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2 years ago

What is the mass of an object that weighs 686N on Earth?

Physics

1 answer:

Oxana [17]2 years ago

8 0

Answer:

70 kg is the mass of the object

Explanation:

This question can be solved with this simple formula:

Weight force = mass . gravity

686 N = mass . 9.8 m/s²

686 N / 9.8 m/s² = mass → 70 kg

Note → 1N = 1 kg . m / s²

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Platinum has a density of 21.4 g/cm3. if thieves were to steal platinum from a bank using a small truck with a maximum payload o

slega [8]

We will convert the 1dm3 in terms of cm3 as follows:

1dm^3 = (10 cm)^3
= 1000 cm^3

The mass of platinum is equal to 900 lb.
Then we will convert the mass in terms of grams as follows:
1 lb = 453.6 g
900 = 900 x 453.6 g
= 408240 g

Then density of platinum is equal to 21.4 g/cm^3
We will calculate the volume of platinum in mass 408240 g as follows:
Volume of platinum = mass of platinum / density of platinum
= 408240 g / 21.4 g/cm^3
= 19076.6 cm^3

The total volume of platinum is 19076.6 cm^3
The volume of platinum in 1 L bar is 1000cm^3
So, to calculate the number of bars we will use the formula as follows;
Number of bars = volume of platinum available / volume of platinum required in 1 L bar
= 19076.6 cm^3 / 1000 cm^3
= 19
So, the number of bars are 19.

4 0

2 years ago

An object of mass 24kg is accelerated up a frictionless place incline at an angle of 37° with horizontal by a constant force, st

RoseWind [281]

a) Average power: 1425 W

b) Instantaneous power at 3.0 sec: 2850 W

Explanation:

The motion of the object along the ramp is a uniformly accelerated motion (because the force applied is constant), so we can use the suvat equation

$s=ut+\frac{1}{2}at^2$

where

s = 18 m is the displacement along the ramp

u = 0 is the initial velocity

t = 3.0 s is the time taken

a is the acceleration of the object along the ramp

Solving for a,

$a=\frac{2s}{t^2}=\frac{2(18)}{(3.0)^2}=4 m/s^2$

Now we can apply Newton's second law to find the net force on the object:

$F=ma=(24 kg)(4 m/s^2)=96 N$

This net force is the resultant of the applied force forward ( $F_a$ ) and the component of the weight acting backward ( $mg sin \theta$ ), so we can find what is the applied force:

$F=F_a - mg sin \theta\\F_a = F+mg sin \theta = 96+(24)(9.8)(sin 37^{\circ})=237.5 N$

where

m = 24 kg is the mass of the object

$g=9.8 m/s^2$ is the acceleration of gravity

Now we can finally find what is the work done by the applied force, which is parallel to the ramp, therefore:

$W=F_a s = (237.6)(18)=4276 J$

where s = 18 m is the displacement.

Therefore the average power needed is:

$P=\frac{W}{t}=\frac{4276}{3}=1425 W$

The instantaneous power at any point of the motion is given by

$P=F_av$

where

$F_a$ is the force applied

v is the velocity of the object

We already calculated the applied force:

$F_a=237.5 N$

While since this is a uniformly accelerated motion, we can find the velocity at the end of the 3.0 seconds using the suvat equation:

$v=u+at=0+(4)(3.0)=12.0 m/s$

And therefore, the instantaeous power at 3.0 sec is:

$P=Fv=(237.5)(12)=2850 W$

Learn more about power:

brainly.com/question/7956557

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8 0

2 years ago

A 0.20-kg object attached to the end of a string swings in a vertical circle (radius = 80 cm). at the top of the circle the spee

gayaneshka [121]

Answer:

Tension in the string at this position: 3.1 N.

Explanation:

Convert the radius of the circle to meters:

$r = 80\;\text{cm} = 0.80\;\text{m}$ .

What's the net force on the object?

The object is in a circular motion. As a result,

$\displaystyle \Sigma F = \frac{m\cdot v^{2}}{r}$ ,

where

$\Sigma F$ is the net force on the object,
$m$ is the mass of the object,
$v$ is the velocity of the object, and
$r$ is the radius of the circular motion.

For this object,

$\displaystyle \Sigma F = \frac{0.20\times {4.5}^{2}}{0.80} = 5.0625\;\text{N}$ .

The output unit of net force should be standard if the unit for mass, velocity, and radius are all standard. The net force shall always point towards the center. In this case the net force points downwards.

What are the forces on this object?

There are two forces on the object at this moment:

Weight, $W$ , which points downwards. $W = m\cdot g = 0.20\times 9.81 = 1.962\;\text{N}$ .
Tension, $T$ , which also points downwards. The size of the tension force needs to be found.