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3 years ago

What became stuck on the turtle's legs before it crossed the highway?

Physics

2 answers:

Kryger [21]3 years ago

7 0

Wild Oat Seeds is the answer (according to answers website)

spin [16.1K]3 years ago

5 0

The correct answer for your question "Wild Oat Seeds"

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please help. Sound travels at 330 m/s. If a lightning bolt strikes the ground 5 m away from you, how long will it take for the s

Olenka [21]

Answer:

0.015 seconds

Explanation:

5/330=0.015

3 0

2 years ago

ASAP

pochemuha

Answer:

Explanation:

The fish falls from vertical rest in a time of

t = √(2h/g) = √(2(2.27)/9.81) = 0.68 s

v = d/t = 5.6 / 0.68 = 8.2 m/s

4 0

3 years ago

A 2-kg bowling ball sits on top of a building that is 40 meters tall.

Dahasolnce [82]

The bowling ball is at rest, so it only has gravitational potential energy.

Ug = mgy
Ug = (2)(9.8)(40) = 784 J

Need any more help?

6 0

3 years ago

Read 2 more answers

In a physics lab, Asha is given a 10.7 kg uniform rectangular plate with edge lengths 67.3 cm by 53.5 cm . Her lab instructor re

Leya [2.2K]

Answer:

$I=2.6363\ kg.m^2$

Explanation:

Given:

dimension of uniform plate, $(0.673\times 0.535)\ m^2$

mass of plate, $m=10.7\ kg$

Now we find the moment of inertia about the center of mass of the rectangular plate is given as:

$I_{cm}=\frac{1}{12} \times m(L^2+B^2)$

where:

$L=$ length of the plate

$B=$ breadth of the plate

$I_{cm}=\frac{1}{12} \times 10.7\times(0.673^2+0.535^2)$

$I_{cm}=0.6591\ kg.m^2$

We know that the center of mass of the rectangular plane is at its geometric center which is parallel to the desired axis XX' .

Now we find the distance between the center of mass and the corner:

$s=\frac{\sqrt{ (0.673^2+0.535^2)}}{2}$

$s=0.4299\ m$

Now using parallel axis theorem:

$I=I_{cm}+m.s^2$

$I=0.6591+10.7\times 0.4299^2$

$I=2.6363\ kg.m^2$

6 0

3 years ago

Consider steady-state conditions for one-dimensional conduction in a plane wall having a thermal conductivity k = 50 W/m · K and

tatuchka [14]

Answer:

solution:

dT/dx =T2-T1/L

q_x = -k*(dT/dx)

dT/dx= (-20-50)/0.35==> -280 K/m

q_x =-50*(-280)*10^3==>14 kW

Case (2)

dT/dx= (-10+30)/0.35==> 80 K/m

q_x =-50*(80)*10^3==>-4 kW

Case (2)

dT/dx= (-10+30)/0.35==> 80 K/m

q_x =-50*(80)*10^3==>-4 kW

Case (3)

q_x =-50*(160)*10^3==>-8 kW

T2=T1+dT/dx*L=70+160*0.25==> 110° C

Case (4)

q_x =-50*(-80)*10^3==>4 kW

T1=T2-dT/dx*L=40+80*0.25==> 60° C

Case (5)

q_x =-50*(200)*10^3==>-10 kW

T1=T2-dT/dx*L=30-200*0.25==> -20° C

note:

all graph are attached

6 0

3 years ago