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3 years ago

Which is heavier? A kilogram of steel? or A kilogram of feathers?

Physics

2 answers:

juin [17]3 years ago

7 0

Answer:

steel

Explanation:

Dahasolnce [82]3 years ago

3 0

Answer: below

Explanation: 1kg of steel is slightly heavier than 1 kg of feathers. 1 kg of feather will displace more air as the density of feather is very less comparitively. More the volume displaced more is the upthrust and less the apparent weight.

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A person weighing 785 newtons on the surface of Earth would weigh 298 newtons on the surface of Mars.

rodikova [14]

Answer:

The gravitational field strength on the surface of Mars = 3.72 m/s²

Explanation:

Gravitational Field Strength: This can be defined as the force per unit mass which is exerted at that point. its direction is the force exerted on a mass in a gravitational field. The S.I unit of gravitational field strength is m/s²

Mathematically, Gravitational field is represented as,

g = F/m ..................... Equation 1.

m = F/g ..................... Equation 2.

Where g = gravitational Field Strength, F = force on the mass, m = mass of the body.

From the question,

Note: That The mass of the object is constant both on the surface of the earth and on the surface of Mars.

On the Surface of the earth,

Given: F = 785 N, g = 9.8 m/s²

Substituting this values into equation 2,

m = 785/9.8

m = 80.10 kg.

On the surface of Mars.

Given: m = 80.10 kg, F = 298.

Substituting into equation 2

g = 298/80.1

g = 3.72 m/s²

Thus the gravitational field strength on the surface of Mars = 3.72 m/s²

3 0

3 years ago

A farmer had 15 sheep, and all but 8 died. How many are left?

Troyanec [42]

So he has 7 sheepleft if i did it correctly
15-8=7

3 0

3 years ago

Read 2 more answers

A merry-go-round with a a radius of R = 1.9 m and moment of inertia I = 209 kg-m2 is spinning with an initial angular speed of ω

RideAnS [48]

Answer:

340.67 kgm²/s

Explanation:

R = Radius of merry-go-round = 1.9 m

I = Moment of inertia = 209 kgm²

$\omega_i$ = Initial angular velocity = 1.63 rad/s

m = Mass of person = 73 kg

v = Velocity = 4.8 m/s

Initial angular momentum is given by

$L=I\omega_i\\\Rightarrow L=209\times 1.63\\\Rightarrow L=340.67\ kgm^2/s$

The initial angular momentum of the merry-go-round is 340.67 kgm²/s

8 0

3 years ago

Could someone please help me with this question?

Elenna [48]

1 because the the mid night summer is dark

3 0

3 years ago

When a board with a box on it is slowly tilted to a larger and larger angle, common experience shows that the box will at some p

Savatey [412]

Answer:

c. The coefficient of kinetic friction is less than the coefficient of static friction

Explanation:

When the box finally does break loose. Then the component of the box weight which is parallel to the board weight parallel component, is equal to the $\mu_gn$ .

$w_{box}=\mu_gn$

For the box to acce;erate thee must be non-zero net force acting on the box parallel to the board. Or we can say,

$w_1>f_g\\w_1>\mu_gn$

Therefore the force of kinetic friction must be less than the force of static friction. Thus,

$\mu_g$

3 0

3 years ago