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2 years ago

Topic : Collisions and Conservation of Momentum

1 answer:

7 0

A 0.2 kg ball moves to the right with a speed of 3 m/s. It hits a 0.5 kg ball that at rest. After the collision, the second ball moves to the right with ...

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2 years ago

A rectangular block floats in pure water with 0.400 in. above the surface and 1.60 in. below the surface. When placed in an aque

labwork [276]

Answer:

specific gravity = 0.8

specific gravity of solution = 2

Explanation:

given data

rectangular block above water = 0.400 in

rectangular block below water = 1.60 in

material floats below water = 0.800 in

solution

first we get here specific gravity of block that is

specific gravity = block vol below ÷ total block vol × specific gravity water ..............1

put here value we get

specific gravity = $\frac{1.60}{1.60+0.400}$ × 1

specific gravity = 0.8

and now we get here specific gravity of solution that is express as

specific gravity of solution = total block vol ÷ block vol below × specific gravity block ........................2

put here value we get

specific gravity of solution = $\frac{1.60+0.400}{0.800}$ × 0.8

specific gravity of solution = 2

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3 years ago

A 1200 kg car traveling east at 4.5 m/s crashes into the side of a 2100 kg truck that is not moving. During the collision, the v

Rasek [7]

Answer:

Explanation:

This is a simple Law of Momentum Conservation problem of the inelastic type. The equation for this is

$[m_1v_1+m_2v_2]_b=[(m_1+m_2)v]_a$ Filling in:

$[1200(4.5)+2100(0)]=[(1200+2100)v]$ which simplifies to

5400 + 0 = 3300v

so v = 1.6 m/s to the east, choice B

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2 years ago

The newton is defined as the:

Orlov [11]

Answer:

Explanation:

because of fhe speed an object when under the influence of earth gravitational field

tama po yan .

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2 years ago

Read 2 more answers

A spring stretches 0.150 m when a 0.30 kg mass is hung from it. The spring is then stretched an additional 0.100 m from this equ

DochEvi [55]

Answer:

a) $k=19.6N/m$

b) $V_m=0.81m/s$

c) $a_m=6.561m/s^2$

d) $K.E=0.096J$

e) $T=0.78sec$ & $F=1.29sec$

f) $mx'' + kx' =0$

Explanation:

From the question we are told that:

Stretch Length $L=0.150m$

Mass $m=0.30kg$

Total stretch length $L_t=0.150+0.100=>0.25$

Generally the equation for Force F on the spring is mathematically given by

$F=-km\\\\k=F/m\\\\k=\frac{m*g}{x}\\\\k=\frac{0.30*9.8}{0.15}$

$k=19.6N/m$

b)Generally the equation for Max Velocity of Mass on the spring is mathematically given by

$V_m=A\omega$

Where

A=Amplitude

$A=0.100m$

And

$\omega=angulat Velocity\\\\\omega=\sqrt{\frac{k}{m}}\\\\\omega=\sqrt{\frac{19.6}{0.3}}\\\\\omega=8.1rad/s$

Therefore

$V_m=A\omega\\\\V_m=8.1*0.1$

$V_m=0.81m/s$

Generally the equation for Max Acceleration of Mass on the spring is mathematically given by

$a_m=\omega^2A$

$a_m=8.1^2*0.1$

$a_m=6.561m/s^2$

Generally the equation for Total mechanical energy of Mass on the spring is mathematically given by

$K.E=\frac{1}{2}mv^2$

$K.E=\frac{1}{2}*0.3*0.8^2$

$K.E=0.096J$

Generally the equation for the period T is mathematically given by

$\omega=\frac{2\pi}{T}$

$T=\frac{2*3.142}{8.1}$

$T=0.78sec$

Generally the equation for the Frequency is mathematically given by

$F=\frac{1}{T}$

$F=1.29sec$

Generally the Equation of time-dependent vertical position of the mass is mathematically given by

$mx'' + kx' =0$

Where

'= signify order of differentiation

7 0

2 years ago