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3 years ago

3.

Physics

1 answer:

Lana71 [14]3 years ago

6 0

Answer:

A. Condensation B. Freezing. C. condensation (not sure) D. Sublimation E. Evaporation. F Evaporation (not sure)

Hope it heps:)

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calculate speed and velocity of the following. the race car was moving for 3.7 hours and during that time it traveled a distance

Ksju [112]

Answer:

v = 135.13 mph

Explanation:

Given that,

The race car was moving for 3.7 hours and during that time it traveled a distance of 500 miles south.

We need to find the speed of the car.

We know that,

Speed = distance/time

So,

$v=\dfrac{500\ miles}{3.7\ h}\\\\v=135.13\ mph$

So, the speed of the car is equal to 135.13 mph.

7 0

3 years ago

A hard-boiled egg of mass 46.0 gg moves on the end of a spring with force constant 25.6 N/mN/m . The egg is released from rest a

soldi70 [24.7K]

Answer:

0.022kg/s

Explanation:

We are given that

Mass of boiled egg=46 g= $\frac{46}{1000} kg=0.046 kg$

$1kg=1000 g$

Constant force=F=25.6 N/m

Initial displacement= $A_1=0.296 m$

Final displacement= $A_2=0.12 m$

Time=t=4.55 s

Damping force= $F_x=-bv_x$

We have to find the magnitude of damping constant b.

We know that the displacement of the oscillator under damping motion is given by

$x=Ae^{-\frac{b}{2m}t}cos(w't+\phi)$

For maximum displacement $cos(w't+\phi)=1$

Therefore , $x=A_2$

Substitute the values

$A_2=A_1e^{-\frac{-b}{2m}t}$

$e^{-\frac{b}{2m}t}=\frac{A_2}{A_1}$

$-\frac{b}{2m}t=ln\frac{A_2}{A_1}$

$lnx=y\implies x=e^y$

Substitute the values

$-\frac{b}{2\times 0.046}\times 4.55=ln\frac{0.12}{0.296}$

$\frac{2\times 0.046}{4.55b}=ln\frac{0.296}{0.12}$

$\frac{2\times 0.046}{4.55}=0.9b$

$b=\frac{2\times 0.46}{4.55\times 0.9}=0.022kg/s$

Hence,the magnitude of damping constant b=0.022kg/s

3 0

3 years ago

The specific weight of sea water is 10.1 kN/m^3. Convert to lbs/in^3.

Viktor [21]

Answer:

0.03719 lbs/in³

Explanation:

Specific weight is given by multiplying the density of an object to the acceleration due to gravity.

$\gamma =\rho g\\\Rightarrow \rho=\frac{gamma}{g}\\\Rightarrow \rho=\frac{10.1\times 10^3}{9.81}\\\Rightarrow \rho=1029.562\ kg/m^3$

$1\ kg=2.20462\ lb$

$1\ m=39.3701\ in$

$\\\Rightarrow 1029.562\ kg/m^3=\frac{1029.562\times 2.20462}{39.3701^3}=0.03719\ lbs/in^3$

So,

$10.1\ kN/m^3=0.03719\ lbs/in^3$

8 0

3 years ago

In an elastic collision, a 400-kg bumper car collides directly from behind with a second, identical bumper car that is traveling

kirza4 [7]

Answer:

v₁ = 3.5 m/s

v₂ = 6.4 m/s

Explanation:

We have the following data:

m₁ = mass of trailing car = 400 kg

m₂ = mass of leading car = 400 kg

u₁ = initial speed of trailing car = 6.4 m/s

u₂ = initial speed of leading car = 3.5 m/s

v₁ = final speed of trailing car = ?

v₂ = final speed of leading car = ?

The final speed of the leading car is given by the following formula:

$v_2=\frac{2m_1}{m_1+m_2}u_1-\frac{m_1-m_2}{m_1+m_2}u_2\\\\v_2=\frac{(2)(400\ kg)}{400\ kg+400\ kg}(6.4\ m/s)-\frac{400\ kg-400\ kg}{400\ kg + 400\ kg}(3.5\ m/s)$

The final speed of the leading car is given by the following formula:

$v_1=\frac{m_1-m_2}{m_1+m_2}u_1+\frac{2m_2}{m_1+m_2}u_2\\\\v_1=\frac{400\ kg-400\ kg}{400\ kg + 400\ kg}(6.4\ m/s)+\frac{(2)(400\ kg)}{400\ kg+400\ kg}(3.5\ m/s)$

4 0

3 years ago

Which of the following statements is true?

rosijanka [135]

The correct answers would be B, and d

3 0

3 years ago

Read 2 more answers