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3 years ago

The large intestine ends in the ___.

Physics

2 answers:

pishuonlain [190]3 years ago

8 0

It ends at the Rectum

Anton [14]3 years ago

7 0

Answer:

THat BUtt hoLE

Explanation:

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In a clear cup there are three substances. Their densities are 3, 1, and 2. From top of the cup to the bottom, what would the or

Vesnalui [34]

Answer:

1, 2 and 3

Explanation:

The most dense substance will settle at the bottom of the cup

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2 years ago

Take a look at the weather map. The front seen there causes short periods of storms and heavy rains. What type of front is this?

erastova [34]

I am pretty sure it is a cold front.

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3 years ago

Read 2 more answers

What is the condition for an object experiencing free fall?

S_A_V [24]

Answer:

The body acts under the influence of gravity.

Explanation:

An object experiencing free fall is acting under the influence of gravitational force and the acceleration due gravity is positive for any falling object. The body is able to fall freely due to the effect of gravity on it. This gravity effect causes the body to get attracted to the earth's gravitational surface due to gravitational pull exerted on the body.

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3 years ago

Initially a car accelerates at 2 m/s2 for x seconds. The car then travels at a velocity of -6 m/s for x seconds. If the car disp

Luda [366]

Answer:

The time travel is

t=8 s

Explanation:

$a= 2 \frac{m}{s^{2} } \\v=-6 \frac{m}{s} \\x=16m$

$x_{f}=x_{o}+v_{o}*t+\frac{1}{2} *a*t^{2}$

$x_{f}=0-6*t+\frac{1}{2} *2*t^{2}$

$t^{2}-6*t-16=0\\ using :\\\frac{-b+/-\sqrt{b^{2}-4*c*a } }{2} \\\frac{-(-6)+/-\sqrt{(-6)^{2}-4*(-16)*(1) } }{2}=\frac{3}{2} +/- \frac{10}{2} \\t_{1} = 2s \\t_{2} = 8s$

Check

$t_{2}=8s$

$x_{f}=x_{o}+v_{o}*t+\frac{1}{2} *a*t^{2}$

$x_{f}=0-6*+\frac{1}{2} *2*8^{2}$

$x_{f}=-48+64\\x_{f}=16$

5 0

3 years ago

To maintain a constant speed, the force provided by a car's engine must equal the drag force plus the force of friction of the r

sweet [91]

Answer:

a). 53.75 N and 101.92 N

b). 381.44 N and 723.25 N

Explanation:

$V= 77 \frac{km}{h}* \frac{1h}{3600 s} *\frac{1000m}{1 km} = 21.38 \frac{m}{s} \\V=106 \frac{km}{h}* \frac{1h}{3600 s} *\frac{1000m}{1 km} = 29.44 \frac{m}{s}$