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2 years ago

The acceleration due to gravity on Earth is 9. 8 m/s2. What is the weight of a 75 kg person on Earth? 9. 8 N 75 N 85 N 735 N.

1 answer:

3 0

The weight of a 75 kg person on Earth, which has an acceleration due to gravity of 9.8m/s², is 735N.

CALCULATE WEIGHT:

Weight is a force, hence, measured in Newton. It is a function of an object's mass in relation to gravity.

Weight of an object can be calculated using the following formula:

W = mg

Where;

W = weight (N)
m = mass (kg)
g = acceleration due to gravity (m/s²)

The weight of the person in this question can be calculated as follows:

W = 75 × 9.8

W = 735N

Therefore, the weight of a 75 kg person on Earth, which has an acceleration due to gravity of 9.8m/s², is 735N.

Learn more about weight at: brainly.com/question/18554478

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What is the approximate terminal velocity of a sky diver before the parachute opens

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Answer:

The approximate terminal velocity of a sky diver before the parachute opens is 320 km/h.

Explanation:

The terminal velocity is the maximum magnitude of velocity that is attained by the diver when he or she falls in the air.
The terminal velocity of the person diving in air before opening parachute is 320 km/h that means the velocity when the person is experiencing free fall is 320 km/h.
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3 years ago

Through which one of the following mediums is the velocity of a sound wave the greatest?

tino4ka555 [31]

Answer: C. Steel

Explanation: When a sound wave travels through a solid body consisting

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3 years ago

Read 2 more answers

A 20.0 cm tall object is placed 50.0 cm in front of a convex mirror with a radius of curvature of 34.0 cm. Where will the image

Neko [114]

Answer:

1.Theimage will be located at -0.13m or -13 cm

2.The height of the image will be 0.052m or 5.2cm

Explanation:

Given that;

Height of object, h=20 cm = 0.2m

Object distance in front of convex mirror, o,= 50 cm =0.5m

Radius of curvature, r, =34 cm =0.34m

Let;

Image distance, i,=?

Image height, h'=?

You know that focal length,f, is half the radius of curvature,hence

f=r/2 = 0.34/2 = 0.17m ( this length is inside the mirror, in a virtual side, thus its is negative)

f= -0.17m

Apply the relationship that involves the focal length;

$=\frac{1}{o} +\frac{1}{i} =\frac{1}{f}$

$=\frac{1}{0.5} +\frac{1}{i} =-\frac{1}{0.17}$

Re-arrange to get i

$\frac{1}{i} =-2-5.88\\\\\\\frac{1}{i} =-7.88\\\\i=-0.13m$

This is a virtual image formed at a negative distance produced through extension of drawing rays behind the mirror if you use rays to locate the image behind the mirror

Apply the magnification formula

magnification, m=height of image÷height of object

$m=\frac{h'}{h} =-\frac{i}{o}$