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Vsevolod [243]
2 years ago
11

Classyfying— what is the name of the force you exert on a sponge when you squeeze it?

Physics
2 answers:
Kruka [31]2 years ago
5 0
Its called compression
aev [14]2 years ago
4 0
 compression
hoped i could help
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Is 2/2 1 or 0? please help lol
Komok [63]

Answer:

1.

Explanation:

Hello!

In this case, for such mathematical operations, we can wee that the slash represents a fraction or a division, say 8 ÷ 4 = 2, 6 ÷ 3 = 2, 20 ÷ 4 = 5, etc. In such a way, since the operation 2/2, represents 2 ÷ 2, it is clear that two is once in 2, therefore, the result is:

2 ÷ 2 = 1.

Best regards!

4 0
2 years ago
An amusement park ride consists of a rotating circular platform 8.26 m in diameter from which 10 kg seats are suspended at the e
VashaNatasha [74]

To solve this problem we will begin by finding the necessary and effective distances that act as components of the centripetal and gravity Forces. Later using the same relationships we will find the speed of the body. The second part of the problem will use the equations previously found to find the tension.

PART A) We will begin by finding the two net distances.

r = \frac{8.26}{2} = 4.13m

And the distance 'd' is

d = lsin\theta

d = 1.14 sin 16.2\°

d = 0.318m

Through the free-body diagram the tension components are given by

Tcos\theta = mg

Tsin\theta = \frac{mv^2}{R}

Here we can watch that,

R = r+d

Dividing both expression we have that,

tan\theta = \frac{v^2}{Rg}

Replacing the values,

tan(16.2) = \frac{v^2}{(4.13+0.318)(9.8)}

v = 4.83371m/s

PART B) Using the vertical component we can find the tension,

Tcos\theta = mg

T = \frac{mg}{cos\theta}

T = \frac{(10+26.2)(9.8)}{cos(16.2)}

T = 369.42N

6 0
3 years ago
A 6.00-kg box sits on a ramp that is inclined at 37.0° above the horizontal. the coefficient of kinetic friction between the box
forsale [732]
We need to see what forces act on the box:

In the x direction:

Fh-Ff-Gsinα=ma, where Fh is the horizontal force that is pulling the box up the incline, Ff is the force of friction, Gsinα is the horizontal component of the gravitational force, m is mass of the box and a is the acceleration of the box.

In the y direction:

N-Gcosα = 0, where N is the force of the ramp and Gcosα is the vertical component of the gravitational force. 

From N-Gcosα=0 we get: 

N=Gcosα, we will need this for the force of friction.

Now to solve for Fh:

Fh=ma + Ff + Gsinα,

Ff=μN=μGcosα, this is the friction force where μ is the coefficient of friction. We put that into the equation for Fh.
G=mg, where m is the mass of the box and g=9.81 m/s²

Fh=ma + μmgcosα+mgsinα

Now we plug in the numbers and get:

Fh=6*3.6 + 0.3*6*9.81*0.8 + 6*9.81*0.6 = 21.6 + 14.1 + 35.3 = 71 N

The horizontal force for pulling the body up the ramp needs to be Fh=71 N.
8 0
3 years ago
calculate the amount of heat (in btu) needed to raise the temp of 2.5lb of glass from 45°f to 350°f. the specific heat capacity
Firlakuza [10]
So you would use the equation Q=cmΔT, where c is the specific heat, m is the mass, and ΔT is change in temperature. Q, or heat added, would equal (0.187)(2.5)(350-45), which simplifies to 142.5875 btu.
7 0
2 years ago
what is the necessary condition within the hydrogen atom for balmer line photons to be absorbed by the hydrogen atom
luda_lava [24]

Explanation :

When an electron jumps from one energy level to another, the energy of atom gets changed.

If a photon gets absorbed, the electron will move to higher energy levels and then fall back to the lower energy levels. Then each time a photon will be absorbed whose energy is given by difference between the initial and final energy levels i.e

In Balmer series, the transition is from higher energy levels to n = 2.

So, the necessary condition for Balmer series is that the electron should be at first excited state or n = 2 level as shown in figure.


6 0
3 years ago
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