All categories

3 years ago

What is the difference between mass and weight? In at least three sentences, explain the difference and give an example.

2 answers:

3 0

Mass and weight are different. Mass is the measurement of how much matter an object has, while weight deals with the pull of gravity. For example, an astronaut floating in space is weightless, due to the limited influence of gravity. But he will still have mass, that is because mass and weight are different :)

Svetlanka [38]3 years ago

3 0

Mass is entirely a property of the object. So the conditions and environment surrounding the object have no effect on it, and it doesn't change.

Weight is a measurement of the interaction between the object and its environment, so its weight changes as it moves from place to place.

Example:

A standard brass cylinder in a laboratory set of calibrated masses is marked "1 kilogram".

When you weigh it on Earth, the scale says "9.8 Newtons" or "2.2 pounds".

If you take the set of calibrated masses to the Moon with you, and weigh the same "1 kilogram" piece, it weighs "1.62 Newtons" or "5.8 ounces" up there.

You might be interested in

Consider a composite cube made of epoxy with fibers aligned along one axis of the cube (the fibers are parallel to four of the t

coldgirl [10]

The question is incomplete. The complete question is :

Consider a composite cube made of epoxy with fibers aligned along one axis of the cube (the fibers are parallel to four of the twelve cube edges). If the cube can only be loaded in axial tension such that the force is uniformly applied over - and is normal to - a cube face, what is the lowest possible positive length change the cube can experience under this tension? The applied tensile force is 102 KN. The unloaded cube edge length is 56 mm. The glass fibers have an elastic modulus of 200 GPa. The epoxy has an elastic modulus of 38 GPa. The cube is comprised of 18 vol% epoxy (the balancing vol % is glass fiber). Hint: The loading axis is intentionally unspecified. Answer Format: Lowest possible length increase (change of length) under tension.

Solution :

Given :

$E_{glass fibre}$ = 200 GPa

$V_{glass fibre} = 82\%$

$E_{epoxy}$ = 38 GPa

$V_{epoxy} = 82\%$

Edge length = 56 mm

Cube is loaded in axial tension such that the force is uniformly applied over a cube face.

$E_{\text{composite}}=\frac{E_{glass fibre} \times E_{epoxy}}{(E_{glass fibre .E_{epoxy}})+(E_{fibre}.V_{glass fibre})}$

$E_{composite} = \frac{200 \times 38}{(200 \times 0.18)+(38\times 0.82)}$

$=113.16$ GPa

Applied stress $=\frac{\text{applied load}}{\text{area}}$

$\sigma=\frac{102 \times 10^3 \ N}{56 \times 56 \times 10^{-6} \ m^2}$

= 32.5 MPa

By Hooke's law

$\sigma = E . \epsilon$

$\sigma = E. \frac{\Delta l}{l}$

$\Delta l = \frac{\sigma}{E}\times l$

Length change, $\Delta l =\frac{32.5 \times 10^6 \ Pa}{113.16 \times 10^9 \ Pa}\times 56 \times 10^{-2} \ m$

$\Delta l = \frac{32.5 \times 56}{113.16} \times 10^{-3} \ mm$

= 0.016 mm

7 0

3 years ago

Help me please I need to turn it in before midnight...

ale4655 [162]

Answer:

The statement which explains how the total time spent in the air is affected as the projectile's angle of launch increases from 25 degrees to 50 degrees is;

C. Increasing the angle from 25° to 50° will increase the total time spent in the air

Explanation:

The equation that can be used to find the total time, T, spent in the air of a projectile is given as follows;

$T = \dfrac{2 \cdot u \cdot sin(\theta) }{g}$

Where;

T = The time of flight of the projectile = The time spent in the air

u = The initial velocity of the projectile

θ = The angle of launch of the projectile

g = The acceleration due to gravity ≈ 9.81 m/s²

Given that sin(50°) > sin(25°), when the angle of launch, θ, is increased from 25 degrees to 50 degrees, we have;

Let T₁ represent the time spent in the air when the angle of launch is 25°, and let T₂ represent the time spent in the air when the angle of launch is 50°, we have;

$T_1 = \dfrac{2 \cdot u \cdot sin(25^{\circ}) }{g}$

$T_2 = \dfrac{2 \cdot u \cdot sin(50^{\circ}) }{g}$

sin(50°) > sin(25°), therefore, we have;

$\dfrac{2 \cdot u \cdot sin(50^{\circ}) }{g} > \dfrac{2 \cdot u \cdot sin(25^{\circ}) }{g}$

Therefore;

T₂ > T₁

Therefore, increasing the angle at which the projectile is launched from 25° to 50° will increase the total time spent in the air.

7 0

3 years ago

a hockey player exerts a force of 5 N over a distance of 0.3 m on a hockey puck. How much work is done on the hockey puck

Katen [24]

Answer: W = 1.5 J

Explanation: Work is the product of force and distance. It can be expressed in the following formula W = Fd.

W = Fd

= 5 N ( 0.3 M )

= 1.5 J

7 0

3 years ago

A crane lifts a 100 kg concrete block to a vertical height of 25 m. Calculate

Nana76 [90]

Answer:

Solon,

total mass (kg)= 100kg

height (h)= 25m

acceleration due to gravity = 9.8m/s²

so,

work done =m*g*h

= 100*9.8*25

= 24,500 joule

5 0

3 years ago

A circular loop with 30 turns and area of 0.25 m2 rotates at 100 rad/s. The axis of rotation of the loop is perpendicular to the

zysi [14]

Answer:

e = 30 V

Explanation:

given,

N = 30 turns

Area = 0.25 m²

angular speed = ω = 100 rad/s

Magnetic field = 0.04 T

maximum induced emf in the loop = ?

e = N B A ω

e = 30 x 0.04 x 0.25 x 100

e = 30 V

hence, the maximum emf induced in the loop is equal to e = 30 V

5 0

3 years ago