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4 years ago

If the force of gravity acting on a box located near earths surface is 12.0 N, What’s it’s mass ?

Physics

1 answer:

Yuri [45]4 years ago

7 0

Force of gravity on an object = (mass of the object) x (acceleration of gravity)

12.0 Newtons = (mass of the object) x (9.8 m/s²)

Mass of the object = (12.0 Newtons) / (9.8 m/s²)

Mass of the object = 1.22 kilograms

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3 years ago

Read 2 more answers

A 0.50-kg ball, attached to the end of a horizontal cord, is rotated in a circle of radius 1.9 m on a frictionless horizontal su

Sedaia [141]

Answer:

$\boxed {\boxed {\sf 18 \ m/s}}$

Explanation:

The ball is moving in a circle, so the force is centripetal.

One formula for calculating centripetal force is:

$F_c= \frac{mv^2}r}$

The mass of the ball is 0.5 kilograms. The radius is 1.9 meters. The centripetal force is 85 Newtons or 85 kg*m/s².

$F_c$ = 85 kg*m/s²
m= 0.5 kg
r= 1.9 m

Substitute the values into the formula.

$85 \ kg*m/s^2 = \frac{0.5 \ kg *v^2}{1.9 \ m}$

Isolate the variable v. First, multiply both sides by 1.9 meters.

$(1.9 \ m)(85 \ kg*m/s^2) = \frac{0.5 \ kg *v^2}{1.9 \ m}*1.9 \ m$

$(1.9 \ m)(85 \ kg*m/s^2) = {0.5 \ kg *v^2}$

$161.5 \ kg*m^2/s^2 = 0.5 \ kg*v^2$

Divide both sides by 0.5 kilograms.

$\frac {161.5 \ kg*m^2/s^2}{0.5 \ kg} = \frac{0.5 \ kg*v^2}{0.5 \ kg}$

$\frac {161.5 \ kg*m^2/s^2}{0.5 \ kg} =v^2$

$323 \ m^2/s^2 = v^2$

Take the square root of both sides of the equation.

$\sqrt {323 \ m^2/s^2} =\sqrt{ v^2$

$\sqrt {323 \ m^2/s^2} =v$

$17.9722007556 \ m/s =v$

The original measurements have 2 significant figures, so our answer must have the same.

For the number we found, 2 sig fig is the ones place. The 9 in the tenth place tells us to round the 7 to an 8.

$18 \ m/s =v$

The maximum speed is approximately <u>18 meters per second.</u>

8 0

3 years ago

A cheerleader lifts his 59.6 kg partner straight up off the ground a distance of 0.749 m before

Airida [17]

m = mass of the partner which the cheerleader lifts = 59.6 kg

h = height to which the partner is lifted by the cheerleader = 0.749 m

g = acceleration due to gravity = 9.8 m/s²

work done by the cheerleader in lifting the partner is same as the potential energy gained by the partner.

W = work done by the cheerleader in lifting the partner

PE = potential energy gained

so W = PE

potential energy is given as

PE = mgh

hence

W = mgh

inserting the values in the above formula

W = 59.6 x 9.8 x 0.749

W = 437.5 J

this is the work done in lifting the partner once.

the cheerleader does this 30 times , hence the total work done is given as

W' = 30 W

W' = 30 x 437.5

W' = 13125 J

5 0

4 years ago

A sled of mass m is given a kick on a frozen pond. The kick imparts to the sled an initial speed of 2.00 m/s . The coefficient o

kobusy [5.1K]

2.04 meters distance is traveled by the sled before stopping.

Mass of the sled = m

The initial speed of the sled = 2 m/s

Coefficient of kinetic friction between sled and ice = 0.100

Let the distance the sled moves before it stops be d.

Gravity = 9.8 m/ s²

Let the initial kinetic energy sled be

$= K _{i}$

$K_{i} = \frac{1}{2} mv ^{2}$

The work done by the frictional force is,

$Work \: done \: by \: frictional \: force =W_{f}$

$W _{f} = μ_{k}mgd$

Work done by frictional force= Initial kinetic energy of the sled

$W_{f} = K_{i}$

$μ_{k}mgd= \frac{1}{2} mv ^{2}$

So, the distance traveled by the sled before stopping is

$d= \frac{1mv ^{2} }{2 \:μ_{k}mg}$

$d= \frac{1v ^{2} }{2 \:μ_{k}g}$

$d= \frac{2^{2} }{2 \times \:0.100 \times 9.8}$

$d= 2.04 \: m$

Therefore, the distance traveled by the sled before stopping is 2.04 meters.

To know more about work done, refer to the below link:

brainly.com/question/13662169

#SPJ4

5 0

2 years ago

21. A positively charged rod will have the strongest attraction for

Ilya [14]

It will have the strongest attraction for a conductor.

8 0

3 years ago