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4 years ago

What is the approximate weight of a 20-kg cannonball on Earth? 2 N 20 N 196 N 1,960 N

Physics

2 answers:

kati45 [8]4 years ago

8 0

Answer:

Weight of the cannonball, W = 196 N

Explanation:

Mass of the cannonball, m = 20 kg

We need to find its weight on earth. The weight of an object on the earth can be calculated by the product of mass and acceleration due to gravity as :

$W=m\times g$

$W=20\times 9.8$

W = 196 N

So, the weight of the cannonball on the Earth's surface is 196 N. Hence, this is the required solution.

Margaret [11]4 years ago

5 0

Since weight is the force an object is exerting on another object, and the formula to calculate force is Force = Mass * Acceleration, the answer to your question is 196 N, since the mass of the cannonball times Earth's gravitational pull equals 196 N.

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3 years ago

Why do we have seasons on Earth? A) As Earth goes around the Sun and Earth's axis remains pointed toward Polaris, the Northern a

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Answer:

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Explanation:

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This results in the change in the intensity of the sunlight received by the Earth's surface, due to which the southern and the northern hemispheres experience opposite seasons.

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3 years ago

Marbles rolling down The ramp and horizontally of your desk and assistant land 70 cm from the base of your desk. Your desk and 7

Amiraneli [1.4K]

Answer:

6.39 meters

Explanation:

t=sqrt(2d/g)

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v=d/t...... v=0.70/0.1488....... v=4.7m/s

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3 years ago

What causes sound energy to be produced? * Speed instruments Vibration Fluency

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3 years ago

Three forces in the x-y plane act on a 6.80 kg mass: 10.10 N directed at 19o, 8.60 N directed at 122o, and 9.80 N directed at 21

g100num [7]

Answer: a = 1.58 m/s²

Explanation: The mass of the object is 6.80kg.

From newton's second law of motion,

Resultant force = mass * acceleration

So we need to get the resultant force first.

Note all angles are measured from the positive x axis in the counter clockwise direction..

The first force is 10.10 N at an angle of 19° to the positive x axis.

Horizontal component of this force = 19× cos 19° =17.964 N

Vertical component of this force = 19 × sin 19° = 6.186N

The second force is 8.6 N, 122° ( 22° to the positive y axis), due to the angle, the force is between the positive y axis and negative x axis ( 2nd quadrant)

Horizontal component of this force = - 8.6×sin 22 = - 3.22 N

Vertical component of this force = 8.6 × cos 22 = 7.974 N

The third force is 9.8 N, 218° ( 38 degree to the negative axis), due to the angle, the force is placed in between the negative x and negative y axis.

Horizontal component of this force = - 9.8×cos 38 = - 7.722 N

Vertical component of this force = - 9.8 × sin 38 = - 6.033 N

Sum of forces on the x axis (fx) = 17.964 - 3.22 - 7.722 = 7.022 N

Sum of forces on the y axis (fy) = 6.186 + 7.974 - 6.033 = 8.127 N

Resultant force = √(fx) ²+(fy)²

Resultant force = √7.022² + 8.127²

Resultant force = √115.356613

Resultant force = 10.74 N

Recall that resultant force = mass * acceleration

10.74 = 6.8 × a

a = 10.74/6.8

a = 1.58 m/s²

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4 years ago