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Ira Lisetskai [31]
3 years ago
9

Jared would describe a square as having four equal sides and four right angles. This is Jared’s __________ of a square. A. model

B. hierarchy C. feature D. concept Please select the best answer from the choices provided A B C D
Physics
1 answer:
zheka24 [161]3 years ago
4 0

Answer:

D

Explanation:

jhusskgtyddutsuudtiff

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Explanation:Consider the wave pattern image reflected about the rigid hook on the wall. · 005(part2of3)10.0 pointsHow many complete waves are emitted in this time
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Why does a box on the seat of a car slide around on the c one the car speeds up slows down or turns a corner?
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Because the box keeps going straight at the same speed, while the seat under it speeds up, slows down, or changes direction.
8 0
3 years ago
A spherical balloon is deflating at 10 cm3/s. At what rate is the radius changing when the volume is 1000π cm3 ? What is the rat
just olya [345]

Answer:

1.dr/dt=0.0096cm/s

2. dA/dt=2.19cm^2/s

Explanation:

A spherical balloon is deflating at 10 cm3/s. At what rate is the radius changing when the volume is 1000π cm3 ? What is the rate of change of surface area at this moment?

for this question, we need to analyze the parameters we know

V=volume of the spherical balloon 1000π cm3

volume of the sphere=\frac{4}{3} \pi r^{3}

1000π=4/3πr^3

dividing both sides by 4

250*3=r^3

r=9.08cm, the radius of the balloon

dv/dt=dv/dr*dr/dt...................................1

dv/dr  ,means

V=\frac{4}{3} \pi r^{3}

dv/dr=4*pi*r^2

dv/dt=10 cm3/s

from equ 1

10=4*pi*9.08^2*dr/dt

10=1036 dr/dt

dr/dt=10/1036

dr/dt=0.0096cm/s

2. to find the rate at which the area is changing we have,

dA/dt=dA/dr*dr/dt

area of a sphere is  4πr^2

differentiate a with respect to r, radius

dA/dr=8πr

dA/dt=8πr*0.0096

dA/dt=8*pi*9.08*0.0096

dA/dt=2.19cm^2/s

is the rate of change of the surface area

7 0
3 years ago
Which of the following is a reason for the start of the space race?
Mariana [72]
Answer #4, because they wanted to show that they are the best
7 0
4 years ago
A 13500 kg jet airplane is flying through some high winds. At some point in time, the airplane is pointing due north, while the
Mekhanik [1.2K]

Answer

given,

mass of the jet airplane = 13500 kg

Force on the plane = 35700 N due north

force from wind  = 15300 N in direction 80.0° south of west.

Force = 35700 \vec{j} N

force by wind = 15300(-cos \theta \vec{i}-sin \theta \vec{j}) N

                       = 15300(-cos 80^0 \vec{i}-sin 80^0 \vec{j}) N

net force on the jet airplane(ma)

          m a = 35700 \vec{j} + 15300(-cos 80^0 \vec{i}-sin 80^0 \vec{j})

          \vec{a} = \dfrac{35700}{13500} \vec{j} + \dfrac{15300}{13500}(-cos 80^0 \vec{i}-sin 80^0 \vec{j})

          \vec{a} = 2.64\vec{j} -0.197 \vec{i} - 1.116 sin 80^0 \vec{j})

           \vec{a} = -0.197 \vec{i} + 1.524 \vec{j}

a = \sqrt{-0.197^2+1.524^2}

a = 1.54 m/s²

\theta = tan^{-1}(\dfrac{-1.524}{0.197})

\theta = -82.63^0

3 0
3 years ago
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