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3 years ago

Estimate the volume of a typical house (2050 feet squared in size and 10 feet tall) answer in units of meters squared

Physics

1 answer:

Aloiza [94]3 years ago

8 0

Answer:

$Volume = 6248.48 m^{3}$

Explanation:

Given:

The area of the house $A = 2050\ ft^{2}$

The height of the house $h=10\ ft$

We need to find the volume of a typical house.

Solution:

We find the volume of the house by multiplying the area of the house and height of the house.

$Volume = Area\times height$

$Volume = A\times h$

Area and height of the house are known, so we substitute these value in above equation.

$Volume = 2050\times 10$

$Volume = 20500\ ft^{3}$

Now we convert the unit from feet to meter.

Divide the volume by 3.2808 for $m^{3}$

$Volume = \frac{20500}{3.2808}$

$Volume = 6248.48\ m^{3}$

Therefore, the volume of the house is $6248.48 m^{3}$

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An unstable particle is created in the upper atmosphere from a cosmic ray and travels straight down toward the surface of the ea

german

Answer:

Q1) Time taking by particle to travel the 40 km wrt. earth = $1.34\times10^{-6}$ sec.

Q2) The distance traveled by particle where the particle created to surface of earth wrt. particle's frame = 3.84 km.

Q3) The time taking by particle to travel from where it is created to the surface of the earth = $1.285\times10^{-5}$ sec.

Explanation:

Given :

Speed of particle wrt. earth $v=0.99537c$

Distance between where particle is created and earth surface = 40 km

we know that,

⇒ $v = \frac{x}{t}$

Where $x = 40\times10^{3} m$ , $v = 0.99537c$ , we know speed of light $c = 3 \times10^{8}$

∴ $t = \frac{x}{v}$

$= \frac{40 \times10^{3} }{0.99537\times3\times10^{8} }$

$t = 1.34\times10^{-6} sec$

∴ Thus, time taking by particle to travel the 40 km wrt. earth $t = 1.34\times10^{-6} sec$

According to the lorentz transformation,

⇒ $l = l_{o} \sqrt{1-\frac{v^2}{c^2} }$

Where $l =$ improper length, $l_{o} =$ proper length (distance measured wrt. rest frame) = 40 km

$l = 40 \sqrt{1-\frac{v^2}{c^2 }$

$l = 40 \times 0.096$

$l = 3.84$ km

∴ Thus, the distance traveled by particle where the particle created to surface of earth wrt. particle's frame = 3.84 km.

According to the time dilation,

$\Delta t = \frac{\Delta t_{o} }{\sqrt{1-\frac{v^2}{c^2} } }$

Where $\Delta t =$ improper time (wrt. earth frame time) $=1.34\times10^{-6} sec$ , $\Delta t _{o} =$ proper time (wrt. particle frame).

$1.34\times10^{-6} = \frac{ \Delta t_{o}}{0.096}$

$\Delta t_{o} = 1.285 \times10^{-5} sec$

Thus, the time taking by particle to travel from where it is created to the surface of the earth $= 1.285 \times10^{-5} sec$ .

6 0

3 years ago

????The Earth is made of mostly_______. Mountains Water Land Carbon

soldier1979 [14.2K]

The answer is water!!!

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3 years ago

What vertical distance Δy does a free-falling particle travel from the moment it starts to the moment it reaches a speed of 7.9

mr_godi [17]

Answer:

3.2 m

Explanation:

The equation to use to solve this problem is:

$v_f^2 = v_i^2 + 2 a \Delta y$

where

$v_f$ is the final velocity

$v_i$ is the initial velocity

a is the acceleration

$\Delta y$ is the distance covered

For the particle in free-fall in this problem, we have

$v_i = 0$ (it starts from rest)

$v_f = 7.9 m/s$

$g=9.8 m/s^2$ (acceleration due to gravity)

By re-arranging the equation, we can find the distance travelled:

$\Delta y = \frac{v_f^2 -v_i^2}{2a}=\frac{(7.9 m/s)^2-0^2}{2(9.8 m/s^2)}=3.2 m$

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4 years ago

A 100-kg tackler moving at a speed of 2.6 m/s meets head-on (and holds on to) an 92-kg halfback moving at a speed of 5.0 m/s. Pa

DIA [1.3K]

Given that,

Mass of trackler, m₁ = 100 kg

Speed of trackler, u₁ = 2.6 m/s

Mass of halfback, m₂ = 92 kg

Speed of halfback, u₂ = -5 m/s (direction is opposite)

To find,

Mutual speed immediately after the collision.

Solution,

The momentum of the system remains conserved in this case. Let v is the mutual speed after the collision. Using conservation of momentum as :

$m_1u_1+m_2u_2=(m_1+m_2)V\\\\V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}\\\\V=\dfrac{100\times 2.6+92\times (-5)}{(100+92)}\\\\V=-1.04\ m/s$

So, the mutual speed immediately after the collision is 1.04 m/s but in opposite direction.

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4 years ago

Water flows at 10 m/s through a pipe with radius 0.025 m. The pipe goes up to the second floor of the building, 2.5 m higher, an

Lera25 [3.4K]

Answer: from the information given, the velocity of the water will decrease but the pipe size will remain the same.

This can be proved with bernoulli's equation.

Explanation: careful analysis of the system using bernoulli's equation of flow is shown in the image attached

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3 years ago