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castortr0y [4]
3 years ago
5

How much does force increase if mass is doubled

Physics
2 answers:
just olya [345]3 years ago
8 0
F=ma
So if m=1 then F=1m
If the mass were to double (m=2) F=2m
So if you are to double the mass, you are essentially doubling the force
katrin2010 [14]3 years ago
3 0
First of all, the question is worded somewhat ambiguously because mass does not necessarily imply that there is a force (i.e. an object could be stationary in a system with no gravity and not have a force, or something of the like). Moreover, simply because an object is twice the mass does not mean the force will change. 

However, we know from Newton's second law that F = ma. Therefore, force will increase or decrease proportionally with mass. This means if an object maintains a contestant acceleration and its mass doubles, the force that it provides will also double. 
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If a force F is exerted on a charged particle when the particle enters a magnetic field B at a velocity v, which of the followin
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Answer:

The correct option is 'c': The Force vector (F) and the magnetic field vector (B) are always perpendicular.

Explanation:

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\overrightarrow{F}=q\overrightarrow{v}\times \overrightarrow{B}

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4 0
3 years ago
A potter spins his wheel at 0.98 rev/s. The wheel has a mass of 4.2 kg and a radius of 0.35 m. He drops a chunk of clay of 2.9 k
Bad White [126]

Answer:

v_{f,w} = 1.791\,\frac{m}{s}, v_{f,c} = 0.972\,\frac{m}{s}

Explanation:

The situation can be modelled by applying the Principle of Angular Momentum Conservation:

I_{w} \cdot \omega_{o} = (I_{w} + I_{c})\cdot \omega_{f}

The final angular speed is:

\omega_{f} = \frac{I_{w}}{I_{w}+I_{c}}\cdot \omega_{o}

\omega_{f} = \left(\frac{\frac{1}{2}\cdot (4.2\,kg)\cdot (0.35\,m)^{2} }{\frac{1}{2}\cdot (4.2\,kg)\cdot (0.35\,m)^{2} + \frac{1}{2}\cdot (2.9\,kg)\cdot (0.19\,m)^{2}}\right)\cdot (0.98\,\frac{rev}{s} )\cdot \left(\frac{2\pi\,rad}{1\,rev}  \right)

\omega_{f} \approx 5.116\,\frac{rad}{s}

The tangential velocities of the wheel and the clay are, respectively:

v_{f, w} = (0.35\,m)\cdot (5.116\,\frac{rad}{s} )

v_{f,w} = 1.791\,\frac{m}{s}

v_{f, c} = (0.19\,m) \cdot (5.116\,\frac{rad}{s} )

v_{f,c} = 0.972\,\frac{m}{s}

5 0
3 years ago
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