Darw a velocity time graph for body moving with uniform acceleration

Three masses are located in the x-y plane as follows: a mass of 6 kg is located at (0 m, 0 m), a mass of 4 kg is located at (3 m

Natali [406]

<h2>Answer:</h2>

D. (1m, 0.5m)

<h2>Explanation:</h2>

The center of mass (or center of gravity) of a system of particles is the point where the weight acts when the individual particles are replaced by a single particle of equivalent mass. For the three masses, the coordinates of the center of mass C(x, y) is given by;

x = (m₁x₁ + m₂x₂ + m₃x₃) / M ----------------(i)

y = (m₁y₁ + m₂y₂ + m₃y₃) / M ----------------(ii)

Where;

M = sum of the masses

m₁ and x₁ = mass and position of first mass in the x direction.

m₂ and x₂ = mass and position of second mass in the x direction.

m₃ and x₃ = mass and position of third mass in the x direction.

y₁ , y₂ and y₃ = positions of the first, second and third masses respectively in the y direction.

From the question;

m₁ = 6kg

m₂ = 4kg

m₃ = 2kg

x₁ = 0m

x₂ = 3m

x₃ = 0m

y₁ = 0m

y₂ = 0m

y₃ = 3m

M = m₁ + m₂ + m₃ = 6 + 4 + 2 = 12kg

Substitute these values into equations (i) and (ii) as follows;

x = ((6x0) + (4x3) + (2x0)) / 12

x = 12 / 12

x = 1 m

y = (6x0) + (4x0) + (2x3)) / 12

y = 6 / 12

y = 0.5m

Therefore, the center of mass of the system is at (1m, 0.5m)

7 0

3 years ago

Spring compressed 10cm by 100N force and held in place with Pin. Pin is pulled and block is pushed Up the incline. Uk(coefficien

otez555 [7]

The compression of 10 cm by a 100 N force on the plane that has a

coefficient of friction of 0.39 give the following values.

The velocity of the block after the Spring extends 7 cm is approximately 1.73 m/s

The height at which the block stops rising is approximately 1.1415 m

The length of the incline is approximately 1.536 m

<h3>How can the velocity and height of the block be calculated?</h3>

Mass of the block, m = 3 kg

$Spring \ constant, K = \dfrac{100 \, N}{0.1 \, m} = \mathbf{ 1000\, N/m}$

Coefficient of kinetic friction, $\mu_k$ = 0.39

Therefore, we have;

Friction force = $\mathbf{\mu_k}$ ·m·g·cos(θ)

Which gives;

Friction force = 0.39 × 3 × 9.81 × cos(48°) ≈ 7.68

Work done by the motion of the block, <em>W</em> ≈ 7.68 × d

The work done = The kinetic energy of the block, which gives;

$\mathbf{\dfrac{1}{2} \times k \cdot x^2 }= 7.68 \cdot d$

The initial kinetic energy in the spring is found as follows;

K.E. = 0.5 × 1000 N/m × (0.1 m)² = 5 J

The initial velocity of the block is therefore;

5 = 0.5·m·v²

v₁ = √(2 × 5 ÷ 3) ≈ 1.83

Work done by the motion of the block, <em>W</em> ≈ 7.68 N × 0.07 m ≈ 0.5376 J

Chane in kinetic energy, ΔK.E. = Work done

ΔK.E. = 0.5 × 3 × (v₁² - v₂²)

Which gives;

ΔK.E. = 0.5 × 3 × (1.83² - v₂²) = 0.5376

Which gives;

The velocity of the block after the Spring extends 7 cm, v₂ ≈ <u>1.73 m/s</u>

The height at which the block will stop moving, <em>h</em>, is given as follows;

$At \ the \ maximum \ height, \ h, \ we \ have ; \ \dfrac{1}{2} \times 1000 \times 0.1^2 = 7.68 \times x$

Which gives;

$Length \ of \ the \ incline \ at \ maximum \ height, \ x_{max} =\dfrac{ 7.68 }{ \dfrac{1}{2} \times 1000 \times 0.1^2 } \approx 1.536$

The distance up the inclined, the block rises, at maximum height is therefore;

$x_{max}$ ≈ 1.536 m

Therefore;

h = 1.536 × sin(48°) ≈ 1.1415

The height at which the block stops rising, h ≈ <u>1.1415 m</u>

From the above solution for the height, the length of the incline is he

distance along the incline at maximum height which is therefore;

Length of the incline, $x_{max}$ = 1.536 m

Learn more about conservation of energy here:

brainly.com/question/7538238

5 0

2 years ago

A 1.2 x 103 kg racecar, with a velocity of 8 m/s, collides with an unsuspecting 80 kg honey badger who is standing

aalyn [17]

Answer: 90 m/s

Explanation:

Given

mass of racecar $M=1.2\times10^3\ kg$

velocity of racecar $u=8\ m/s$

mass of still honeybadger $m=80\ kg$

after collision race car is traveling at a speed of $v_1=2\ m/s$

conserving linear momentum

$Mu+m\times0=Mv_1+ mv_2\quad[v_2=\text{velocity of honeybadger after colllision}]$

$1.2\times10^3\times8+0=1.2\times10^3\times2+80\times v_2$

$1.2\times10^3(8-2)=80v_2\\v_2=\frac{7.2\times10^3}{80}=90\ m/s$

4 0

3 years ago

PLEASEE HELPPP IM GONNA FAIL NEED THIS BEFORE 9:30 MIDDLE SCHOOL SCIENCE

dezoksy [38]

Answer:

I think your soppose to multiply

Explanation:

7 0

2 years ago

What is the acceleration of a car going 50 mph that slows down to rest over 10 seconds?

VladimirAG [237]

Answer:

If the velocity is constant, then there is no acceleration. That is, the value of the acceleration is 0.

Explanation:

6 0

3 years ago