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Marizza181 [45]
3 years ago
14

Henry can lift a 200 N load 20 m up a ladder in 40 s. Ricardo can lift twice the load up one-half the distance in the same amoun

t of time. Which comparison between Ricardo and Henry is correct?
Physics
2 answers:
prohojiy [21]3 years ago
4 0
Henry will lift 200 N load 20 m up a ladder in 40 s.  While the Ricardo will take 400 N load in 80 seconds. So, For Henry to take 400 N load it will take him 80 seconds in two attempts. And,also, he will have to cover 40 m of distance. 
madreJ [45]3 years ago
4 0

Answer;

-Ricardo exerted a greater force, did the same amount of work, and had the same power output as Henry.

Explanation;

For Ricardo to lift twice the load lifted by Henry for the one-half the distance in the same amount of time, he will need more force as the load is heavier.

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Salt water is denser than fresh water. a ship floats in both fresh water and salt water. compared to the fresh water, the volume
gavmur [86]
The ship floats in water due to the buoyancy Fb that is given by the equation:

Fb=ρgV, where ρ is the density of the liquid, g=9.81 m/s² is the acceleration of the force of gravity and V is volume of the displaced liquid.

The density of fresh water is ρ₁=1000 kg/m³.

The density of salt water is in average ρ₂=1025 kg/m³.

To compare the volumes of liquids that are displaced by the ship we can take the ratio of buoyancy of salt water Fb₂ and the buoyancy of fresh water Fb₁.

The gravity force of the ship Fg=mg, where m is the mass of the ship and g=9.81  m/s², is equal to the force of buoyancy Fb₁ and Fb₂ because the mass of the ship doesn't change:
 
Fg=Fb₁ and Fg=Fb₂. This means Fb₁=Fb₂.

Now we can write:

Fb₂/Fb₁=(ρ₂gV₂)/(ρ₁gV₁), since Fb₁=Fb₂, they cancel out:

1/1=1=(ρ₂gV₂)/(ρ₁gV₁), g also cancels out:

(ρ₂V₂)/(ρ₁V₁)=1, now we can input ρ₁=1000 kg/m³ and ρ₂=1025 kg/m³

(1025V₂)/(1000V₁)=1

1.025(V₂/V₁)=1

V₂/V₁=1/1.025=0.9756, we multiply by V₁

V₂=0.9756V₁

Volume of salt water V₂ displaced by the ship is smaller than the volume of sweet water V₁ because the force of buoyancy of salt water is greater than the force of fresh water because salt water is more dense than fresh water.  


7 0
3 years ago
A train travels 98 kilometers in 4 hours, and then 61 kilometers in 3 hours. What is its average speed?
IrinaK [193]
Speed (velocity)=distance/time

V1=98km/4hr=24.5km/hr
V2=61km/3hr=20.3km/hr

Average speed (velocity)=total velocity/ number

Average speed (velocity)=44.8km/hr/2=22.4

So the average speed is 22.4km/hr
6 0
3 years ago
Astronauts often undergo special training in which they are subjected to extremely high centripetal accelerations. One device ha
egoroff_w [7]

The centripetal acceleration of an object is given by the relation,

Ac =V^2/R

where Ac = centripetal acceleration = 98 m/s^2

R = radius of rotation = 15 m

V = speed of astronaut

Hence, \frac{V^2}{15} =98

solving this we get, V = 38.34 m/s

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Question 5 of 20 Engineers are using computer models to study train collisions to design safer train cars. They start by modelin
aniked [119]

Answer:

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Consider a uniformly charged sphere of radius Rand total charge Q. The electric field Eout outsidethe sphere (r≥R) is simply tha
AlexFokin [52]

1) Electric potential inside the sphere: \frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})

2) Ratio Vcenter/Vsurface: 3/2

3) Find graph in attachment

Explanation:

1)

The electric field inside the sphere is given by

E=\frac{1}{4\pi \epsilon_0}\frac{Qr}{R^3}

where

\epsilon_0=8.85\cdot 10^{-12}F/m is the vacuum permittivity

Q is the charge on the sphere

R is the radius of the sphere

r is the distance from the centre at which we compute the field

For a radial field,

E(r)=-\frac{dV(r)}{dr}

Therefore, we can find the potential at distance r by integrating the expression for the electric field. Calculating the difference between the potential at r and the potential at R,

V(R)-V(r)=-\int\limits^R_r  E(r)dr=-\frac{Q}{4\pi \epsilon_0 R^3}\int r dr = \frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)

The potential at the surface, V(R), is that of a point charge, so

V(R)=\frac{Q}{4\pi \epsilon_0 R}

Therefore we can find the potential inside the sphere, V(r):

V(r)=V(R)+\Delta V=\frac{Q}{4\pi \epsilon_0 R}+\frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})

2)

At the center,

r = 0

Therefore the potential at the center of the sphere is:

V(r)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})\\V(0)=\frac{3Q}{8\pi \epsilon_0 R}

On the other hand, the potential at the surface is

V(R)=\frac{Q}{4\pi \epsilon_0 R}

Therefore, the ratio V(center)/V(surface) is:

\frac{V(0)}{V(R)}=\frac{\frac{3Q}{8\pi \epsilon_0 R}}{\frac{Q}{4\pi \epsilon_0 R}}=\frac{3}{2}

3)

The graph of V versus r can be found in attachment.

We observe the following:

- At r = 0, the value of the potential is \frac{3}{2}V(R), as found in part b) (where V(R)=\frac{Q}{4\pi \epsilon_0 R})

- Between r and R, the potential decreases as -\frac{r^2}{R^2}

- Then at r = R, the potential is V(R)

- Between r = R and r = 3R, the potential decreases as \frac{1}{R}, therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 (\frac{1}{3}V(R))

Learn more about electric fields and potential:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

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3 years ago
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