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3 years ago

Which describes how to generate a real image of an object using a concave mirror?

2 answers:

7 0

A concave mirror is curved inward in the middle, more like a cave. Because the mirror is curved inward, the angle of the light surface can be focused similar to that of the camera. They can form real images that are projected out in front of the mirror at the place where light focuses. When the object is located at the center of the curvature the image formed will also be at the curvature. The image will be inverted and the magnification value is equal to 1 which will become a real image because the ray of light converges at the location of the formed image.

Karolina [17]3 years ago

7 0

Answer:

The object must be farther from the mirror than the focal point

Explanation:

gradpoint lesson

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A box slides down a 30.0° ramp with an acceleration of 1.20 m/s^2. Determine the coefficient of kinetic friction between the box

Zielflug [23.3K]

m = mass of the box

N = normal force on the box

f = kinetic frictional force on the box

a = acceleration of the box

μ = coefficient of kinetic friction

perpendicular to incline , force equation is given as

N = mg Cos30 eq-1

kinetic frictional force is given as

f = μ N

using eq-1

f = μ mg Cos30

parallel to incline , force equation is given as

mg Sin30 - f = ma

mg Sin30 - μ mg Cos30 = ma

"m" cancel out

a = g Sin30 - μ g Cos30

inserting the values

1.20 = (9.8) Sin30 - (9.8) Cos30 μ

μ = 0.44

4 0

3 years ago

ome metal oxides can be decomposed to the metal and oxygen under reasonable conditions. 2 Ag2O(s) → 4 Ag(s) + O2(g) Thermodynami

yuradex [85]

Answer : The values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $62.2kJ,132.67J/K\text{ and }22.66kJ$ respectively.

Explanation :

The given balanced chemical reaction is,

$2Ag_2O(s)\rightarrow 4Ag(s)+O_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{product}$

$\Delta H^o=[n_{Ag}\times \Delta H_f^0_{(Ag)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]-[n_{Ag_2O}\times \Delta H_f^0_{(Ag_2O)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[4mole\times (0kJ/mol)+1mole\times (0kJ/mol)}]-[2mole\times (-31.1kJ/mol)]$

$\Delta H^o=62.2kJ=62200J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{product}$

$\Delta S^o=[n_{Ag}\times \Delta S_f^0_{(Ag)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]-[n_{Ag_2O}\times \Delta S_f^0_{(Ag_2O)}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

Now put all the given values in this expression, we get:

$\Delta S^o=[4mole\times (42.55J/K.mole)+1mole\times (205.07J/K.mole)}]-[2mole\times (121.3J/K.mole)]$

$\Delta S^o=132.67J/K$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is 298 K.

$\Delta G^o=(62200J)-(298K\times 132.67J/K)$

$\Delta G^o=22664.34J=22.66kJ$

Therefore, the values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $62.2kJ,132.67J/K\text{ and }22.66kJ$ respectively.

6 0

3 years ago

If a transformer has a primary with 100 V and fifty coils, and a secondary that yields 20 V, how many coils are on the secondary

melomori [17]

Ideally, if all the magnetic of one winding cuts the other winding, and there isn't any loss in the transformer core or the resistance of the wire, then the voltage across each winding is proportional to the number of turns in its coil.

If you apply 100 V to a winding of 50 turns, then a winding that yields 20 volts
must be wound with

(20/100) of 50 turns = 10 turns

7 0

3 years ago

When an object oscillating in simple harmonic motion is at its maximum displacement from the equilibrium position, which of the

ziro4ka [17]

Answer:

E. Zero Maximum

Explanation:

At the point of maximum displacement, the speed is zero while the restoring force is maximum. In fact:

- The restoring force is given by $F=kx$ , where k is the spring constant and x is the displacement - at the point of maximum displacement, x is maximum, so F is maximum as well

- the total energy of the system is sum of kinetic energy and elastic potential energy:

$E=K+U=\frac{1}{2}mv^2+\frac{1}{2}kx^2$

where m is the mass of the system and v is the speed. Since E (the total energy) is constant due to the law of conservation of energy, we have that when K increases, U decreases, and viceversa. As a result, when x increases, v decreases, and viceversa. At the point of maximum displacement, x is maximum, so v will have its minimum value (which is zero, since the system is changing direction of motion).

4 0

3 years ago

Sí un auto viaja a 8m/s determine Tiempo en llegar a 200km de distancia Distancia que recorre en 40 minutos

GrogVix [38]

Answer:

a. Tiempo = 25000 segundos

b. Distancia = 19200 metros

Explanation:

Dados los siguientes datos;

Velocidad = 8 m/s

Distancia = 200 km a metros = 200 * 1000 = 200,000

Para encontrar el tiempo para cubrir la distancia anterior;

Tiempo = distancia/velocidad

Tiempo = 200000/8

Tiempo = 25000 segundos

b. Para encontrar la distancia recorrida en 40 minutos;

Tiempo = 40 minutos a segundos = 40 * 60 = 2400 segundos

Distancia = velocidad * tiempo

Distancia = 8 * 2400

Distancia = 19200 metros

7 0

3 years ago