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Darya [45]
3 years ago
12

chris and molly wins $2400 in a competition they share the money in the ratio2:3 how much money do they each receive ?​

Physics
1 answer:
nlexa [21]3 years ago
5 0

Answer:

chris gets 960 and molly gets 1440

Explanation:

add the ratio up and divide

2+3=5

2400/5=480

480x2=960

480x3=1440

960+1440= 2400

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A submerged submarine alters its _____ to rise or fall in the water.
Dima020 [189]
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Submarines use <span>buoyancy by filling ballast tanks up with water. When they are filled with water, they are more dense than the surrounding water, so they are able to sink. If they want to rise, they fill these tanks up with air so that the density is less than the water it surrounds. 
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3 0
3 years ago
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A converging lens has a focal length of 20 cm. An object 1 cm tall is placed 10 cm from the center of the lens. What is the heig
SCORPION-xisa [38]

Answer: 2 cm

Explanation:

Given , for a converging lens

Focal length : f=20\ cm

Height of object : h=1\ cm

Object distabce from lens : u=-10\ cm

Using lens formula: \dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}, we get

\dfrac{1}{20}=\dfrac{1}{v}+\dfrac{1}{10}, where v = image distance from the lens.

On solving aboive equation , we get

\dfrac{1}{v}=\dfrac{1}{20}-\dfrac{1}{10}=\dfrac{1-2}{20}=\dfrac{-1}{20}\Rightarrow\ v=-20\ cm

Formula of Magnification : m=\dfrac{v}{u}=\dfrac{h'}{h} , where h' is the height of image.

Put value of u, v and h in it , we get

\dfrac{-20}{-10}=\dfrac{h'}{1}\\\\\Rightarrow\ h'=2\ cm

Hence, the height of the image is 2 cm.

3 0
3 years ago
I need help with question #8 please!!
otez555 [7]

Answer: 3.1158

Explanation: THERE

3 0
2 years ago
Two teams of nine members each engage in a tug of war. Each of the first team’s members has an average mass of 68 kg and exerts
olga2289 [7]

Answer:

a)  a = - 0.106 m/s^2  (←)

b) T = 12215.1064 N

Explanation:

If

F₁ = 9*1350 N = 12150 N   (→)

F₂ = 9*1365 N = 12285 N  (←)

∑Fx = M*a = (M₁  +M₂)*a           (→)

F₁ - F₂ = (M₁  +M₂)*a        

→       a = (F₁ - F₂) / (M₁  +M₂ ) = (12150-12285)N/(9*68+9*73)Kg

→       a = - 0.106 m/s^2            (←)

(b) What is the tension in the section of rope between the teams?

If we apply  ∑Fx = M*a   for the team 1

F₁ - T = - M₁*a  ⇒   T = F₁ + M₁*a  

⇒   T = 12150 N + (9 * 68 Kg)*(0.106 m/s^2)

⇒ T = 12215.1064 N

If we choose the team 2 we get

- F₂ + T = - M₂*a  ⇒   T = F₂ - M₂*a  

⇒   T = 12285 N - (9 * 73 Kg)*(0.106 m/s^2)

⇒ T = 12215.1064 N

4 0
3 years ago
What is the gauge pressure of the water right at the point p, where the needle meets the wider chamber of the syringe? neglect t
Helen [10]

Missing details: figure of the problem is attached.

We can solve the exercise by using Poiseuille's law. It says that, for a fluid in laminar flow inside a closed pipe,

\Delta P =  \frac{8 \mu L Q}{\pi r^4}

where:

\Delta P is the pressure difference between the two ends

\mu is viscosity of the fluid

L is the length of the pipe

Q=Av is the volumetric flow rate, with A=\pi r^2 being the section of the tube and v the velocity of the fluid

r is the radius of the pipe.

We can apply this law to the needle, and then calculating the pressure difference between point P and the end of the needle. For our problem, we have:

\mu=0.001 Pa/s is the dynamic water viscosity at 20^{\circ}

L=4.0 cm=0.04 m

Q=Av=\pi r^2 v= \pi (1 \cdot 10^{-3}m)^2 \cdot 10 m/s =3.14 \cdot 10^{-5} m^3/s

and r=1 mm=0.001 m

Using these data in the formula, we get:

\Delta P = 3200 Pa

However, this is the pressure difference between point P and the end of the needle. But the end of the needle is at atmosphere pressure, and therefore the gauge pressure (which has zero-reference against atmosphere pressure) at point P is exactly 3200 Pa.

8 0
3 years ago
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