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2 years ago

9

A particle R moves with a velocity of 6 m/s due East and another particle S moves at 8 m/s due South. What is the magnitude of t

he velocity of R w.r.t S?

1 answer:

cluponka [151]2 years ago

6 0

Answer:

10

Explanation:

We can look at this problem as a triangle, r is the hypotenuse so if we take the square root of 6^2+8^2 we get 10

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Any help is appreciated

sdas [7]

image distance,di=10 cm

object distance,do=20cm

magnification, m=di/do

=10/20

=0.5

since the image is virtual, magnification is negative.

therefore m=-0.5

8 0

2 years ago

Inthisexperiment,aspringforcewasusedtokeep moving object travelingin a circularpath.The size of a spring force should be proport

poizon [28]

Answer:

By calculation, it can be shown that;

K = $\frac{F_{angular}}{2\times x\times c}$

Whereby for constant K, as ${F_{angular}}$ increases, x also increases.

Explanation:

The experiment set up consisted of the use of a spring force to maintain the object in circular path.

The energy in the spring is given by

$\frac{1}{2}\cdot k\cdot x^2$ .

Rotational kinetic energy = $\frac{1}{2}$ ·I·ω²

Inertia, I = $\frac{1}{2}$ ·m·r²

ω = $\frac{v}{r}$

Substituting gives

Rotational kinetic energy = $\frac{1}{2}$ ·

= $\frac{1}{4}$ ·m· v²

Equating both equations gives

K = $\frac{2\times m\times v^2}{4\times x^{2} }$ = $\frac{1\times m\times v^2}{2\times x^{2} }$

Within the proportionality limit, x ∝ r

therefore we can write x = c·r which gives

$\frac{ m\times v^2}{2\times x\times c\times r }$ = $\frac{v^{2} }{r} \times\frac{m}{2\times x\times c}$

Since $\frac{v^{2} }{r}$ = angular acceleration, α, then

m× $\frac{v^{2} }{r}$ = Angular force

Therefore K = $\frac{F_{angular}}{2\times x\times c}$

Therefore as Force, F increases, x also increases and the size of a spring force should be proportional to the amount of stretch in the spring.

3 0

3 years ago

Match the term to the RIGHT definition

Levart [38]

Earth = A
Mars = B
Mercury = D
Venus = C

4 0

2 years ago

Read 2 more answers

The energy expenditure value of traveling by car is 3.6 mj/passenger-kilometer. The value for traveling by train is 1.1 mj/passe

andrezito [222]

Answer:

Using lighter material in car construction, improving energy efficiency by enhancing engine design or replacing the engine with more efficient technologies.

Explanation:

Using lighter materials in the car construction, reducing the potential energy required to accelerate and to move the car, as well as energy losses due to rolling friction. There is evidence of such benefits by replacing steel and aluminium parts with components made of composite materials.

Improving the design of internal combustion engines to minimize energy losses and accordingly, improving energy efficiency. A more radical approach is replacing internal combustion engines with electric engines, which offer higher efficiencies. Such conclusions can be easily inferred from model based on Work-Energy Theorem and Principle of Energy Conservation:

$\eta_{engine} \cdot U_{engine} = \frac{1}{2} \cdot m_{car} \cdot v^{2} + \mu_{r} \cdot m_{car} \cdot g \cdot \Delta s$

7 0

3 years ago

a closed tank is partially filled with glycerin. if the air pressure in the tank is 6 lb/in.2 and the depth of glycerin is 10 ft

vlabodo [156]

Answer:

<u><em>note:</em></u>

<u><em>solution is attached due to error in mathematical equation. please find the attachment</em></u>

4 0

3 years ago