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3 years ago

What makes a grenade explode?

1 answer:

6 0

The striker throws the safety lever away from the grenade body as it rotates to detonate the primer. The primer explodes and ignites the fuse The fuse burns down to the detonator, which explodes the main charge

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The elementary particle called a muon is unstable and decays in about 2.20μs2.20μs , as observed in its rest frame, into an elec

vredina [299]

Answer:

5.865 μs

Explanation:

t₀ = Time taken to decay a muon = 2.20 μs

c = Speed of Light in vacuum = 3×10⁸ m/s

v = Velocity of muon = 0.927 c

t = Lifetime observed

Time dilation

$t=\frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}\\\Rightarrow t=\frac{2.2\times 10^{-6}}{\sqrt{1-\frac{(0.927c)^2}{c^2}}}\\\Rightarrow t=\frac{2.2\times 10^{-6}}{\sqrt{1-0.927^2}}\\\Rightarrow t=\frac{2.2\times 10^{-6}}{\sqrt{0.140671}}\\\Rightarrow t=\frac{2.2\times 10^{-6}}{0.3750}\\\Rightarrow t=5.865\times 10^{-6}\ seconds$

∴Lifetime observed for muons approaching at 0.927 the speed of light is 5.865 μs

3 0

3 years ago

A block of gelatin is 120mm by 120mm by 40mm whrn unstressed. A force of 49N is applied tangentially to the upper surface causin

Inessa05 [86]

Answer:

The shearing stress is 10208.3333 Pa

The shearing strain is 0.25

The shear modulus is 40833.3332 Pa

Explanation:

Given:

Block of gelatin of 120 mm x 120 mm by 40 mm

F = force = 49 N

Displacement = 10 mm

Questions: Find the shear modulus, Sm = ?, shearing stress, Ss = ?, shearing strain, SS = ?

The shearing stress is defined as the force applied to the block over the projected area, first, it is necessary to calculate the area:

A = 40*120 = 4800 mm² = 0.0048 m²

The shearing stress:

$Ss=\frac{F}{A} =\frac{49}{0.0048} =10208.3333Pa$

The shearing strain is defined as the tangent of the displacement that the block over its length:

$SS=tan\theta =\frac{Displacement}{L} =\frac{10}{40} =0.25$

Finally, the shear modulus is the division of the shearing stress over the shearing strain:

$Sm=\frac{10208.3333}{0.25} =40833.3332Pa$

6 0

4 years ago

Solve 3x+1 <img src="https://tex.z-dn.net/?f=%20%5Cleqslant%201" id="TexFormula1" title=" \leqslant 1" alt=" \leqslant 1" ali

Rzqust [24]

3x + 1 ≤ 1

Subtract 1 from each side: 3x ≤ 0

Divide each side by 3 : x ≤ 0

8 0

3 years ago

John(body mass=160pounds) is taking off for a long jump. The average ground reaction force Fg at takeoff is 1400 N pointing forw

slega [8]

Answer:

The free body diagram of John is shown in the attached figure (in the FBD john's mass is supposed to be concentrated at his center of mass and FBD is made of center of mass)

b) As shown in the FBD the ground reaction forces are:

i) In X direction $F_{x}=1400cos(35^{o})=1146.81N$

ii) In Y direction $F_{y}=1400sin(35^{o})=803.0N$

c) The respective accelerations in x and y direction's is calculated by newton's second law as indicated under

$\sum F_{x}=ma_{x}\\\\\therefore a_{x}=\frac{\sum F_{x}}{m}=\frac{1146.8N}{72.57kg}=15.80m/s^{2}\\\\\sum F_{y}=ma_{y}\\\\\therefore a_{y}=\frac{\sum F_{y}}{m}=\frac{803.00-72.57\times 9.81}{72.57}=1.255m/s^{2}$

4 0

3 years ago

A brick is resting on the surface of a flat board. as one end of the board is slowly raised, what change, if any, is there in th

vovangra [49]

The normal force decreases, this is the frictional force. It will be counteracted by the force which accelerates the brick to slide downward opposite to the end where the board is raised. As the angle increases the force acting upon the brick opposite to the normal force will decrease.

5 0

4 years ago