Ask question

Ask question

All categories

3 years ago

15

Four objects are situated along the y axis as follows: a 1.99-kg object is at 2.99 m, a 2.96-kg object is at 2.57 m, a 2.43-kg o

bject is at the origin, and a 3.96-kg object is at -0.502 m. Where is the center of mass of these objects

1 answer:

Dominik [7]3 years ago

3 0

Answer:

The center of mass for the object is $y_c = 1.063 \ m$ from the origin

Explanation:

From the question we are told that

The mass of the first object is $m_1 = 1.99 \ kg$

The position of first object with respect to origin $y_1 = 2.99 \ m$

The mass of the second object is $m_2 = 2.96 \ kg$

The position of second object with respect to origin $y_2 = 2.57 \ m$

The mass of the third object is $m_3 = 2.43 \ kg$

The position of third object with respect to origin $y_3 = 0 \ m$

The mass of the fourth object is $m_3 = 3.96 \ kg$

The position of fourth object with respect to origin $y_3 = -0.502 \ m$

Generally the center of mass of the object along the x-axis is zero because all the mass lie on the y axis

Generally the location of the center mass of the object is mathematically represented as

$y_c = \frac{m_1 * y_1 + m_2 * y_2 + m_3 * y_3 + m_4 * y_4}{m_1 + m_2 + m_3 + m_4}$

=> $y_c = \frac{1.99 * 2.99 + 2.96 * 2.57 + 2.43 * 0 + 3.96 * (-0.502)}{1.99+ 2.96 + 2.43 + 3.96}$

=> $y_c = 1.063 \ m$

You might be interested in

Consider the following four objects: a hoop, a solid sphere, a flat disk, a hollow sphere. Each of the objects has mass M and ra

Svetlanka [38]

Answer:

Hoop.

Explanation:

The angular acceleration performed at a given torque:

$\alpha = \frac{\tau}{I}$

The moments of inertia of each element are described below:

Hoop

$I = M\cdot R^{2}$

Solid sphere

$I = \frac{2}{5}\cdot M \cdot R^{2}$

Flat disk

$I = \frac{1}{2}\cdot M \cdot R^{2}$

Hollow sphere

$I = \frac{2}{3}\cdot M \cdot R^{2}$

The greater the moment of inertia, the greater the torque to obtain the same angular acceleration. Therefore, the hoop requires the largest torque to receive the same angular acceleration.

8 0

4 years ago

What is the mass in kg of a leopard in a tree if the tree branch is 36 m up and the leopard's gravitational potential energy is

ElenaW [278]

Answer:

83.3kg

Explanation:

GPE = m × g × h

GPE = mass of leopard × 10 × 36m

29988J = 360 × mass

mass = 83.3kg

5 0

3 years ago

What is a substance?

Effectus [21]

Answer:

A. a uniform mixture that can't be separated

8 0

3 years ago

Read 2 more answers

The earth has a vertical electric field at the surface, pointing down, that averages 100 N/C. This field is maintained by variou

zlopas [31]

Answer:

q = 3.6 10⁵ C

Explanation:

To solve this exercise, let's use one of the consequences of Gauss's law, that all the charge on a body can be considered at its center, therefore we calculate the electric field on the surface of a sphere with the radius of the Earth

r = 6 , 37 106 m

E = k q / r²

q = E r² / k

q = $\frac{100 \ (6.37 \ 10^6)^2}{9 \ 10^9}$

q = 4.5 10⁵ C

Now let's calculate the charge on the planet with E = 222 N / c and radius

r = 0.6 r_ Earth

r = 0.6 6.37 10⁶ = 3.822 10⁶ m

E = k q / r²

q = E r² / k

q = $\frac{222 (3.822 \ 10^6)^2}{ 9 \ 10^9}$

q = 3.6 10⁵ C

4 0

3 years ago

You set your stationary bike on a high 80-N friction-like resistive force and cycle for 30 min at a speed of 8.0 m/s . Your body

stellarik [79]

A) The change in internal chemical energy is $1.15\cdot 10^7 J$

B) The time needed is 1 minute

Explanation:

First of all, we start by calculating the power output of you and the bike, given by:

$P=Fv$

where

F = 80 N is the force that must be applied in order to overcome friction and travel at constant speed

v = 8.0 m/s is the velocity

Substituting,

$P=(80)(8.0)=640 W$

The energy output is related to the power by the equation

$P=\frac{E}{t}$

where:

P = 640 W is the power output

E is the energy output

$t = 30 min \cdot 60 = 1800 s$ is the time elapsed

Solving for E,

$E=Pt=(640)(1800)=1.15\cdot 10^6 J$

Since the body is 10% efficient at converting chemical energy into mechanical work (which is the output energy), this means that the change in internal chemical energy is given by

$\Delta E = \frac{E}{0.10}=\frac{1.15\cdot 10^6}{0.10}=1.15\cdot 10^7 J$

B)

From the previous part, we found that in a time of

t = 30 min

the amount of internal chemical energy converted is

$E=1.15\cdot 10^7 J$

Here we want to find the time t' needed to convert an amount of chemical energy of

$E'=3.8\cdot 10^5 J$

So we can setup the following proportion:

$\frac{t}{E}=\frac{t'}{E'}$

And solving for t',

$t'=\frac{E't}{E}=\frac{(3.8\cdot 10^5)(30)}{1.15\cdot 10^7}=1 min$

Learn more about power and energy:

brainly.com/question/7956557

#LearnwithBrainly

3 0

3 years ago