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neonofarm [45]
4 years ago
8

Which kinds of objects emit visible light in the electromagnetic spectrum?1)all objects2)radioactive objects3)relatively cold ob

jects4)relatively hot objects
Physics
1 answer:
Lostsunrise [7]4 years ago
8 0
Relatively hot objects emit visible light.
Some examples: 
==> the wire coils in the toaster; 
==> the spoon that you stuck in the flame on the stove;
==> the fine wire in the lightbulb when current goes through it.

VERY radioactive objects also do that.  But if you're actually
standing there watching an object that's THAT radioactive,
then you're in big trouble.
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What statement applies to the horizontal rows or periods in the periodic table? A. The properties are all the same. B. The eleme
Sliva [168]
The right answer for the question that is being asked and shown above is that: "C. <span>. The properties change going across each row. " the </span>statement that applies to the horizontal rows or periods in the periodic table is that t<span>he properties change going across each row. </span>
7 0
3 years ago
What is the upper block's acceleration if the coefficient of kinetic friction between the block and the table is 0.13?
IgorLugansk [536]

Let us assume that pulley is mass less.

Let the tension produced at both ends of the pulley is T.

We are asked to calculate the acceleration of the block.

Let the masses of two bodies are denoted as m_{1} \ and\ m_{2}\ respectively

Let\ m_{1} =1 kg\ and\ m_{2} =2 kg

As per this diagram, the body having mass 1 kg is moving downward and the body having mass 2 kg is moving on the surface of the table.

Let the acceleration of each block is a .

For body having mass 1 kg:

The net force acting on 1 kg body will be-

                             m_{1} g-T=m_{1} a        [1]

Here tension in the rope will be vertically upward and weight of the body will be in vertical downward direction.

For body having mass 2 kg:

The coefficient of kinetic friction [\mu]=0.13

Hence\ the\ frictional\ force\ F=\mu N

                                                     F=\mu m_{2} g

Hence the net force acting on the body having mass 2 kg-

                                  T-\mu m_{2} g=m_{2} a  [2]

Here the tension of the rope is towards right i.e along the direction of motion of the 2 kg block and frictional force is towards left.

Combining 1 and 2 we get-

                           m_{1} g-T=m_{1}a             [1]

                           T-\mu m_{2}g= m_{2} a   [2]

                           ---------------------------------------------------

                           [m_{1} -\mu m_{2} ]g=[m_{1} +m_{2} ]a

                           a=\frac{m_{1}-\mu m_{2}} {m_{1}+ m_{2}}*g

                           a=\frac{1-[2*0.13]}{1+2} *9.8\ m/s^2

                           a=\frac{0.74}{3} *9.8\ m/s^2

                           a=2.417 m/s         [ans]

6 0
3 years ago
Read 2 more answers
In regards to Pressure ( Ch. Static Fluids - Introductory Physics)
ehidna [41]

Explanation:

P₁ = P₂ + ρgh

g is the acceleration due to gravity

ρ is the density of the fluid

h is the depth of the fluid

P₁ is the pressure at that depth

P₂ is the pressure at the surface

P₁ and P₂ can either be absolute pressures or gauge pressures, but they must match.

For example, if you wanted to find the <em>absolute</em> pressure at the bottom of an <em>open</em> tank, you would use P₂ = Patm = 14.7 psi or 101.3 kPa.

If instead you wanted to find the <em>gauge</em> pressure, you would use P₂ − Patm = 0 psi or 0 kPa.

If the tank is sealed and pressurized, you would use the P₂ of the tank.

3 0
3 years ago
Sobre un recipiente lleno de agua se sumerge completamente un objeto que desplaza un volumen de 0.096 metros cúbicos de agua hac
Licemer1 [7]

Answer:

Fb = 941.76 [N]

Explanation:

La fuerza de flotacion se define como el producto de la densidad del liquido por la aceleración gravitacional y por el volumen desplazado

Fb = Ro*g*V

donde:

Fb = fuerza de flotacion [N] (unidades del Newtons)

Ro = densidad del liquido = 1000 [kg/m³]

g = aceleracion gravitacional = 9.81 [m/s²]

V = volumen desplazado = 0.096 [m³]

Reemplazando:

Fb = 1000*9.81*0.096

Fb = 941.76 [N]

7 0
3 years ago
The potential energy of a catapult was completely converted into kinetic energy by releasing a small stone with a mass of 20 gra
DerKrebs [107]

1,000 grams = 1 kilogram
20 grams = 0.02 kilogram

Kinetic energy = (1/2) (mass) x (speed)²

                           (1/2) (0.02) x (15)² =

                                 (0.01)  x  (225)  =  2.25 joules
 
4 0
4 years ago
Read 2 more answers
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