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miv72 [106K]
3 years ago
10

Suppose the average mass of each of 20,000 asteroids in the solar system is 1017 kg. Compare the total mass of these asteroids t

o the mass of Earth. Assuming a spherical shape and a density of 3000 kg/m3, estimate the diameter of an asteroid having this average mass.
Physics
1 answer:
blondinia [14]3 years ago
6 0

Answer:

The mass of the asteroids is 0.000334896182184 times the mass of the Earth.

39929.4542466 m

Explanation:

Total mass of the asteroids

m_a20000\times 10^{17}=2\times 10^{21}\ kg

m_e = Mass of Earth = 5.972\times 10^{24}\ kg

The ratio is

\dfrac{m_a}{m_e}=\dfrac{2\times 10^{21}}{5.972\times 10^{24}}\\\Rightarrow \dfrac{m_a}{m_e}=0.000334896182184

The mass of the asteroids is 0.000334896182184 times the mass of the Earth.

Volume is given by

V=\dfrac{m}{\rho}\\\Rightarrow \dfrac{4\pi}{3\times 8} d^3=\dfrac{m}{\rho}\\\Rightarrow d^3=\dfrac{3\times 8}{4\pi}\dfrac{m}{\rho}\\\Rightarrow d=(\dfrac{3\times 8}{4\pi}\dfrac{m}{\rho})^{\dfrac{1}{3}}\\\Rightarrow d=(\dfrac{3\times 8}{4\pi}\dfrac{10^{17}}{3000})^{\dfrac{1}{3}}\\\Rightarrow d=39929.4542466\ m

The diameter is 39929.4542466 m

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2.
larisa [96]

Answer:

Aceleración, a = 2 m/s²

Explanation:

Dados los siguientes datos;

Velocidad inicial = 108 km/h

Tiempo = 10 segundos

Velocidad final = 36 km/h

To find the average acceleration;

Conversión:

36 km/h to meters per seconds = 36*1000/3600 = 10 m/s

108 km/h to meters per seconds = 108*1000/3600 = 30 m/s

I. Para encontrar la aceleración, usaríamos la primera ecuación de movimiento;

V = U + at

Dónde;

V es la velocidad final.

U es la velocidad inicial.

a es la aceleración.

t es el tiempo medido en segundos.

Sustituyendo en la fórmula, tenemos;

30 = 10 + a*10

30 = 10 + 10a

10a = 30 - 10

10a = 20

Aceleracion = \frac{20}{10}

Aceleración, a = 2 m/s²

3 0
3 years ago
A car starts from rest with an acceleration of 5 ft/s. What is its velocity after it has gone 600 ft?
Soloha48 [4]

Answer:

First, let’s correct the question. Acceleration is the rate of change in velocity. Its unit therefore is ft/sec/sec. If S is the distance traveled for a given duration, S = Vot + (1/2)at^2 where Vo is the initial velocity, a is the acceleration and t is the time. For Vo = 0, a = 6m/sec/sec and t = 3 sec. The distance traveled is S = 0 + (1/2) x 6 x 3^2 = 27 meters

3 0
2 years ago
Ted Williams hits a baseball with an initial velocity of 120 miles per hour (176 ft/s) at an angle of θ = 35 degrees to the hori
lys-0071 [83]

Answer: the maximum heigth of the stadium at ist back wall is 151.32 ft

Explanation:

1. use the position (x) equation in parobolic movement to find the time (t)

565 ft = [frac{176 ft}{1 s\\}[/tex] * cos (35°)  * t

t= 3.92 s

2. use the position (y) equation in parabolic movement to find de maximun heigth  the ball reaches at 565 ft from the home plate.

y= [[frac{176 ft}{1 s\\}[/tex] * sen (35°) * 3.92 s] - \frac{32.2 ft/s^{2} *3.92 s^{2}  }{2}

y= 148.32 ft

3. finally add the 3 ft that exist between the home plate and the ball

148.32 ft + 3 ft = 151.32

6 0
3 years ago
If the temperature is held constant during this process and the final pressure is 683 torrtorr , what is the volume of the bulb
Anna [14]

Answer:

Explanation:

Let the volume of the unknown bulb = X L

The volume of the system , after opening valve = (X + 0.72 L )

Use Boyles law gas equation,

P1V1 = P2V2 ( at temperature is constant )

Given:

P1 = 1.2 atm

P2 = 683 torr

Converting mmHg to atm,

1 atm = 760 mmHg(torr)

683 mmHg = 683/760

= 0.8987 atm

1.2X = 0.8987*(X + 0.720)

1.2X = 0.8987X + 0.6471

0.3013X = 0.6471

X = 2.15 L

5 0
3 years ago
A crowbar 2727 in. long is pivoted 66 in. from the end. What force must be applied at the long end in order to lift a 600600 lb
NikAS [45]

To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.

\sum \tau = F*d

If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,

F*(27-6)= 6*600

F = \frac{6*600}{21}

F= 171.42 lb

So, Force that must be applied at the long end in order to lift a 600lb object to the short end is 171.42lb

4 0
3 years ago
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