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never [62]
3 years ago
11

What type of relationship exists between acceleration and mass?

Physics
1 answer:
likoan [24]3 years ago
7 0
Here, you can derive that by numerical method, as follows:
F = m.a
m = F/a

So, here we can see when we decrease one, other increase by same effect; we can say they are "Indirectly Proportional" to each other!

Hope this helps!
You might be interested in
A bar slides vertically between two conducting rails without friction in a magnetic field. The rails are connected via a resisto
Gala2k [10]

Answer:

The energy dissipated is  E =  9.8 J

Explanation:

From the question we are told that

    The distance covered at constant velocity d =  2 m  

      The velocity is  v = 1.5 \ m/s

Generally at constant velocity the magnetic  force on the bar is mathematically represented as

         F  =  m * g

substituting values

        F  =   0.5 *  9.8

        F  =   4.9 \ N

The energy dissipated is mathematically evaluate as

         E =  F * d

 substituting the value

         E =  4.9  * 2

        E =  9.8 J

     

3 0
4 years ago
5. A 6.0-kilogram mass is moving with a speed of 2.0 m/s. What is the kinetic energy of the mass?
Ann [662]

Answer:

K.E. = ½ × mv²

= ½ × 6 × (2)²

= ½ × 6 × 4

= 3 × 4

= 12 J

3 0
3 years ago
I’m not sure if 46 is A, could someone pls check
ki77a [65]

i'm not sure. I will find out for you and let you know as soon as possible.

7 0
4 years ago
One Newton is equivalent to<br> A. 1 kg/s2<br> B. 1 kg*m/s<br> C. 1 kg*m/s2<br> D. 1 kg/s
tiny-mole [99]

Answer:

B

Explanation:

7 0
3 years ago
A proton with a speed of 3.000×105 m/s has a circular orbit just outside a uniformly charged sphere of radius 7.00 cm. What is t
Rama09 [41]

To solve this problem we will apply the principles of energy conservation. The kinetic energy in the object must be maintained and transformed into the potential electrostatic energy. Therefore mathematically

KE = PE

\frac{1}{2} mv^2 = \frac{kq_1q_2}{r}

Here,

m = mass (At this case of the proton)

v = Velocity

k = Coulomb's constant

q_{1,2} = Charge of each object

r= Distance between them

Rearranging to find the second charge we have that

q_2 = \frac{\frac{1}{2} mv^2 r}{kq_1}

Replacing,

q_2 = \frac{\frac{1}{2}(1.67*10^{-27})(3*10^5)^2(7*10^{-2})}{(9*10^9)(1.6*10^{-19})}

q_2 = 3.6531nC

Therefore the charge on the sphere is 3.6531nC

4 0
3 years ago
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