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Novay_Z [31]
3 years ago
6

5) help plz I need help I don't know if I am right so if I am not can you plz help me

Physics
2 answers:
seraphim [82]3 years ago
7 0

the answer is waning Gibbous

The Waning Gibbous is an intermediary Moon phase. It starts right after the Full Moon, and it lasts until the Third Quarter.

inessss [21]3 years ago
7 0

Answer:

the answer is waning gibbous

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when an object is placed near a concave mirror, at what position does it forms a magnified and erect image.​
alexira [117]

Answer:

Between the principal focus and the pole of the mirror

4 0
3 years ago
Sam is recklessly driving 60 mph in a 30 mph speed zone when he suddenly sees the police. he steps on the brakes and slows to 30
barxatty [35]
For this problem, we use the derived equations for rectilinear motion at constant acceleration. The equations used for this problem are:

a = (v - v₀)/t
2ax = v² - v₀²
where
a is the acceleration
x is the distance
v is the final velocity
v₀ is the initial velocity
t is the time

The solution is as follows;

a = (60mph - 30 mph)/(3 s * 1 h/3600 s)
a = 36,000 mph²

2(36,000 mph²)(x) = 60² - 30²
Solving for x,
x = 0.0375 miles
5 0
3 years ago
Breathing on a cold window can cause it to fog up. which change of state causes this
Dominik [7]
When you breathe on a cold window and it fogs up, that change of state is called: Condensation.

Condensation is: The process in which molecules of a gas slow down, come together, and form a liquid water that collects as droplets on a cold surface when humid air is in contact with it.

I know this because I learned it in science class in school.

Hope I helped!

- Debbie
3 0
3 years ago
A lawn roller in the form of a thin-walled, hollow cylinder with mass M is pulled horizontally with a constant horizontal force
olga nikolaevna [1]

Answer:

a_{cm}=\frac{F}{2M}\\\\F_{fr}=\frac{F}{2}

Explanation:

Given the mass as M, the rotational inertia of the mower is;

I_{cm}=MR^2

-The roller doesn't slip while rolling;

v_{cm}=wR, a_{cm}=\alpha R

\sum F_x=Ma_x, -F+F_{fr}=-Ma_{cm}\\\\a_{cm}=\frac{F-Fr}{M}    \ \ \ \ \ \ \ \ eqtn1\\\\\sum \tau_{cm}=I_{cm}\alpha, F_{fr}(R)}=(MR^2)(\frac{a_{cm}}{R}), ->F_{fr}=Ma_c_m\ \ \  \ \ \ \ eqtn2\\\\a_{cm}=\frac{F-Ma_{cm}}{M}, ->F=2Ma_{cm}\\\\a_{cm}=\frac{F}{2M}\\\\\\\therefore F_{fr}=M(\frac{F}{2M})\\\\F_{fr}=\frac{F}{2}

6 0
3 years ago
"A block of metal weighs 40 N in air and 30 N in water. What is the buoyant force on the block due to the water? The density of
Alja [10]

Answer:

buoyant force on the block due to the water= 10 N

Explanation:

We know that

buoyant force(F_B) on a block= weight of the block in air (actual weight) - weight of block in water.

Given:

A block of metal weighs 40 N in air and 30 N in water.

F_B =  40-30= 10 N

therefore,  buoyant force on the block due to the water= 10 N

6 0
3 years ago
Read 2 more answers
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