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kherson [118]
3 years ago
14

Define an element and give 5 examples of elements that are important to life .

Physics
1 answer:
fenix001 [56]3 years ago
8 0

Answer:

A pure substance consisting only of atoms with the same number of protons in their nuclei-these appear on the periodic table

Oxygen

Hydrogen

Carbon

Sulfur

Phosphate

Nitrogen

Magnesium

Calcium

Potassium

Chlorine

(I know that these are more examples than needed, but you can use any)

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A 2.0-kg ball rolls to the right at 3.0 m/s. A 4.0-kg ball rolls to the left at 2.0 m/s . What is the momentum of the system aft
charle [14.2K]

Answer:

Final momentum after a head on collision is -2kgm/s

Explanation:

         One ball moves to the right and the other moves opposite  and momentum is a vector quantity so that considering the direction

Initial momenta are        P₁=2x3=6kgm/s        P₂=4x(-2)=-8kgm/s      

Final momentum is the vector sum of P(final)= 6-8= -2 kgm/s

4 0
3 years ago
A uniform solid disk with a mass of 24.3 kg and a radius of 0.364 m is free to rotate about a frictionless axle. Forces of 90.0
a_sh-v [17]

Answer:

a. -12.7 Nm

b. -7.9 rad/s^2

Explanation:

I have attached an illustration of a solid disk with the respective forces applied, as stated in this question.

Forces applied to the solid disk include:

F_1 = 90.0N\\F_2 = 125N

Other parameters given include:

Mass of solid disk, M = 24.3kg

and radius of solid disk, r = 0.364m

a.) The formula for determining torque (T), is T = r * F

Hence the net torque produced by the two forces is given as a summation of both forces:

T = T_{125} + T_{90}\\= -r(125)sin90 + r(90)sin90\\= 0.364(-125 + 90)\\= -12.7 Nm

b.)  The angular acceleration of the disk can be found thus:

using the formula for the Moment of Inertia of a solid disk;

I_{disk} = {\frac{1}{2}}Mr^2

where M = Mass of solid disk

and r = radius of solid disk

We then relate the torque and angular acceleration (\alpha) with the formula:

T = I\alpha \\-12.7 = ({\frac{1}{2}}Mr^2)\alpha \\\alpha  = -{\frac{12.7}{1.61}} = -7.9 rad/s^2

4 0
4 years ago
The weights in atwoods machine, starting at rest, attain a velocity of 2ft/sec in one sec. Find the ratio of the masses
Orlov [11]
Refer to the figure shown below.
Let m₁ and m₂ e the two masses.
Let a = the acceleration.
Let T =  tension over the frictionless pulley.

Write the equations of motion.
m₂g - T = m₂a            (1)
T - m₁g = m₁a            (2)

Add equations (1) and (2).
m₂g - T + T - m₁g = (m₁ + m₂)a
(m₂ - m₁)g = (m₁ + m₂)a

Divide through by m₁.
(m₂/m₁ - 1)g = (1 + m₂/m₁)a

Define r = m₂/m₁ as the ratio of the two masses. Then
(r - 1)g = (1 +r)a
r(g-a) = a + g
r = (g - a)/(g + a)

With  = 2 ft/s from rest, the acceleration is
a = 2/32.2 = 0.062 ft/s²
Therefore
r = (32.2 - 0.062)/(32.2 + 0.062) = 0.9962

Answer:
The ratio of masses is 0.9962 (heavier mass divided by the lighter mass).

8 0
3 years ago
How is force related to math
dybincka [34]

Answer:

Newton's second law of motion describes the relationship between force and acceleration. They are directly proportional. If you increase the force applied to an object, the acceleration of that object increases by the same factor. In short, force equals mass times acceleration.

Explanation:

8 0
3 years ago
Read 2 more answers
620 J of heat is added to the cylinder of an engine, which causes the gas inside to expand. As a result, the piston of the engin
Maru [420]

Answer:

<u>400</u> J work is done BY the engine.

The internal energy of the gas is <u>620</u> J

Explanation:

The given information are;

The heat added to the cylinder = 620 J

The force applied by the piston of the engine = 8.0 kN = 8,000 N

The distance over which the force moves (the piston) = 5.0 cm = 0.05 m

The work done (by the engine) = Force × Distance = 8,000 N × 0.05 m = 400 J

The internal energy is the sum of the kinetic and potential energy of the system

Therefore, given that the internal energy, U, is the sum total of the energy in the system

∴ U = The heat supplied to the system = 620 J

Which gives;

<u>400</u> J work is done BY the engine.

The internal energy of the gas is <u>620</u> J.

6 0
3 years ago
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