6) A stone is thrown vertically upward and returns to its starting position in o 5 s,What was its initial velocity?

A two-phase, liquidâ€“vapor mixture of h2o, initially at x 5 30% and a pressure of 100 kpa, is contained in a pistonâ€“ cylinder

Andrew [12]

A boiling pot of water (the water travels in a current throughout the pot), a hot air balloon (hot air rises, making the balloon rise) , and cup of a steaming, hot liquid (hot air rises, creating steam) are all situations where convection occurs.
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7 0

2 years ago

At the moment a hot cake is put in a cooler, the difference between the cakes and the cooler's

hram777 [196]

Answer: D(t)= 50(4/5)^t

Explanation: If 1/5 of the temperature difference is lost each minute, that means 4/5 of the difference remains each minute. So each minute, the temperature difference is multiplied by a factor of 4/5 (or 0.8).

If we start with the initial temperature difference, 50° Celsius, and keep multiplying by 4/5, this function gives us the temperature difference t minutes after the cake was put in the cooler.

5 0

2 years ago

Read 2 more answers

A rod of length Lo moves iwth a speed v along the horizontal direction. The rod makes an angle of (θ)0 with respect to the x' ax

Colt1911 [192]

Answer:

From the question we are told that

The length of the rod is $L_o$

The speed is v

The angle made by the rod is $\theta$

Generally the x-component of the rod's length is

$L_x = L_o cos (\theta )$

Generally the length of the rod along the x-axis as seen by the observer, is mathematically defined by the theory of relativity as

$L_xo = L_x \sqrt{1 - \frac{v^2}{c^2} }$

=> $L_xo = [L_o cos (\theta )] \sqrt{1 - \frac{v^2}{c^2} }$

Generally the y-component of the rods length is mathematically represented as

$L_y = L_o sin (\theta)$

Generally the length of the rod along the y-axis as seen by the observer, is also equivalent to the actual length of the rod along the y-axis i.e $L_y$

Generally the resultant length of the rod as seen by the observer is mathematically represented as

$L_r = \sqrt{ L_{xo} ^2 + L_y^2}$

=> $L_r = \sqrt{[ (L_o cos(\theta) [\sqrt{1 - \frac{v^2}{c^2} }\ \ ]^2+ L_o sin(\theta )^2)}$

=> $L_r= \sqrt{ (L_o cos(\theta)^2 * [ \sqrt{1 - \frac{v^2}{c^2} } ]^2 + (L_o sin(\theta))^2}$

=> $L_r = \sqrt{(L_o cos(\theta) ^2 [1 - \frac{v^2}{c^2} ] +(L_o sin(\theta))^2}$

=> $L_r = \sqrt{L_o^2 * cos^2(\theta) [1 - \frac{v^2 }{c^2} ]+ L_o^2 * sin(\theta)^2}$

=> $L_r = \sqrt{ [cos^2\theta +sin^2\theta ]- \frac{v^2 }{c^2}cos^2 \theta }$

=> $L_o \sqrt{1 - \frac{v^2}{c^2 } cos^2(\theta ) }$

Hence the length of the rod as measured by a stationary observer is

$L_r = L_o \sqrt{1 - \frac{v^2}{c^2 } cos^2(\theta ) }$

Generally the angle made is mathematically represented

$tan(\theta) = \frac{L_y}{L_x}$

=> $tan {\theta } = \frac{L_o sin(\theta )}{ (L_o cos(\theta ))\sqrt{ 1 -\frac{v^2}{c^2} } }$

=> $tan(\theta ) = \frac{tan\theta}{\sqrt{1 - \frac{v^2}{c^2} } }$

Explanation:

7 0

2 years ago

A projectile enters a resisting medium at x = 0 with an initial velocity v0 = 910 ft/s and travels 5 in. before coming to rest.

evablogger [386]

Answer:

a = - 1.987 × 10⁶ ft/s²

t = 6.84 × 10⁻⁴ s

Explanation:

v₀ = 910 ft/s

x = 5 in.

relation v = v₀ - k x

v = 0 as body comes to rest

0 = 900 - 5k/12

k = 2184 s⁻¹

acceleration

$\frac{\mathrm{d} v}{\mathrm{d} t} = -k\frac{\mathrm{d} x}{\mathrm{d} t}$

where

(A) a = -k × v

at v= 910 ft/s

a = - 1.987 × 10⁶ ft/s²

(B) at x = 3.9 in.

v = 910 - 3.9(2184)/12

v = 200.2 m/s

$\frac{\mathrm{d} v}{\mathrm{d} t} = -k\frac{\mathrm{d} x}{\mathrm{d} t}$

$\frac{dv}{v} = -kdt$

$\int\limits^{200.2}_{900} {\frac{1}{v} }dv = -k\int\limits^t_0 dt$

$ln(200.2)-ln(900) = -kt$

t = 6.84 × 10⁻⁴ s

3 0

2 years ago

A fishing boat accidentally spills 3.0 barrels of diesel oil into the ocean. each barrel contains 42 gallons. if the oil film on

9966 [12]

Number of barrels are 3.0. Each barrel contains 42 gallons of oil. Thus, total volume of oil will be 42×3=126 gallons.

Converting gallons into m^{3}

1 gallon=0.00378 m^{3}

Thus, 126 gallons=0.4769 m^{3}

Thickness of oil film is 2.5\times 10^{2} nm, converting it into meters as follows:

1 nm=10^{-9} m

Thus,

2.5\times 10^{2} nm=1.5\times 10^{-7}m

Now, volume V of oil is related to area A and thickness T as follows:

V=A×T

rearranging,

A=\frac{V}{T}=\frac{0.4769 m^{3}}{2\times 10^{-7}m}=2.38\times 10^{6}m^{2}

Thus, square meters of oil will be 2.38\times 10^{6}m^{2}

7 0

2 years ago