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4 years ago

The ___________ energy in a mechanical system is determined by adding the potential and kinetic energy together.

Physics

2 answers:

hoa [83]4 years ago

3 0

Answer:

mechanical energy

Explanation:

i believe thats the answer hope that helps

Valentin [98]4 years ago

3 0

Answer:

Mechanical energy

Explanation:

The mechanical energy in a mechanical system is the summation of its kinetic energy and its potential energy.

Mechanical Energy= Potential Energy +Kinetic energy

Potential energy is defined as the energy stored in a stable object and kinetic energy is defined as the energy of a moving object or energy required in the motion of a object.

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What does temperature actually measure?

Alborosie

Temperature is measured with a thermometer

8 0

4 years ago

A narrow beam of light containing orange (610 nm) and blue (470 nm) wavelengths goes from polystyrene to air, striking the surfa

Eva8 [605]

Answer: The angle between the two colours when they emerge is 0.4°

Explanation:

The refracting angle for different colors knowing its refractive index can be calculated using Snell's law.

Ni × sin α = Nr × sinβ

Where Ni is the refractive index for light in incident medium

α is the angle the incident makes with normal

Nr is the refractive index for light in the refractive medium

β is the angle the refracted makes with the normal

Making β the subject

β = arcSin ( Ni × sinα)/Nr

For Orange color

The value of refractive index in polystyrene medium is 1.490 ,that is Ni= 1.490

α is 30° and Nr is the refractive index of air = 1

β = arcSin(1.490×sin30°)/1

β= 48.15°

For Blue color

The value of refractive index in polystyrene medium is 1.499,that is Ni= 1.499

α is 30° and Nr is the refractive index of air = 1

β = arcSin(1.499×sin30°)/1

β= 48.55°

The angle between the two colours is the difference in the angles of their refracted rays

48.55-48.15=0.4°

6 0

3 years ago

A 3.0-kg block starts at rest at the top of a 37° incline, which is 5.0 m long. Its speed when it reaches the bottom is 2.0 m/s.

Mama L [17]

Answer: $f_{r}$ = 16.49N

Explanation: The object is placed on an inclined plane at an angle of 37° thus making it weight have two component,

$W_{x}$ = horizontal component of the weight = mgsinФ

$W_{y}$ = vertical component of weight = mgcosФ

Due to the way the object is positioned, the horizontal component of force will accelerate the object thus acting as an applied force.

by using newton's law of motion, we have that

mgsinФ - $f_{r}$ = ma

where m = mass of object=5kg

a = acceleration= unknown

Ф = angle of inclination = 37°

g = acceleration due to gravity = 9.8 $m/s^{2}$

$f_{r}$ = frictional force = unknown

we need to first get the acceleration before the frictional force which is gotten by using the equation below

$v^{2} = u^{2} + 2aS$

where v = final velocity = 2m/s

u = initial velocity = 0m/s (because the object started from rest)

a= unknown

S= distance covered = length of plane = 5m

$2^{2} = 0^{2} + 2*a*5\\\\4= 10 *a\\\\a = \frac{4}{10} \\a = 0.4m/s^{2}$

we slot in a into the equation below to get frictional force

mgsinФ - $f_{r}$ = ma

3 * 9.8 * sin 37 - $f_{r}$ = 3* 0.4

17.9633 - $f_{r}$ = 1.2

$f_{r}$ = 17.9633 - 1.2

$f_{r}$ = 16.49N

4 0

4 years ago

GIVING BRAINLIEST PLS HELP!!!

MA_775_DIABLO [31]

Answer:

it's the first one

Explanation:

Have a great day hope it helps

3 0

3 years ago

A yo‑yo with a mass of 0.0800 kg and a rolling radius of =2.70 cm rolls down a string with a linear acceleration of 5.70 m/s2.

N76 [4]

Explanation:

Given that,

Mass, m = 0.08 kg

Radius of the path, r = 2.7 cm = 0.027 m

The linear acceleration of a yo-yo, a = 5.7 m/s²

We need to find the tension magnitude in the string and the angular acceleration magnitude of the yo‑yo.

(a) Tension :

The net force acting on the string is :

ma=mg-T

T=m(g-a)

Putting all the values,

T = 0.08(9.8-5.7)

= 0.328 N

(b) Angular acceleration,

The relation between the angular and linear acceleration is given by :

$\alpha =\dfrac{a}{r}\\\\\alpha =\dfrac{5.7}{0.027}\\\\=211.12\ m/s^2$

(c) Moment of inertia :

The net torque acting on it is, $\tau=I\alpha$ , I is the moment of inertia

Also, $\tau=Fr$

So,

$I\alpha =Fr\\\\I=\dfrac{Fr}{\alpha }\\\\I=\dfrac{0.328\times 0.027}{211.12}\\\\=4.19\times 10^{-5}\ kg-m^2$

Hence, this is the required solution.

3 0

3 years ago