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3 years ago

Why are some fibers spun together to form yarns?

Physics

1 answer:

dangina [55]3 years ago

3 0

Answer:

Spun yarns are produced by placing a series of individual fibres or filaments together to form a continuous assembly of overlapping fibres, usually bound together by twist. This operation is required in wool spinning because different parts of the fleece contain fibres of different length and thickness.

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The acceleration due to gravity is higher on juniper Than on earth. The mass and weight of rock on juniper compared to that on e

soldier1979 [14.2K]

Answer:

2.5 times higher then that on the Earth

Explanation:

Gravity is higher on Jupiter then on Earth because Jupiter is much bigger, because of it's mass compared to Earth the gravity on Jupiter is about 2.4 - 2.5 times higher then Earths surface gravity which means a rock on Jupiter would be around "2.4 - 2.5 times as heavier then it would be on Earth."

Hope this helps.

3 0

3 years ago

Two bicycle tires are set rolling with the same initial speed of 4.0 m/s along a long, straight road, and the distance each trav

umka2103 [35]

Answer:

The coefficient of rolling friction will be "0.011".

Explanation:

The given values are:

Initial speed,

$v_i = 4.0 \ m/s$

then,

$v_f=\frac{4.0}{2}$

$=2.0 \ m/s$

Distance,

s = 18.2 m

The acceleration of a bicycle will be:

⇒ $a=\frac{v_f^2-v_i^2}{2s}$

On substituting the given values, we get

⇒ $=\frac{(2.0)^2-(4.0)^2}{2\times 18.2}$

⇒ $=\frac{4-8}{37}$

⇒ $=\frac{-4}{37}$

⇒ $=0.108 \ m/s^2$

As we know,

⇒ $f=ma$

and,

⇒ $\mu_rmg=ma$

⇒ $\mu_r=\frac{a}{g}$

On substituting the values, we get

⇒ $=\frac{0.108}{9.8}$

⇒ $=0.011$

7 0

3 years ago

Zero, a hypothetical planet, has a mass of 5.3 x 1023 kg, a radius of 3.3 x 106 m, and no atmosphere. A 10 kg space probe is to

Andrej [43]

(a) $3.1\cdot 10^7 J$

The total mechanical energy of the space probe must be constant, so we can write:

$E_i = E_f\\K_i + U_i = K_f + U_f$ (1)

where

$K_i$ is the kinetic energy at the surface, when the probe is launched

$U_i$ is the gravitational potential energy at the surface

$K_f$ is the final kinetic energy of the probe

$U_i$ is the final gravitational potential energy

Here we have

$K_i = 5.0 \cdot 10^7 J$

at the surface, $R=3.3\cdot 10^6 m$ (radius of the planet), $M=5.3\cdot 10^{23}kg$ (mass of the planet) and m=10 kg (mass of the probe), so the initial gravitational potential energy is

$U_i=-G\frac{mM}{R}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{3.3\cdot 10^6 m}=-1.07\cdot 10^8 J$

At the final point, the distance of the probe from the centre of Zero is

$r=4.0\cdot 10^6 m$

so the final potential energy is

$U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{4.0\cdot 10^6 m}=-8.8\cdot 10^7 J$

So now we can use eq.(1) to find the final kinetic energy:

$K_f = K_i + U_i - U_f = 5.0\cdot 10^7 J+(-1.07\cdot 10^8 J)-(-8.8\cdot 10^7 J)=3.1\cdot 10^7 J$

(b) $6.3\cdot 10^7 J$

The probe reaches a maximum distance of

$r=8.0\cdot 10^6 m$

which means that at that point, the kinetic energy is zero: (the probe speed has become zero):

$K_f = 0$

At that point, the gravitational potential energy is

$U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{8.0\cdot 10^6 m}=-4.4\cdot 10^7 J$

So now we can use eq.(1) to find the initial kinetic energy:

$K_i = K_f + U_f - U_i = 0+(-4.4\cdot 10^7 J)-(-1.07\cdot 10^8 J)=6.3\cdot 10^7 J$

3 0

3 years ago

A 7.0 kg bowling ball has a moment of inertia of 2.8x10-2 kg m2, and a radius of 0.10 m. If it rolls down the lane at an angular

slega [8]

Hi there!

Angular momentum is equivalent to:

$\large\boxed{L = I\omega}$

L = angular momentum (kgm²/s)

I = moment of inertia (kgm²)

ω = angular velocity (rad/sec)

Plug in the given values for moment of inertia and angular speed:

$L = (0.028)(40) = \boxed{1.12 kgm^2/s}$

8 0

2 years ago

If the box is a distance 1.81 m from the rear of the truck when the truck starts, how much time elapses before the box falls off

alexira [117]

If the box is a distance 1.81 m from the rear of the truck when the truck starts, ... Force of Friction = mu_s * Normal Force( M * G) ... The box starts moving! ... Now that the box is moving, the bed of the truck pulls at it with 17.4 ... out how long it will take the box to reach the back of the truck. ... T^2 = 2 * 1.81 / .64

4 0

4 years ago