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2 years ago

Where is the most and the least KENETIC ENERGY

Physics

1 answer:

My name is Ann [436]2 years ago

6 0

Answer:

Where is kinetic energy the least?, and most?

Explanation:

Least : When the Moon is farthest from Earth in its nearly elliptical orbit, its speed is least. Its kinetic energy has become least, and its potential energy is greatest.

Further : Kinetic energy, form of energy that an object or a particle has by reason of its motion. If work, which transfers energy, is done on an object by applying a net force, the object speeds up and thereby gains kinetic energy.

May be different for you, but i really hope I helped out.

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One boat tows another boat by means of a tow line, which is under a constant tension of 465 N. The boats move at a constant spee

Ludmilka [50]

Answer:

Work done, W = 141174 Joules

Explanation:

It is given that,

Constant tension acting on the boat, T = F = 465 N

Speed of the boat, v = 4.6 m/s

Time, t = 1.1 min = 66 seconds

Let W is the work done by the tension. It is equal to the product of force and displacement. It is given by :

$W=F\times d$

Since, $d=vt$

$W=F\times v\times t$

$W=465\ N\times 4.6\ m/s\times 66\ s$

W = 141174 Joules

So, the work is done by the tension is 141174 Joules. Hence, this is the required solution.

6 0

3 years ago

I need help on this and the first person who will answer this correctly gets BRANLIST

Nikolay [14]

Answer:

i think it's b but i don't know for sure

Explanation:

3 0

2 years ago

A wire is 1.1 m long and 0.91 mm in cross-sectional area. It carries a current of 4.4 A when a 1.8 V potential difference is app

Vlad [161]

Answer:

The conductivity of the material from which the conductor is made is 2.95 x10⁶ [S. m⁻¹] .

Explanation:

L= 1.1m

S= 0.91 mm²= 9.1 x10⁻⁷ m²

I= 4.4 A

V= 1.8V

$\frac{V}{I} =R\\\\\frac{1.8V}{4.4A} =R= 9/22 ohms$

R=9/22Ω

$R=\frac{L}{σ . S} \\\\Conductivity=sigma=\frac{L}{R.S} \\\\Conductivity= \frac{1.1m}{9/22 ohms . 9.1 E10 (-7) square meters}$

Conductivity=2.95 x10⁶ [S. m⁻¹]

3 0

2 years ago

Assume that the car at point A and the one at point E are traveling along circular paths that have the same radius. If the car a

vazorg [7]

Answer:

The magnitude of car A acceleration is 4 times that of car E

Explanation:

Assuming they are traveling at a constant speed. Their (centripetal) acceleration is as the following

$a = \frac{v^2}{r}$

where v is the (linear) velocity and r is the radius. We can use this to calculate their ratio

$a_A / a_E = \frac{v_A^2/r_A}{v_E^2/r_E} = \left(\frac{v_A}{v_E}\right)^2\frac{r_E}{e_A}$

Since $v_A = 2V_E \rightarrow \frac{v_A}{v_E} = 2$ and $r_A = r_E$

$a_A / a_E = 2^2 = 4$

So the magnitude of car A acceleration is 4 times that of car E

6 0

2 years ago

An object is placed near a concave mirror having a radius of curvature of magnitude 60 cm. How far should you place the object f

Viefleur [7K]

Answer:

u = 18 cm

Explanation:

given,

radius of curvature = 60 cm

magnification of mirror = 2.5

distance of object = ?

R = 2 f

f = R/2

f = 60/2 = 30 cm

$m = -\dfrac{v}{u}$

$2.5 = -\dfrac{v}{u}$

v = -2.5 u

now,

Using mirror formula

$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$

$\dfrac{1}{30} = \dfrac{1}{-2.5u} + \dfrac{1}{u}$

$\dfrac{1}{30} = \dfrac{0.6}{u}$

u = 0.6 x 30

u = 18 cm

distance of object be equal to u = 18 cm

6 0

3 years ago