use to imagine the models, as we say "In the vacuum" even if this is not remotely true.

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An initially uncharged 3.47-μF capacitor and a 6.43-kΩ resistor are connected in series to a 1.50-V battery that has negligible
harkovskaia [24]

Answer: a) io=233.28 A ( initial current); b) τ=R*C= 22.31 ms; c) 81.7 ms

Explanation:  In order to explain this problem we have to use, the formule for the variation of the current in a RC circuit:

I(t)=io*Exp(-t/τ)

and also we consider that io=V/R=(1.5/6.43*10^3)

=233.28 A

then the time constant for the RC circuit is τ=R*C=6.43*10^3*3.47*10^-6

=22.31 ms

Finally the time to reduce the current to 2.57% of its initial value is obtained from:

I(t)=io*Exp(-t/τ)  for I(t)/io=0.0257=Exp(-t/τ) then

ln(0.0257)*τ =-t

t=-ln(0.0257)*τ=81.68 ms

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3 years ago
A very long train is rolling at 4 m/s along a straight track. An observer is standing on the ground very dangerously close to th
hichkok12 [17]

Answer:

A. \vec{r}=(6\frac{m}{s})t\ \ \hat{i}

B.  t = 50 s

Explanation:

A. The vectorial equation of the person who is getting closer to the other person is:

\vec{r}=\vec{v}t

r: position vector

v: speed vector = 6m/s i  (if you consider the motion as a horizontal motion)

Then, you replace and obtain:

\vec{r}=(6\frac{m}{s})t\ \ \hat{i}

B. The time is:

t=\frac{d}{v}

d: distance to the observer = 300m

v: speed of the person on the car = 6.00 m/s

t=\frac{300m}{6m/s}=50s

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A hockey puck is sliding at a constant rate of 2m/s on a frictionless surface. How fast will the puck be moving after 10 sec
AnnyKZ [126]
2m/s because the hockey puck is traveling at a constant speed ( acceleration is 0 ). Unless something acts on the hockey puck it will travel 2 m/s forever.
5 0
3 years ago
The driver of a 1000 kg car traveling at a speed of 16.7 m/s applies the brakes. If the brakes provide a force of - 8000 N to st
RSB [31]

Answer:

Given:

m=1000kg

u= 16.7m/s

v=0m/s

F=8000N

Required:

s=?

Solution:

F=m × a

8000N=1000kg × a

a=8m/s^2

Since it decelerate a= -8m/s^2

v^2 = u^2 + 2as

s=v^2 - u^2 / 2a

s= 0 - (16.7m/s)^2 / 2 × -8m/s^2

s= -278.89/-16

s= 17.43m

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6 0
2 years ago
A 250 g block of ice is removed from the refrigerator at -8.0°C. How much thermal energy does the ice absorb as it warms to room
zlopas [31]

Answer:

Q = 114895 J

Explanation:

To find the thermal energy gained by the ice you use the following formula:

Q=mc(T_2-T_1)+H_f\ m

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T2: final temperature = 22°C

T1: initial temperature = -8.0°C

Hf: heat of fusion of water = 3.34*10^5 J/kg

c: specific heat of water = 4186 J/kg

By replacing the values of the parameters you have:

Q=(0.250kg)(4186J/kg\°C)(22+8)\°C+(3.34*10^5 J/kg)(0.250kg)\\\\Q=114895\ J

where you have considered that ice melts completely

3 0
4 years ago
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