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Darya [45]
3 years ago
9

A 5.00 gram bullet moving at 600 m/s penetrates a tree trunk to a depth of 4.00 cm.

Physics
1 answer:
Crazy boy [7]3 years ago
6 0
A) Work energy relation;
 Work =ΔKE ; work done = Force × distance, while, Kinetic energy = 1/2 MV²
 F.s = 1/2mv²
F× 4×10^-2 = 1/2 × 5 ×10^-3 × (600)²
 F = 900/0.04
    = 22500 N
Therefore, force is 22500 N

b) From newton's second law of motion;
 F = Ma
Thus; a = F/m
             = 22500/(5×10^-3)
             = 4,500,000 m/s²
But v = u-at
  0 = 600- 4500,000 t
t = 1.33 × 10^-4 seconds
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Answer:

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Explanation:

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The force exerted by the box on the floor is given as follows;

Pressure, P = \dfrac{Force, F}{Area, A}

∴ F = P × A

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∴ F = 16 N/m² × 2.4 m² = 38.4 N

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7 0
3 years ago
Suppose that the average speed (vrms) of carbon dioxide molecules (molar mass 44.0 g/mol) in a flame is found to be 2.67 105 m/s
34kurt

Answer:

T = 1.26 \times 10^8 K

Explanation:

As we know that rms speed of ideal gas is given by the formula

v_{rms} = \sqrt{\frac{3RT}{M}}

here we know that

v_{rms} = 2.67 \times 10^5 m/s

molecular mass of gas is given as

M = 44 g/mol = 0.044 kg/mol

now from above formula we have

2.67\times 10^5 = \sqrt{\frac{3(8.31)T}{0.044}}

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You hold glider AA of mass 0.125 kgkg and glider BB of mass 0.375 kgkg at rest on an air track with a compressed spring of negli
andrew11 [14]

Answer:

The magnitude of the velocity of glider B is 0.2m/s and the direction is the negative direction  

Explanation:

Inelastic Collision

Given data

mass of glider A m1= 0.125kg

initial velocity u1=0

final velocity v1= 0.600 m/s

mass of glider B m2= 0.375kg

initial velocity u2=0

final velocity v2=?

We know that the expression for the conservation of momentum is given as

m1u1+m2u2=m1v1+m2v2

since u1=u2=u=0m/s

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0=0.075+0.375v2

0.375v2=-0.075

v2=-0.075/0.375

v2=-0.2m/s

 The magnitude of the velocity of glider B is 0.2m/s and the direction is the negative direction      

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3 years ago
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