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Deffense [45]
3 years ago
9

Suppose you hit a steel nail with a 0.500-kg hammer, initially moving at 15.0 m/s and brought to rest in 2.80 mm. How much is th

e nail compressed if it is 2.50 mm in diameter and 6.00-cm long
Physics
1 answer:
katrin [286]3 years ago
6 0

Complete Question

Suppose you hit a steel nail with a 0.500-kg hammer, initially moving at 15.0 m/s and brought to rest in 2.80 mm. How much is the nail compressed if it is 2.50 mm in diameter and 6.00-cm long.What Average force is excreted on the Nail

Answer:

F=2*10^{4}N

Explanation:

From the question we are told that:

Mass m=0.500kg

Initial Velocity V=15.0m/s

Distance x=2.80mm=>0.00280m

Diameter d=2.50mm=>0.00250m

Length l=6.00cm=>0.6m

Generally the equation for Force is mathematically given by

 F=\frac{mv^2}{2d}

 F=\frac{0.500*15^2}{2.80*10^{-3}}

 F=2*10^{4}N

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In Challenge Example 11.9 (p. 280), after the explosion, suppose that the m1 fragment shot directly north at 12 m/s and the m3 f
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3 years ago
Normal atmospheric pressure is 1.013 105 Pa. The approach of a storm causes the height of a mercury barometer to drop by 27.1 mm
Burka [1]

Answer:

The atmospheric pressure is 0.97622\times10^{5}\ Pa.

Explanation:

Given that,

Atmospheric pressure P_{atm}= 1.013\times10^{5}\ Pa    

drop height h'= 27.1 mm

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Using formula of pressure

p = \rho g h

Put the value into the formula

1.013\times10^{5}=13.59\times10^{3}\times9.8\times h

h =\dfrac{1.013\times10^{5}}{13.59\times10^{3}\times9.8}

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We need to calculate the atmospheric pressure

Using formula of atmospheric pressure

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Put the value into the formula

P= 13.59\times10^{3}\times9.8\times0.733

P=0.97622\times10^{5}\ Pa

Hence, The atmospheric pressure is 0.97622\times10^{5}\ Pa.

7 0
3 years ago
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