The amount of matter contained in a given volume for a substance is called

football player A has a mass of 110 kg is running on the field is velocity of 2 m/s football player B has a mass of 120 kg and a

valkas [14]

Complete Question:

Football player A has a mass of 110 kg, and he is running down the field with a velocity of 2 m/s. Football player B has a mass of 120 kg and is stationary. What is the total momentum after the collision?

Answer:

Total momentum = 220 Kgm/s.

Explanation:

Given the following data;

For footballer A

Mass, M1 = 110kg

Velocity, V1 = 2m/s

For footballer B

Mass, M1 = 120kg

Velocity, V1 = 0m/s since he's stationary.

To find the total momentum;

Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.

Mathematically, momentum is given by the formula;

$Momentum = mass * velocity$

a. To find the momentum of A;

$Momentum \; A = 110 * 2$

Momentum A = 220 Kgm/s.

b. To find the momentum of B;

$Momentum \; B = 120 * 0$

Momentum B = 0 Kgm/s.

c. To find the total momentum of the two persons;

$Total \; momentum = Momentum \; A + Momentum \; B$

Substituting into the equation, we have;

$Total \; momentum = 220 + 0$

Total momentum = 220 Kgm/s. 

7 0

3 years ago

In the combo circuit diagrammed, R1 = 19.2 Ω, R2 = 20.7 Ω, and R3 = 25.8 Ω. Find the equivalent resistance of the circuit.

garik1379 [7]

Answer:

Equivalent resistance: 13.589 Ω

Explanation:

R series = R1 + R2 + R3 ...

$\frac{1}{R_{eq} } = \frac{1}{R1} +\frac{1}{R2} +\frac{1}{R3} ...$

Find the equivalent resistance of the right branch of the circuit:

$R_{eq} = R_{2} +R_{3} \\R_{eq} = 20.7 + 25.8 = 46.5 ohms$

$\frac{1}{R_{eq} } = \frac{1}{19.2} +\frac{1}{46.5}\\\\\frac{1}{R_{eq} } = 0.0735887097\\\\R_{eq} = 13.5890411$

6 0

2 years ago

A what is a negatively charged Atoms

mr Goodwill [35]

Answer:

Negatively charged or positively charged atom is generally termed as ANION/CATION. Short Explanation: If an atom loses electrons or gains protons, it will have a net positive charge and is called a Cation. If an atom gains electrons or loses protons, it will have a net negative charge and is called an Anion.

Explanation:

7 0

2 years ago

How far from the castle wall does the launched rock hit the ground?

lina2011 [118]

King Arthur's knights use a catapult to launch a rock from their vantage point on top of the castle wall, 14 m above the moat. The rock is launched at a speed of 27 m/s and an angle of 32degrees above the horizontal.

4 0

2 years ago

Read 2 more answers

Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.

SIZIF [17.4K]

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

(b) m1 = 915kg;

m2 = 605kg;

m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

$F_{pull}=m_{T}.a$

$m_{T}=\frac{F_{pull}}{a}$

$m_{T}=\frac{3615}{1.516}$

$m_{T}=2384.5$

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

For $m_{1}$ :

The only force acting On the $m_{1}$ box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

$m_{1} = \frac{F_{12}}{a}$

$m_{1} = \frac{1387}{1.516}$

$m_{1}$ = 915kg

For $m_{2}$ :

There are two forces acting on $m_{2}$ : tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

$m_{2} = \frac{F_{23}-F_{12}}{a}$

$m_{2} = \frac{2304-1387}{1.516}$

$m_{2}$ = 605kg

For $m_{3}$ :

$m_{3} = m_{T} - (m_{1}+m_{2})$

$m_{3} = 2384.5-1520.0$

$m_{3}$ = 864.5kg

8 0

3 years ago