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pashok25 [27]
3 years ago
14

Help me pls with physics!! -

Physics
1 answer:
Citrus2011 [14]3 years ago
4 0

Answer:

good luck

Explanation:

46 i think

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A block of mass M on a horizontal surface is connected to the end of a massless spring of spring constant k. The block is pulled
slamgirl [31]

Answer:

Minimum coefficient of kinetic friction between the surface and the block is \mu_k=\frac{kx}{2Mg} .

Explanation:

Given:

Mass of the block = M

Spring constant = k

Distance pulled = x

According to the question:

<em>We have to find the minimum co-efficient of kinetic friction between the surface and the block that will prevent the block from returning to its equilibrium with non-zero speed.  </em>

So,

From the FBD we can say that:

⇒ Normal force, N=Mg                                   <em>...equation(i)</em>

⇒ Elastic potential energy, PE = \frac{kx^2}{2}               <em>  ...equation (ii)</em>

⇒ Frictional force, f = \mu_kN                                <em> ...equation (iii)</em>

⇒ Plugging (i) in (iii).

⇒ f=\mu_kMg

Now,

⇒ As we know that the energy lost due to friction is equivalent to PE .

⇒ PE=fx                     <em>...considering PE as</em> mgh or f(x) .

   Arranging the equation.

⇒ \frac{kx^2}{2}=\mu_k Mg (x)

⇒ \frac{kx}{2}=\mu_k Mg                 <em>...eliminating x from both sides.</em>

⇒ \frac{kx}{2Mg}=\mu_k                    <em>...dividing both sides wit Mg.</em>

Minimum coefficient of kinetic friction between the surface and the block is \frac{kx}{2Mg}=\mu_k .

4 0
4 years ago
If 2 balls had the same volume but ball a has twice as much mass as babil which one will have the greater density
Georgia [21]
The ball with greater mass has more density
3 0
3 years ago
How much Hydrogen (H) atoms are in this 5NH4Cl?
Jobisdone [24]

Answer:

5

Explanation:

8 0
3 years ago
When the Northern Hemisphere is tilted ________the Sun, the sunlight is _________ direct than it is in the Southern Hemisphere.
matrenka [14]

Answer:

its 2

Explanation:

4 0
2 years ago
Read 2 more answers
How fast can a 4000 kg truck travel around a 70 m radius turn without skidding if its tires share a 0.6 friction coefficient wit
aleksley [76]

Answer:

Velocity of truck will be 20.287 m /sec

Explanation:

We have given mass of the truck m = 4000 kg

Radius of the turn r = 70 m

Coefficient of friction \mu =0.6

Centripetal force is given  F=\frac{mv^2}{r}

And frictional force is equal to F_{frictional}=\mu mg

For body to be move these two forces must be equal

So \frac{mv^2}{r}=\mu mg

v=\sqrt{\mu rg}=\sqrt{0.6\times 70\times 9.8}=20.287m/sec

7 0
3 years ago
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