Describe how a neutral material becomes attracted to a negatively charged object brought near it.

Sasha lifts a couch 8.2 meters from the ground floor of her house to the attic. If the couch has a mass of 120 kg, what is the g

gavmur [86]

As we know that gravitational potential energy is given by

$U = mgH$

here we have

m = mass = 120 kg

$g = 9.81 m/s^2$

h = height = 8.2 m

now from above formula

$U = 120kg (9.81 m/s^2) (8.2 m)$

$U = 9653.04 J$

so above is the gravitational potential energy of the couch

4 0

3 years ago

Read 2 more answers

Which dartboard represents high accuracy and low precision?

Virty [35]

Dart board 3 represents high accuracy but low precision

8 0

3 years ago

A bird flies at a speed of 2.3 m/s if it has 14 j of kinetic energy what is the mass

Sidana [21]

Kinetic Energy = 1/2 * mv²

Kinetic Energy = 14 J, v = 2.3 m/s , m = ?

14 = 1/2 * m* 2.3²

14 = 0.5*m*2.3*2.3

m = 14 / (0.5*2.3*2.3)

m = 5.29 kg.

Mass = 5.29 kg.

7 0

3 years ago

Read 2 more answers

The specific heat of substance A is greater than that of substance B. Both A and B are at the same initial temperature when equa

Sonja [21]

Answer:

$m_A c_{pA} (T_{fA} -T) = m_B c_{pB} (T_{fB}- T)$

For this case, if we try to find the final temperature of A and B, we see that we will obtain an expression in terms of specific heats and masses, from the information given we know the relationship between specific heats, but we don't know the relationship that exists among the masses, then the best option for this case is:

d) More information is needed

(The relation between the masses is not given)

Explanation:

For this case we know the following info:

$c_{pA} > c_{pB}$

Where c means specific heat for the substance A and B.

We also know that the initial temperatures for both sustances are equal:

$T_{iA}= T_{iB}$

We assume that we don't have melting or vaporization in the 2 substances. So we just have presence of sensible heat given by this formula:

$Q = m c_p \Delta T$

And for this case we know that Both A and B are at the same initial temperature when equal amounts of energy are added to them, so then we have this:

$Q_A = Q_B$

And if we replace the formula for sensible heat we got:

$m_A c_{pA} \Delta T_A = m_B c_{pB} \Delta T_B$

And if we replace for the change of the temperature we got:

$m_A c_{pA} (T_{fA} -T_{iA}) = m_B c_{pB} (T_{fB}- T_{iB})$

And since $T_{iA}= T_{iB}= T$ we have this:

$m_A c_{pA} (T_{fA} -T) = m_B c_{pB} (T_{fB}- T)$

For this case, if we try to find the final temperature of A and B, we see that we will obtain an expression in terms of specific heats and masses, from the information given we know the relationship between specific heats, but we don't know the relationship that exists among the masses, then the best option for this case is:

d) More information is needed

(The relation between the masses is not given)

4 0

3 years ago

You are investigating how objects move when they are dropped from different heights. To collect your data, you drop a 1 kg weigh

ASHA 777 [7]

The time of motion of the 5 kg object will be the same as 1 kg since both objects are dropped from the same height.

The given parameters;

<em>Mass of the first object, m1 = 1 kg</em>

<em>Mass of the second object, m2 = 5 kg</em>

The final velocity of the objects during the downward motion is calculated as follows;

$v_f = v_0 + gt\\\\v_f = 0 + gt\\\\\v_f = gt$

The time of motion of the object from the given height is calculated as;

$h = v_0 t + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\gt^2 = 2h\\\\t^2 = \frac{2h}{g} \\\\t = \sqrt{\frac{2h}{g} }$

The time of motion of each object is independent of mass of the object.

Thus, the time of motion of the 5 kg object will be the same as 1 kg since both objects are dropped from the same height.

Learn more about time of motion here: brainly.com/question/2364404

3 0

2 years ago