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Salsk061 [2.6K]
3 years ago
8

The acceleration due to gravity on the moon is 1. 6 m/s2, about a sixth that of Earth’s. Which accurately describes the weight o

f an object on the moon? An object on the moon is 1. 6 times lighter than on Earth. An object on the moon is 1. 6 times heavier than on Earth. An object on the moon is six times lighter than on Earth. An object on the moon is six times heavier than on Earth.
Physics
1 answer:
snow_tiger [21]3 years ago
4 0

Weight is the force applied on the planet by its mass. The weight of an object is 6 times lighter than that of the earth. Option C is correct.

What is Wight?

  • It is the product of the mass and the gravitational acceleration.
  • Weight is the force applied on the planet by its mass.

The formula of weight is,

w = mg

Where,

w- weight

m - mass

g- gravitational acceleration

In the question given here the gravitational acceleration of the moon is 6 times less than that of the earth.

Therefore, the weight of an object is 6 times lighter than that of the earth.

Learn more about gravitational acceleration:

brainly.com/question/10888182

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A 4.0×1010kg asteroid is heading directly toward the center of the earth at a steady 16 km/s. To save the planet, astronauts str
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Answer:

a. t = 69.4 hr = 2.89 days

b. theta = 91.67*10^-3 degrees

c.  deflection_angle = 0.134 degrees

Explanation:

a).

The asteroid impacts the earth in t

t = x/v = (4.0*10^6 km)/(16 km/sec)

t = 2.5 * 10^5 sec

t = 69.4 hr = 2.89 days

b).

tan(theta) = 6400 km/(4.0*10^6 km)

tan(theta) = 1.6*10^-3

theta = arctan(1.6*10^-3)

theta = 1.6*10^-3 radians  (for small angles, tan(theta) ~= theta)

theta = 91.67*10^-3 degrees

c).

v_minimum = 6400 km/(2.5 * 10^5 sec)

v_minimum = 25.6 m/s

Using F = m*a, we can calculate the acceleration of the asteroid due to the rocket's thrust:

5.0*10^9 N = 4.0*10^10 kg * a

a = (5.0*10^9 N)/(4.0*10^10 kg)  

a = 0.125 m/s^2

The transverse velocity after 300 seconds of this acceleration is:

v_transverse = a*t = 0.125 m/s^2 * 300 s

v_transverse = 37.5 m/s = 37.5*10^-3 km/s

tan(deflection_angle) = v_transverse/(20 km/s)

tan(deflection_angle) = (37.5*10^-3 km/s)/(16 km/s) = 2.34^-3

deflection_angle = arctan(2.34*10^-3)  

deflection_angle = 2.34*10^-3 radians = 0.134 degrees

v_transverse/(16 km/s) > (6400km)/(5.0*10^6 km)  

(note that the right hand side if this inequality is tan(theta) calculated above)

v_transverse > 23.704 m/

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3 years ago
a plane inclined at an angle of 30° to the horizontal has an efficiency of 50%. the force parallel to the plane required to push
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Answer:

Correct Answer: Option B

Explanation

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A cannon is mounted on a tower above a wide, level field. The barrel of the cannon is 20 m above the ground below. A cannonball
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<u>Answer:</u>

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  So it will take 2.02 seconds to reach ground.

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