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Korolek [52]
3 years ago
10

Determine the acceleration that result when a 12-N net force is applied to a 3-kg object and then to a 6-kg object

Physics
1 answer:
user100 [1]3 years ago
3 0

For 3 kg mass: <em>F = \lpha ma,a = \frac{F}{m} = \frac{12N}{3kg} = 4\lpha m / s^{2}</em>

For 6 kg mass: <em>F = \lpha ma,a = \frac{F}{m} = \frac{12N}{6kg} = 2\lpha m / s^{2}</em>


Hope this helped, have a great day!~

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Compare and contrast electrical and magnetic forces
erik [133]

Answer:

the eletrical forces are created when an object is both moving charges and stationary charges.

the magnetic forces are created when an act on only moving charges.

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. A student claims that if lighting strikes a metal flagpole, the force exerted by the Earth’s magnetic field on the current in
Ratling [72]

Answer:

Explanation:

Given that, current generated from lightning range from

10⁴ A < I < 10^5 A

We know that,

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F = iLB

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B = 10^-5 T

So, let assume the worst case of a 15m flag pole

L = 15m

Then,

F = iLB

F = 10^5 × 10 × 10^-5

F = 15 N

Therefore, 15N is fairly strong so it will come to the material that was use for the material of the flag pole.

Therefore, it is possible that the student is right depending on the material of the flag pole.

7 0
3 years ago
If two runners take the same amount of time to run a mile, they have the same __________.
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None of the choices is correct.

If two runners take the same amount of time to run a mile,
they have the same average speed.  But their velocities
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Speed is   (distance covered) divided by (time to cover the distance).

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3 0
3 years ago
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Which statement is true for a sound wave entering an area of warmer air
Reika [66]
That waves travel faster than the wave lenght!
8 0
3 years ago
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A crate rests on a flatbed truck which is initially traveling at 17.9 m/s on a level road. The driver applies the brakes and the
Mamont248 [21]

Answer:

The minimum coefficient of friction required is 0.35.  

Explanation:

The minimum coefficient of friction required to keep the crate from sliding can be found as follows:

-F_{f} + F = 0      

-F_{f} + ma = 0      

\mu mg = ma

\mu = \frac{a}{g}

Where:

μ: is the coefficient of friction

m: is the mass of the crate

g: is the gravity

a: is the acceleration of the truck

The acceleration of the truck can be found by using the following equation:

v_{f}^{2} = v_{0}^{2} + 2ad

a = \frac{v_{f}^{2} - v_{0}^{2}}{2d}

Where:  

d: is the distance traveled = 46.1 m

v_{f}: is the final speed of the truck = 0 (it stops)      

v_{0}: is the initial speed of the truck = 17.9 m/s

a = \frac{-(17.9 m/s)^{2}}{2*46.1 m} = -3.48 m/s^{2}        

If we take the reference system on the crate, the force will be positive since the crate will feel the movement in the positive direction.  

\mu = \frac{a}{g}  

\mu = \frac{3.48 m/s^{2}}{9.81 m/s^{2}}

\mu = 0.35

Therefore, the minimum coefficient of friction required is 0.35.  

I hope it helps you!

4 0
3 years ago
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