An object with a mass of 4.0 kg has a force of 12.0 newtons applied to it. What is the resultant acceleration of the object? Help me?? :(

3m/s²

Explanation:

Force applied to an object can be calculated thus;

F = ma

Where;

F = force applied (Newtons)

m = mass of substance (kg)

a = acceleration (m/s²)

According to the information provided in this question, F = 12.0 newtons, m = 4.0 kg, a = ?

Derived from F = m.a

We have; a = F/m

a = 12/4

a = 3

The resultant acceleration of the object is 3m/s².

Problem 1-

The acceleration = 5 m/s²

Problem 2 -

The acceleration = 3. m/s²

Problem 3 -

The acceleration = 2.5 m/s²

Problem 4-

The force applied = 7.5 N

Problem 5-

The mass of the object,  = 0.5 kg.

Explanation:

Problem 1-

The mass of the ball, m = 0.5 kg

The force (applied) with which it was hit, F = 2.5 N

The acceleration, a = Force/Mass = 2.5 N/(0.5 kg) = 5 m/s²

The acceleration = 5 m/s²

Problem 2 -

The mass of the box, m = 15 kg

The force (applied) with which it was hit, F = 50 N

The acceleration, a = Force/Mass = 50 N/(15 kg) = 10/3 = 3. m/s²

The acceleration = 3. m/s²

Problem 3 -

The mass of the object, m = 2.0 kg

The force (applied) with which it was hit, F = 5.0 N

The acceleration, a = Force/Mass = 5.0 N/(2.0 kg) = 2.5 m/s²

The acceleration = 2.5 m/s²

Problem 4-

The mass of the object, m = 3.0 kg

The acceleration, a, of the object = 2.5 m/s²

The force applied to it, F = Mass × Acceleration = 3.0 kg × 2.5 m/s² = 7.5 N

The force applied = 7.5 N

Problem 5-

The acceleration, a, of the object = 12 m/s²

The force applied, F = 6.0 N

The mass of the object, m = Force/Acceleration = 6.0 N/(12 m/s²) = 0.5 kg

The mass of the object,  = 0.5 kg.

Problem 1-

The acceleration = 5 m/s²

Problem 2 -

The acceleration = 3. m/s²

Problem 3 -

The acceleration = 2.5 m/s²

Problem 4-

The force applied = 7.5 N

Problem 5-

The mass of the object,  = 0.5 kg.

Explanation:

Problem 1-

The mass of the ball, m = 0.5 kg

The force (applied) with which it was hit, F = 2.5 N

The acceleration, a = Force/Mass = 2.5 N/(0.5 kg) = 5 m/s²

The acceleration = 5 m/s²

Problem 2 -

The mass of the box, m = 15 kg

The force (applied) with which it was hit, F = 50 N

The acceleration, a = Force/Mass = 50 N/(15 kg) = 10/3 = 3. m/s²

The acceleration = 3. m/s²

Problem 3 -

The mass of the object, m = 2.0 kg

The force (applied) with which it was hit, F = 5.0 N

The acceleration, a = Force/Mass = 5.0 N/(2.0 kg) = 2.5 m/s²

The acceleration = 2.5 m/s²

Problem 4-

The mass of the object, m = 3.0 kg

The acceleration, a, of the object = 2.5 m/s²

The force applied to it, F = Mass × Acceleration = 3.0 kg × 2.5 m/s² = 7.5 N

The force applied = 7.5 N

Problem 5-

The acceleration, a, of the object = 12 m/s²

The force applied, F = 6.0 N

The mass of the object, m = Force/Acceleration = 6.0 N/(12 m/s²) = 0.5 kg

The mass of the object,  = 0.5 kg.

Answer:

Step-by-step explanation:

The input force is 50 N. But it will not create not any change. No mechanical advantage is observed.

Answer:

Step-by-step explanation:

It is clear by the equation 2(27+3×35.5)= 267 gm of AlCl3 reacts with 6× 80 = 480 gm of Br2 . So 29.2 gm reacts = 480× 29.2/267= 52.6 gm

Calcium (Ca)(On the periodic table, ionization energy increases as you go up and to the right of the periodic table)

glycoproteins

Explanation:

A positive reaction for Molisch's test is given by almost all carbohydrates (exceptions include tetroses & trioses). It can be noted that even some glycoproteins and nucleic acids give positive results for this test (since they tend to undergo hydrolysis when exposed to strong mineral acids and form monosaccharides).

Answer:

Taking into accoun the ideal gas law, The volume of a container that contains 24.0 grams of N2 gas at 328K and 0.884 atm is 26.07 L.

An ideal gas is a theoretical gas that is considered to be composed of point particles that move randomly and do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:  

P×V = n×R×T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas. The universal constant of ideal gases R has the same value for all gaseous substances.

Explanation:

In this case, you know:

P= 0.884 atm

V= ?

n= 0.857 moles (where 28 g/mole is the molar mass of N₂, that is, the amount of mass that the substance contains in one mole.)

R=0.082

T= 328 K

Replacing in the ideal gas law:

0.884 atm×V= 0.857 moles× 0.082 ×328 K

Solving:

V= 26.07 L

The volume of a container that contains 24.0 grams of N2 gas at 328K and 0.884 atm is 26.07 L.