A decrease in the magnitude of velocity is called

Gases are more prone to expansion and contraction than liquid . The biggest change in the volume in your thermometer was probabl

KiRa [710]

Answer:

In the air

Explanation:

There are three states of matter:

- Solids: in solids, the particles are tightly bond together by strong intermolecular forces, so they cannot move freely - they can only vibrate around their fixed position

- Liquids: in liquids, particles are more free to move, however there are still some intermolecular forces keeping them close to each other

- Gases: in gases, particles are completely free to move, as the intermolecular forces between them are negligible

For this reason, it is generally easier to compress/expand the volume of a gas with respect to the volume of a liquid.

In this problem, we are comparing water (which is a liquid) with air (which is a gas). From what we said above, this means that the change in volume is larger in the air rather than in the water.

8 0

2 years ago

An 89 kg man drops from rest on a diving board −3.1 m above the surface of the water and comes to rest 0.5 s after reaching the

OLga [1]

To solve this problem we will use the linear motion kinematic equations, for which the change of speed squared with the acceleration and the change of position. The acceleration in this case will be the same given by gravity, so our values would be given as,

$m= 89 kg\\x = 3.1 m\\t = 0.5s\\a = g = 9.8m/s^2$

Through the aforementioned formula we will have to

$v_f^2-v_i^2 = 2ax$

The particulate part of the rest, so the final speed would be

$v_f^2 = 2gx$

$v_f=\sqrt{2(9.8)(3.1)}$

$v_f = 7.79m/s$

Now from Newton's second law we know that

$F = ma$

Here,

m = mass

a = acceleration, which can also be written as a function of velocity and time, then

$F = m\frac{dv}{dt}$

Replacing we have that,

$F = (89)\frac{7.79}{0.5}$

$F = 1386.62N$

Therefore the force that the water exert on the man is 1386.62

3 0

3 years ago

An object is placed 4.0 cm to the left of a convex lens with a focal length of +8.0 cm . Where is the image of the object?

Serjik [45]

The image of the object is 8cm to the left of the lens (D)

What is the image of an object?

The image of an object is said to be the location where light rays from that object intersect with a mirror by reflection.

It is calculated thus:

1÷v = 1÷f - 1÷u

<h3>How to calculate the image of an object</h3>

From the formula

1÷v = 1÷f - 1÷u

<h3>Where </h3>

V = image distance fromthe object

U = object

f = focal length

Substitute the values

1÷v = 1÷8 - 1÷ 4

1÷v = - 1÷8

Make v the subject of formula

v = -8cm

Therefore, the image of the object is 8cm to the left of the lens (D)

Learn more on focal length here:

brainly.com/question/25779311

#SPJ1

6 0

2 years ago

A large rocket has a mass of 2.00×10⁶ kg at takeoff, and its engines produce a thrust of 3.50×10⁷ N. Find its initial accelerati

Kazeer [188]

Answer:

17.5 m/s²

1.90476 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

Force

$F=ma\\\Rightarrow a=\frac{F}{m}\\\Rightarrow a=\frac{3.5\times 10^7}{2\times 10^6}\\\Rightarrow a=17.5\ m/s^2$

Initial acceleration of the rocket is 17.5 m/s²

$v=u+at\\\Rightarrow \frac{120}{3.6}=0+17.5t\\\Rightarrow t=\frac{\frac{120}{3.6}}{17.5}=1.90476\ s$

Time taken by the rocket to reach 120 km/h is 1.90476 seconds

Change in the velocity of a rocket is given by the Tsiolkovsky rocket equation

$\Delta v=v_{e}\ln \frac{m_0}{m_f}$

where,

$m_0$ = Initial mass of rocket with fuel

$m_f$ = Final mass of rocket without fuel

$v_e$ = Exhaust gas velocity

Hence, the change in velocity increases as the mass decreases which changes the acceleration

4 0

2 years ago

If NEPA charges 5k per kWh, what is the cost of

tiny-mole [99]

Answer:

24k

Explanation:

We multiply by 200V by 24

8 0

2 years ago