Which best describes internet wikis as a source of scientific information

A huge rotating cloud of particles in space gravitate together to form an increasingly dense ball. As ir shrinks in size, the cl

Trava [24]

Answer:

rotates faster

Explanation:

A huge rotating cloud of particles in space gravitate together to form an increasingly dense ball As it shrinks in size, the cloud rotates faster. Because Angular momentum is conserved, so when it shrinks the moment of inertia decreases, then angular speed must increase. So it rotates fast.

4 0

3 years ago

PLEASEEEEEE HELP

emmainna [20.7K]

Answer:

8.89 m/s² west

Explanation:

Assume east is +x. Given:

v₀ = 120 m/s

v = 0 m/s

t = 13.5 s

Find: a

v = at + v₀

0 m/s = a (13.5 s) + 120 m/s

a = -8.89 m/s²

a = 8.89 m/s² west

8 0

3 years ago

Gravity on earth is 9.8 m/s squared, and gravity on the moon is 1.6 m/s squared. So if the mass of an object on earth is 40 kilo

garik1379 [7]

The mass of an object on Earth is the same as its mass on the Moon. The weight is different.

Weight = m * g

Weight ( Moon ) = 40 kg * 1.6 m/s² = 64 N

If the mass of an object on Earth is 40 kg, its mass on the Moon is 40 kg and its weight on the Moon is 64 N.

7 0

3 years ago

The drag force that resists the motion of a car traveling at 80 km h^- 1 is 300 N.

kobusy [5.1K]

The power require to keep the car traveling is 6,666 W.

The power of the engine at the given efficiency is 3,999.6 W.

<h3>What is Instantaneous power?</h3>

This the product of force and velocity of the given object.

The power require to keep the car traveling is calculated as follows;

P = Fv

$P = 300\ N \ \times \ \frac{80 \ kmh^{-1}}{3.6 \ km h^{-1}/m/s} \\\\
P = 300 \ N \times 22.22 \ m/s\\\\
P = 6,666 \ W$

The power of the engine at the given efficiency is calculated as follows;

$E = \frac{P_{out}}{P _{in}} \times 100\%\\\\
60\% = \frac{P_{out}}{6,666} \times 100\%\\\\
0.6 = \frac{P_{out}}{6,666} \\\\
P_{out} = 3,999.6 \ W$

Learn more about efficiency here: brainly.com/question/15418098

8 0

2 years ago

A -3.00 nc point charge is at the origin, and a second -5.50 nc point charge is on the x-axis at x = 0.800 m. find the electric

Liula [17]

The electric field produced by a single-point charge is given by

$E(r)=k\frac{q}{r^2}$

where

k is the Coulomb's constant

q is the charge

r is the distance from the charge

To find the electric field at x=0.200 m, we need to find the electric field produced by each charge at that point, and then find their resultant.

1) The first charge is $q=-3.00 nC=-3.00 \cdot 10^{-9} C$ , and it is located at x=0, so its distance from the point x=0.200 m is

$r=0.200 m-0=0.2 m$

Therefore, the electric field is

$E_1=(8.99 \cdot 10^9 Nm^2C^{-2})\frac{(3.0 \cdot 10^{-9} C)}{(0.2 m)^2}=675 N/C$

And since the charge is negative, the direction of the field is toward the charge, so toward negative x direction.

2) The second charge is $q=-5.50 nC=-5.5 \cdot 10^{-9}C$ and it is located at x=0.800 m, so its distance from the point is

$r=0.800 m-0.200 m=0.6 m$

Therefore, the electric field is

$E_2 = (8.99 \cdot 10^9 Nm^2C^{-2})\frac{(5.5 \cdot 10^{-9} C)}{(0.6 m)^2}=137.5 N$

And since the charge is negative, the direction of the field is toward the charge, so toward positive x-direction.

3) The total electric field at x=0.200 m will be given by the difference between the two fields (because they are in opposite directions). Taking the x-positive direction as positive direction, we have

$E=E_2 -E_1 =137.5 N/C/C-675 N/C=-537.5 N/C$

and the sign tells us that the field is directed toward negative x-direction.

7 0

3 years ago