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Murrr4er [49]
4 years ago
15

All objects____ energy. Eliminate Conduct Radiate Insulate

Physics
1 answer:
Trava [24]4 years ago
7 0

the answer is c. radiate

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A spring on Earth has a 0.500 kg mass suspended from one end and the mass is displaced by 0.3 m. What will the displacement of t
Alisiya [41]

Answer:

The displacement of the same mass on the same spring on the Moon is 0.05 m.

Explanation:

Given;

mass suspended from one end of the spring, m = 0.500 kg

displacement on the spring on Earth, x = 0.3 m

Apply Newton's second law of motion;

F = ma = mg

where;

m is mass on the spring

g is acceleration due to gravity

Also, apply Hook's law;

F = Kx

where;

K is force constant

x is extension or diplacement of the spring

Combine the two equations from the two laws;

mg = kx

when the spring in on Earth;

0.5 x 9.8 = 0.3k

4.9 = 0.3k

k = 4.9 / 0.3

k = 16.333 N/m

when the spring is on moon;

mg = kx

mass is the same = 0.5 kg

acceleration due to gravity on moon = ¹/₆ that of Earth =  ¹/₆  x 9.8 m/s²

0.5 (¹/₆  x 9.8) = 16.333 x

0.8167 = 16.333 x

x = 0.8167 / 16.333

x = 0.05 m

Therefore, the displacement of the same mass on the same spring on the Moon is 0.05 m.

4 0
3 years ago
Light travels at a speed of about 3.0 108 m/s. (a) how many miles does a pulse of light travel in a time interval of 0.1 s, whic
babunello [35]

Speed of light is given as

c = 3 * 10^8 m/s

time interval is given as

\Delta t = 0.1 s

so the distance covered by the light is given as

d = v * \Delta t

d = 3 * 10^8 * 0.1 = 3 * 10^7 meter

now as we know that

1 mile = 1609 meter

so the distance moved by light is

d = \frac{3 * 10^7}{1609} = 18645.12 miles

now for the comparision of this distance with diameter of earth

as we know that radius of earth is

R = 6.38 * 10^6 m

so the diameter of earth will be

d = 2R = 12.76 * 10^6 m

now the ratio of diameter with the distance that light move will be

\frac{distance}{diameter} = \frac{3 * 10^7}{12.76 * 10^6}

\frac{distance}{diameter} = 2.35

<em>so it is 2.35 times more than the diameter of earth</em>

4 0
3 years ago
Describe how impacts between objects in the solar system have helped create notable features in the solar system such as craters
konstantin123 [22]
When 'The big bang' happened lots of large pieces of molten rock was flying around the solar system. As the rocks crashed together they got bigger and as the got bigger they attracted more rocks. Some scientists think that a large piece of molten rock hit the still developing Earth and created the Moon. This impact also caused the Earths angled spin. The Moon got trapped in Earth's orbit and has stayed ever since. Small astroids have hit the Moon causing craters. The Earth doesn't get hit as much because of our thicker atmosphere. Hope this helps!
4 0
3 years ago
A 1-kg collar (located at point (2,2) from the origin) is pulled along a vertical, frictionless bar with a force of 10 N applied
faltersainse [42]

Answer:

The acceleration of the collar is 10 m/s²

Explanation:

Given;

mass of the collar, m = 1 kg

applied force on the bar, F = 10 N

The acceleration of the collar can be calculated by applying Newton's second law of motion;

F = ma

where;

F is the applied force

m is mass of the object

a is the acceleration

a = F / m

a = 10 / 1

a = 10 m/s²

Therefore, the acceleration of the collar is 10 m/s²

3 0
3 years ago
Que fuerza se obtendrá en el émbolo mayor de una prensa hidraúlica cuya área es de 167 cm2, cuando el émbolo menor de área igual
stiv31 [10]

Responder:

<h2>5.368N, </h2>

Explicación:

Según el principio pascal, establece que la presión aplicada en un punto sobre un líquido en un recipiente cerrado es igual a igual a la presión en cualquier otro punto del líquido.

Matemáticamente Presión ejercida por el pistón más pequeño = Presión ejercida por el pistón más grande.

La presión es la relación entre la fuerza y ​​su área de sección transversal.

P = Fuerza / Área de sección transversal

Sea P1 la presión sobre el pistón más pequeño y P2 la presión ejercida por el pistón más grande.

Como P1 = P2 entonces;

F1 / A1 = F2 / A2

Dado F1 = 450N, A1 = 14 cm², A2 = 167 cm² y F2 =?

Sustituyendo el valor conocido en la fórmula para obtener el requerido, tenemos;

\frac{450}{14} = \frac{F2}{167}\\  \\Cross\ multiplying\\\\14*F_2 = 450*167\\\\14F_2 = 75,150\\\\F_2 = \frac{75,150}{14} \\\\F_2 = 5,367.9N

F_2 \approx 5, 368N

Por lo tanto, la fuerza que se obtendrá en el pistón más grande de una prensa hidráulica cuya área es de 167 cm² es aproximadamente 5,368N,

4 0
4 years ago
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