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vesna_86 [32]
3 years ago
15

What sauce goes best with chicky nuggies

Physics
1 answer:
Sever21 [200]3 years ago
6 0

Answer:

honey mustard or chick fil a sauce their special sauce

Explanation:

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What must occur for work to be done on an object?
stiv31 [10]
The answer is D since work done must have displacement..if theres force but no displacement there will be no work done based on W=Fs
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3 years ago
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A platinum sphere with radius 0.0139 m is totally immersed in mercury. Find the weight of the sphere, the buoyant force acting o
Inessa [10]

Answer:

A) W = 0.59 N

B) Buoyant Force = 0.37N

C) Apparent Weight = 0.22N

Explanation:

Volume of a sphere is givwn by;

V= (4/3)πr³

We are given radius r = 0.0139 m

Thus, V = (4/3)π(0.0139³)

V = 2.81237 x 10^(-6) m³

A) Weight of sphere, W= mg

where mass, m can be expressed as; m = Volume x density of platinum

m = 2.81237 x 10^(-6) x 2.14 × 10⁴ = 6.0185 x 10^(-2) kg

So, W = 6.0185 x 10^(-2) x 9.8

W = 0.59 N

B) Buoyant force = mg

where mass of displaced mercury m can be expressed as; m = Volume x density of mercury

m = 2.81237 x 10^(-6) x 1.36 × 10⁴

m = 3.825 x 10^(-2) kg

Thus, Buoyant force = 3.825 x 10^(-2) x 9.8 = 0.37N

C) Apparent weight = Weight of sphere - Buoyant force

Thus, apparent weight = 0.59 - 0.37 = 0.22N

8 0
3 years ago
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Which sub-atomic particle is positively charged and found in the nucleus?
IrinaK [193]

The proton. There are three main types of subatomic particles.  Protons, Neutrons, and Electrons.  Protons are positively charged and found in the nucleus.   Neutrons are not charged at all, and found in the nucleus, and electrons are negatively charged and found outside the nucleus in regions known as orbitals.

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A 2.7-cm-tall object is 20 cm to the left of a lens with a focal length of 10 cm . A second lens with a focal length of 48 cm is
irina1246 [14]

Answer

given,

height of object = 2.7 cm

distance left of lens (u₁)= 20 cm

focal length of lens(f₁)= 10 cm  

the distance of image

\dfrac{1}{f_1}=\dfrac{1}{u_1}+\dfrac{1}{v_1}

\dfrac{1}{v_1}=\dfrac{1}{f_1}-\dfrac{1}{u_1}

\dfrac{1}{v_1}=\dfrac{1}{10}-\dfrac{1}{20}

   v₁ = 20 cm

magnification of first lens

m_1= -\dfrac{v}{u}

m_1=-\dfrac{20}{20}

    m₁ = -1

distance of object from the second lens

   u₂ = 52-20 = 32 cm

   f₂ = 48 cm

now,

\dfrac{1}{f_2}=\dfrac{1}{u_2}+\dfrac{1}{v_2}

\dfrac{1}{v_2}=\dfrac{1}{f_2}-\dfrac{1}{u_2}

\dfrac{1}{v_2}=\dfrac{1}{48}-\dfrac{1}{52}

   v₁ = 624 cm

magnification of first lens

m_1= -\dfrac{v}{u}

m_1=-\dfrac{624}{52}

    m₁ = -12

total magnification

m = m₁ m₂

m = (-1)(-12)

m = 12

height of image

m =-\dfrac{h'}{h}

12=-\dfrac{h'}{2.7}

h' = -32.4 cm

a) distance between image and second lens is equal to 624 cm

b) height of image is equal to 32.4 cm

       

4 0
3 years ago
All of the following are examples of physical properties except which one?
Thepotemich [5.8K]

Answer:

the answer is c

Explanation:

melting = become liquefied by heat

solubility = a new bond formation

conductivity = electric transmitter

flammability = burnable

8 0
3 years ago
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