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uysha [10]
3 years ago
6

Which needs less heat to increase its temperature,

Physics
2 answers:
rosijanka [135]3 years ago
6 0
Sand is what it said and I think I agree

Julli [10]3 years ago
5 0

Answer:

sand

Explanation:

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Your clothing tends to cling together after going through the dryer. Why? Would you expect more or less clinging if all your clo
lys-0071 [83]

Answer and Explanation:

The clothing after spinning in the dryer cling together. This is because in the dryer they are rubbed against each other and due to this rubbing, electrons are transferred from one to the other clothes and acquire charge as a result of charging by friction thus producing static electricity.

As the material of the clothes in the dryer is different, clinging will be more.

The sticking of these clothes together is known as Static cling.

In case, the clothing are of same material, the static electricity produced as a result of frictional charging would be less and hence less static cling would occur.

4 0
4 years ago
Atoms of what two elements are in every organic compound?
Georgia [21]
Carbon atoms and hydrogen I hope this helps
8 0
3 years ago
A big olive (* - 0.50 kg) lies at the origin of an xy coordinate system, and a big BrazlI nut (M - 1.5^kg) lie^s at the point (1
Afina-wow [57]

The <em>estimated</em> displacement of the center of mass of the olive is \overrightarrow{\Delta r} = -0.046\,\hat{i} -0.267\,\hat{j}\,[m].

<h3>Procedure - Estimation of the displacement of the center of mass of the olive</h3>

In this question we should apply the definition of center of mass and difference between the coordinates for <em>dynamic</em> (\vec r) and <em>static</em> conditions (\vec r_{o}) to estimate the displacement of the center of mass of the olive (\overrightarrow{\Delta r}):

\vec r - \vec r_{o} = \left[\frac{\Sigma\limits_{i=1}^{2}r_{i,x}\cdot(m_{i}\cdot g + F_{i, x})}{\Sigma \limits_{i =1}^{2}(F_{i,x}+m_{i}\cdot g)} ,\frac{\Sigma\limits_{i=1}^{2}r_{i,y}\cdot(m_{i}\cdot g + F_{i, y})}{\Sigma \limits_{i =1}^{2}(F_{i,y}+m_{i}\cdot g)} \right]-\left(\frac{\Sigma\limits_{i=1}^{2}r_{i,x}\cdot m_{i}\cdot g}{\Sigma \limits_{i= 1}^{2} m_{i}\cdot g}, \frac{\Sigma\limits_{i=1}^{2}r_{i,y}\cdot m_{i}\cdot g}{\Sigma \limits_{i= 1}^{2} m_{i}\cdot g}\right) (1)

Where:

  • r_{i, x} - x-Coordinate of the i-th element of the system, in meters.
  • r_{i,y} - y-Coordinate of the i-th element of the system, in meters.
  • F_{i,x} - x-Component of the net force applied on the i-th element, in newtons.
  • F_{i,y} - y-Component of the net force applied on the i-th element, in newtons.
  • m_{i} - Mass of the i-th element, in kilograms.
  • g - Gravitational acceleration, in meters per square second.

If we know that \vec r_{1} = (0, 0)\,[m], \vec r_{2} = (1, 2)\,[m], \vec F_{1} = (0, 3)\,[N], \vec F_{2} = (-3, -2)\,[N], m_{1} = 0.50\,kg, m_{2}  = 1.50\,kg and g = 9.807\,\frac{kg}{s^{2}}, then the displacement of the center of mass of the olive is:

<h3>Dynamic condition\vec{r} = \left[\frac{(0)\cdot (0.50)\cdot (9.807)+(0)\cdot (0) + (1)\cdot (1.50)\cdot (9.807) + (1)\cdot (-3)}{(0.50)\cdot (9.807) + 0 + (1.50)\cdot (9.807)+(-3)}, \frac{(0)\cdot (0.50)\cdot (9.807) + (0)\cdot (3) + (2)\cdot (1.50)\cdot (9.807) +(2) \cdot (-2)}{(0.50)\cdot (9.807) + (3)+(1.50)\cdot (9.807)+(-2)}  \right]\vec r = (0,704, 1.233)\,[m]</h3>

<h3>Static condition</h3><h3>\vec{r}_{o} = \left[\frac{(0)\cdot (0.50)\cdot (9.807) + (1)\cdot (1.50)\cdot (9.807)}{(0.50)\cdot (9.807) + (1.50)\cdot (9.807)}, \frac{(0)\cdot (0.50)\cdot (9.807) + (2)\cdot (1.50)\cdot (9.807)}{(0.50)\cdot (9.807)+(1.50)\cdot (9.807)}  \right]</h3><h3>\vec r_{o} = \left(0.75, 1.50)\,[m]</h3><h3 /><h3>Displacement of the center of mass of the olive</h3>

\overrightarrow{\Delta r} = \vec r - \vec r_{o}

\overrightarrow{\Delta r} = (0.704-0.75, 1.233-1.50)\,[m]

\overrightarrow{\Delta r} = (-0.046, -0.267)\,[m]

The <em>estimated</em> displacement of the center of mass of the olive is \overrightarrow{\Delta r} = -0.046\,\hat{i} -0.267\,\hat{j}\,[m]. \blacksquare

To learn more on center of mass, we kindly invite to check this verified question: brainly.com/question/8662931

3 0
2 years ago
What is another name for a "causal loop"?
OleMash [197]
B I believe but hey we also have Google to double check
8 0
4 years ago
Read 2 more answers
A boxer punches a sheet of paper in midair and brings it from rest up to a speed of 30 m/s in 0.060 s .
zimovet [89]

Answer:

Force exerted, F = 1.5 N

Explanation:

It is given that, a boxer punches a sheet of paper in midair and brings it from rest up to a speed of 30 m/s in 0.060 s.

i.e. u = 0

v = 30 m/s

Time taken, t = 0.06 s

Mass of the paper, m = 0.003 kg

We need to find the force the boxer exert on it. The force can be calculated using second law of motion as :

F=m\times a

F=m\times (\dfrac{v-u}{t})

F=0.003\times (\dfrac{30}{0.06})

F = 1.5 N

So, the force the boxer exert on the paper is 1.5 N. Hence, this is the required solution.

6 0
3 years ago
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