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4 years ago

6

The removal of an embedded gas from a solid object, as happens when formaldehyde in new carpets and furniture is released into t

he air, is a process known as

1 answer:

Sphinxa [80]4 years ago

8 0

Answer;

offgassing

Explanation;

The removal of the embedded gas from a solid object, as happens when formaldehyde in new carpets & furniture is released into the air, is a process known as offgassing.

It occurs when new, manufactured items in our homes release volatile organic compounds (VOCs) and other chemicals. These compounds include, paints, glue, finishes, among others, they release gases into air.

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Two point charges are separated by 10 cm, with an attractive force between them of 15 N. Find the force between them when they a

suter [353]

Answer:

(a) the force is 8.876 N

(b) the magnitude of each charge is 4.085 μC

Explanation:

Part (a)

Given;

coulomb's constant, K = 8.99 x 10⁹ N.m²/C²

distance between two charges, r = 10 cm = 0.1 m

force between the two charges, F = 15 N

when the distance between the charges changes to 13 cm (0.13 m)

force between the two charges, F = ?

Apply Coulomb's law;

$F = \frac{Kq_1q_2}{r^2} \\\\let \ Kq_1q_2 = C\\\\F =\frac{C}{r^2} \\\\C = Fr^2\\\\F_1r_1^2 = F_2r_2^2\\\\F_2 =\frac{F_1r_1^2}{r_2^2} \\\\F_2 = \frac{15*0.1^2}{0.13^2} \\\\F_2 = 8.876 \ N$

Part (b)

the magnitude of each charge, if they have equal magnitude

$F = \frac{KQ^2}{r^2}$

where;

F is the force between the charges

K is Coulomb's constant

Q is the charge

r is the distance between the charges

$F = \frac{KQ^2}{r^2} \\\\Q = \sqrt{\frac{Fr^2}{K} } \\\\Q = \sqrt{\frac{15*(0.1)^2}{8.99*10^9} } = 4.085 *10^{-6} \ C\\\\Q = 4.085 \ \mu C$

4 0

4 years ago

A hoop and a disk with uniform mass distribution have the same radius but the total masses are not known. Can they both roll dow

ser-zykov [4K]

Answer:

Explanation:

radius of hoop and the radius of disk is same = R

Let the mass of hoop is M and the mass of disk is M'.

As they reach the bottom of teh surface in same time so they travel equal distance thus, they have same acceleration.

The acceleration is given by

$a=\frac{gSin\theta }{1+\frac{I}{MR^{2}}}$

As the acceleration is same so that the moment of inertia is also same.

Moment of inertia of disk = moment of inertia of hoop

1/2 x mass of disk x R² = mass of hoop x R²

So, mass of disk = 2 x mass of hoop

Option (c) is correct.

5 0

3 years ago

7. How long does it take a ball rolling down a hill to change its speed from 3 m/sec to 34.5 m/sec

lys-0071 [83]

The time elapsed is 9 seconds

Explanation:

The motion of the ball is a uniformly accelerated motion (a motion with constant acceleration), so we can use the following suvat equation:

$v=u+at$

where :

v is the final velocity of the ball

u is the initial velocity

a is the acceleration

t is the time elapsed

For the ball in this problem, we have:

u = 3 m/s is the initial velocity

v = 34.5 m/s is the final velocity

$a=3.5m/s^2$ is the acceleration

Solving for t, we find the time taken for this change in velocity:

$t=\frac{v-u}{a}=\frac{34.5-3}{3.5}=9 s$

Learn more about acceleration:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

8 0

3 years ago

Figure 1 shows the kinetic energy as a function of time for a 2kg object that is released from rest and falls toward Earth’s sur

garri49 [273]

<u><em>Answer:</em></u>

The answer is 1400 J, according to my Physics teacher.

<u><em>Explanation:</em></u>

You need to take into account everything that is listed in the question; it's important to remember that the question is asking about the change in gravitational potential energy of the object-object-Earth system from 0s to 10s, not 0s to 20s. :)

7 0

3 years ago

Jack (mass 59.0 kg ) is sliding due east with speed 8.00 m/s on the surface of a frozen pond. He collides with Jill (mass 47.0 k

Phantasy [73]

Answer:

Part(A): The magnitude of Jill's final velocity is $\bf{6.59~m/s}$ .

Part(B): The direction is $\bf{42.7^{0}}$ south to east.

Explanation:

Given:

Mass of Jack, $m_{1} = 59.0~Kg$

Mass of Jill, $m_{2} = 47..0~Kg$

Initial velocity of Jack, $v_{1i} = 8.00~m/s$

Initial velocity of Jill, $v_{2i} = 0$

Final velocity of Jack, $v_{1f} 5.00~m/s$

The final angle made by Jack after collision, $\alpha = 34.0^{0}$

Consider that the final velocity of Jill be $v_{2f}$ and it makes an angle of $\beta$ with respect to east, as shown in the figure.

Conservation of momentum of the system along east direction is given by

$~~~~&& m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f} \cos \alpha + m_{2}v_{2f}^{x}\\&or,& v_{2f}^{x} = \dfrac{m_{1}(v_{1i} - v_{1f} \cos \alpha)}{m_{2}}$

where, $v_{2f}^{x}$ is the component of Jill's final velocity along east. The direction of this component will be along east.

Substituting the value, we have

$v_{2f}^{x} &=& \dfrac{(59.0~Kg)(8.00~m/s - 5.00 \cos 34.0^{0}~m/s)}{47.0~Kg}\\~~~~~&=& 4.84~m/s$

Conservation of momentum of the system along north direction is given by

$~~~~&& v_{2f}^{y} + v_{1f} \sin \alpha = 0\\&or,& v_{2f}^{y} = - v_{1f} \sin \alpha = (8.00~m/s) \sin 34^{0} = 4.47~m/s$

where, $v_{2f}^{y}$ is the component of Jill's final velocity along north. The direction of this component will be along the opposite to north.

Part(A):

The magnitude of the final velocity of Jill is given by

$v_{2f} &=& \sqrt{(v_{2f}^{x})^{2} + (v_{2f}^{y})^{2}}\\~~~~~&=& 6.59~m/s$

Part(B):

The direction is given by

$\beta &=& \tan^{-1}(\dfrac{4.47~m/s}{4.84~m/s})\\~~~~&=& 42.7^{0}$

4 0

4 years ago