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2 years ago

(8th grade HELP)

Physics

1 answer:

Shalnov [3]2 years ago

4 0

What is the question your asking

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With your bare feet, stand with one foot on a rug, and the other foot on a tile, linoleum, or wood floor. Which one feels colder

Orlov [11]

It is tile cause tile is made out of stone which is a good conductor

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2 years ago

An aquarium 8 m long, 4 m wide, and 4 m deep is full of water. The hydrostatic pressure on the bottom of the aquarium is ____ Ne

MA_775_DIABLO [31]

Answer:

Explanation:

Given

length=8 m

width=4 m

deep=4 m

Area of base $A=4\times 8=32 m^2$

hydro static Pressure on bottom $=\rho gh$

$P=10^3\times 9.8\times 4=39.2 kPa$

Force on bottom $=P\times A=1254.4 KN$

Now pressure at the side wall

$F=\rho gA \bar{x}$

where $\bar{x}$ is the mean length of wall = 4 m

$F=10^3\times 9.8\times 4\times 8\times 2=627.2 KN$

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3 years ago

The angular acceleration of a wheel is given in rad/s2 by 32 t 3 - 11t 4, where t is in seconds. if the wheel starts from rest a

disa [49]

Answer:

$t=3.64s$

Step by step explanation

Step 1

In this step we use the definition of acceleration to determine the expression to integrate. Note that the acceleration is the derivative of velocity with respect to time.

$a=\frac{dv}{dt} \\\implies adt=dv\\\implies dv=(32t^3-11t^4)dt$

Step 2

Perform integration on the expression from step 1. This calculation is performed as shown below.

$v=\int(32t^3-11t^4)dt\\v=\frac{32}{4}t^4- \frac{11}{5}t^5 +c\\\8t^4- \frac{11}{5}t^5 +c$

Step 3

In this step we use the condition that at $t=0$ , the velocity $v=0$ to find the exact form of the velocity function. We substitute this point into the velocity function we found in step 2.

$0=8(0)^4-\frac{11}{5} (0)^5+c\\\implies c=0$

Step 4

We now use the function in step 3 to find out when the velocity is zero again.

$v=0=8t^5-\frac{11}{5}t^4\\\implies 0=8t^4-\frac{11}{5} t^5\\\implies t^4(8-\frac{11}{5} t)\\\implies t=0, 8-\frac{11}{5}t=0\\\\\implies t=0,t=\frac{40}{11} =3.64$

The velocity is zero again when $t=3.64$

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2 years ago

How are rocks formations formed in Hawaii?

seropon [69]

Answer:mostly likely from volcanos

Explanation:

when the volcano erupts the magma becomes lava and cools making igneous rocks

7 0

2 years ago

An inquisitive physics student and mountain climber climbs a 47.0-m-high cliff that overhangs a calm pool of water. He throws tw

e-lub [12.9K]

Answer:

a) Only the first root is physically reasonable. Therefore, both stones hit the water in 2.866 seconds, b) The initial velocity of the second stone is -16.038 meters per second, c) The speed of the first stone is 30.227 meters per second and the speed of the second stone is 34.338 meters per second.

Explanation:

a) The time after the release after the release of the first stone can be get from the following kinematic formula for the first rock:

$y_{1} = y_{1,o} + v_{1,o} \cdot t +\frac{1}{2}\cdot g \cdot t^{2}$

Where:

$y_{1}$ - Final height of the first stone, measured in meters.

$y_{1,o}$ - Initial height of the first stone, measured in meters.

$v_{1,o}$ - Initial speed of the first stone, measured in meters per second.

$t$ - Time, measured in seconds.

$g$ - Gravity constant, measured in meters per square second.

Given that $y_{1,o} = 47\,m$ , $y_{1} = 0\,m$ , $v_{1,o} = -2.12\,\frac{m}{s}$ and $g = -9.807\,\frac{m}{s^{2}}$ , the following second-order polynomial is built:

$-4.984\cdot t^{2} - 2.12\cdot t + 47 = 0$

Roots of the polynomial are, respectively:

$t_{1} \approx 2.866\,s$ and $t_{2}\approx -3.291\,s$

Only the first root is physically reasonable. Therefore, both stones hit the water in 2.866 seconds.

b) As the second stone is thrown a second later than first one, its height is represented by the following kinematic expression:

$y_{2} = y_{2,o} + v_{2,o}\cdot (t-t_{o}) + \frac{1}{2}\cdot g \cdot (t-t_{o})^{2}$

$y_{2}$ - Final height of the second stone, measured in meters.

$y_{2,o}$ - Initial height of the second stone, measured in meters.

$v_{2,o}$ - Initial speed of the second stone, measured in meters per second.

$t$ - Time, measured in seconds.

$t_{o}$ - Initial absolute time, measured in seconds.

$g$ - Gravity constant, measured in meters per square second.

Given that $y_{2,o} = 47\,m$ , $y_{2} = 0\,m$ , $t_{o} = 1\,s$ , $t = 2.866\,s$ and $g = -9.807\,\frac{m}{s^{2}}$ , the following expression is constructed and the initial speed of the second stone is: