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Charra [1.4K]
3 years ago
12

A 10.1 g bullet leaves the muzzle of a rifle with a speed of 425 m/s. What constant force is exerted on the bullet while it is t

raveling down the 0.6 m length of the barrel of the rifle?
Physics
1 answer:
PIT_PIT [208]3 years ago
5 0

Answer

Force will be 1520.2604 N

Explanation:

We have given mass of the bullet m = 10.1 gram = 0.0101 kg

Distance S= 0.6 m

Final velocity v = 425 m/sec

Initial velocity u will be 0 m/sec

From third equation of motion v^2=u^2+2as

425^2=0^2+2\times a\times 0.6

a=150520.833m/sec^2

We know that force is given by F = ma

So force = 0.0101×150520.833 = 1520.2604 N

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Find the potential inside and outside a uniformly charged solid sphere whose radius is R and whose total charge is q. Use infini
amid [387]

Answer:

Recall that the electric field outside  a uniformly charged solid sphere  is exactly the same as if the charge were all at a point in the centre of the  sphere:

E_{outside} =\frac{1}{4\pi(e_{0})}\frac{Q}{r^{2} } r^{'}

lnside the sphere, the electric field also acts like a point charge, but only for the proportion of the charge further inside than the point r:

E_{inside} =\frac{1}{4\pi(e_{0})}\frac{Q}{R^{2} } \frac{r}{R} r^{'}

To find the potential, we integrate the electric field on a path from infinity (where of course, we take the direct path so that we can write the it as a 1 D integral):

V(r>R)=\int\limits^r_\infty {\frac{1}{4\pi(e_{0)} }\frac{Q}{r^2}  } \, dr=\frac{q}{4\pi(e_{0)} } \frac{1}{r} \\V(r

=\frac{q}{4\pi e_{0} } [\frac{1}{R} -\frac{r^{2}-R^{2}  }{2R^{3} } ]

∴NOTE: Graph is attached

8 0
3 years ago
A piece of styrofoam has a charge of 0.002 mC and is placed 0.5 m from a grain of salt with a charge of 0.03 nC. How much electr
aleksklad [387]

Answer:

2.16×10⁻⁶ N

Explanation:

Applying,

F = kqq'/r² (coulomb's Law)....................... Equation 1

Where F = electrostatic force, k = coulomb's constant, q = charge on the styrofoam, q' = charge on the grain of salt, r = distance between the charges.

From the question,

Given: q = 0.002 mC = 2.0×10⁻⁶ C, q' = 0.03 nC = 3.0×10⁻¹¹ C, r = 0.5 m

Constant: k = 8.99×10⁹ Nm²/C²

Substitute these values into equation 1

F = (2.0×10⁻⁶)(3.0×10⁻¹¹)(8.99×10⁹)/0.5²

F = 2.16×10⁻⁶ N

5 0
2 years ago
A coin mass 0.005 kg is dropped from a height of 3 m. How much mechanical energy does it have right before it hits the ground ?
lesya692 [45]
Mechanical energy equals the sum of potential and kinetic energy. During the process, all PE converts into KE, assuming air resistance is neglected. So, the mechanical energy does not change and is equal to the initial potential energy.
ME
=mgh
=0.005 x 9.81 x 3
=0.147J
6 0
3 years ago
The s.I unit of R so that the equation velocity= R × density S dimensionally correct R is a constant
Leviafan [203]

Answer:

R = m⁴/kg . s

Explanation:

In this case, the best way to solve this is working with the units in the expression.

The units of velocity (V) are m/s

The units of density (d) are kg/m³

And R is a constant

If the expression is:

V = R * d

Replacing the units and solving for R we have

m/s = kg/m³ * R

m * m³ / s = kg * R

R = m * m³ / kg . s

<h2>R = m⁴ / kg . s</h2>

This should be the units of R

Hope this helps

5 0
3 years ago
If potential energy at point A is 12 joules, what is the potential energy at D?
Irina-Kira [14]

Answer:

In D: 3J

Explanation:

Potential energy: Ep=mgh where m is the mass, h altitude.

In point A: h=20cm=0.2m

Epa=12=0.2×mg. Thus mg=12/0.2=60N

For point D: hd=5cm=0.05m

Epd=mg×0.05=60×0.05=3J

5 0
3 years ago
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