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algol [13]
2 years ago
14

PLease help with this this is pretty hard for me i kinda get it but not fully

Physics
1 answer:
vodka [1.7K]2 years ago
5 0

Explanation:

Force = mass x acceleration

F= ma (formula)

15N = o.5kg x a

Divide both sides by o.5

a = 30 (greater acceleration)

15N = 4.5kg x a

Divide both sides by 4.5

a= 3.33 (great acceleration)

15N = 0.05kg x a

Divide both sides by 0.05

a = 300 (greatest acceleration)

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An astronaut brings a cube from the Earth to the Moon.
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Answer:

Answer D. Inertial mass constant, weight decreases.

Explanation:

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An insulating plastic rod is charged by rubbing it with a wool cloth, and then brought to an initially neutral conducting metall
barxatty [35]

Answer:

A) is repelled by the sphere.

Explanation:

When a charged insulated rod is touched with an insulated conducting sphere , some  charge on the rod gets transferred to the sphere . So they become similarly charged . We all know that there is repulsion when two similarly charged object are brought near to each other . Hence here too there will be repulsion between the rod and the sphere when the rod is brought near the sphere.

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4 years ago
A hockey puck has a momentum of 3.8 kg•m/s [E]. If its
agasfer [191]

Answer:

1.6 \times 10 { }^{ - 1} kg \\

Explanation:

I hope, it helped you.

4 0
3 years ago
To practice Problem-Solving Strategy 27.1: Magnetic Forces. A particle with mass 1.81×10−3 kgkg and a charge of 1.22×10−8 CC has
Roman55 [17]

Answer:

The magnitude of the acceleration  is a = 0.33 m/s^2

The direction is - \r k i.e the negative direction of the z-axis

Explanation:

 From  the question we are that

       The mass of the particle m = 1.8*10^{-3} kg

         The charge on the particle is q = 1.22*10^{-8}C

         The velocity is \= v = (3.0*10^4 m/s ) j

        The the magnetic field is  \= B = (1.63T)\r i + (0.980T) \r j

The charge experienced  a force which is mathematically represented as

         

                    F = q (\= v * \= B)

    Substituting value

         F = 1.22*10^{-8} (( 3*10^4 ) \r j \ \ X \ \  ( 1.63 \r i  + 0.980 \r j )T)

            = 1.22 *10^{-8} ((3*10^4 * 1.63)(\r j \ \  X \ \  \r i) + (3*10^4 * 0.980) (\r j \ \ X \ \  \r  j))

            = (-5.966*10^4 N) \r k

Note :

           i \ \ X \ \ j = k \\\\j \ \  X  \ \ k = i\\\\k  \ \ X \ \ i = j\\\\j \ \ X \ \ i = -k \\\\k \ \  X  \ \ j = -i\\\\i  \ \ X \ \ k = - j\\

Now force is also mathematically represented as

        F = ma

Making a the subject

      a = \frac{F}{m}

   Substituting values

     a =\frac{(-5.966*10^4) \r k}{1.81*10^{-3}}

        = (-0.33m/s^2)\r k

        = 0.33m/s^2 * (- \r k)

6 0
4 years ago
Consider a uniform sphere, which has a mass of 4.80 kg and a radius of 22.0 cm. A tangential force of 11.2 N is applied to the o
Tcecarenko [31]

Answer:

The moment of inertia of this sphere is 0.0929\ kg-m^2.                  

Explanation:

It is given that,

Mass of the sphere, m = 4.8 kg

Radius of the sphere, r = 22 cm = 0.22 m

Tangential force, F = 11.2 N

The moment of inertia of the uniform sphere is given by :

I=\dfrac{2}{5}mr^2

I=\dfrac{2}{5}\times 4.8\ kg\times (0.22\ m)^2

I=0.0929\ kg-m^2

So, the moment of inertia of this sphere is 0.0929\ kg-m^2. Hence, this is the required solution.              

8 0
3 years ago
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