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2 years ago

What number will be shown by hydrometer if sunk in pure water ?

Physics

2 answers:

adoni [48]2 years ago

8 0

1000kg/meters cubed if it’s sunk in pure water

serious [3.7K]2 years ago

7 0

The hydrometer displays the density of the liquid you stick it in, COMPARED TO WATER.

If you stick it in pure water, it should read 1.00 . This is a good way to check your hydrometer and see whether it's reading correctly ("calibrated").

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A trebuchet was a hurling machine built to attack the walls of a castle under siege. A large stone could be hurled against a wal

Studentka2010 [4]

(a) 18.9 m/s

The motion of the stone consists of two independent motions:

- A horizontal motion at constant speed

- A vertical motion with constant acceleration ( $g=9.8 m/s^2$ ) downward

We can calculate the components of the initial velocity of the stone as it is launched from the ground:

$u_x = v_0 cos \theta = (25.0)(cos 41.0^{\circ})=18.9 m/s\\u_y = v_0 sin \theta = (25.0)(sin 41.0^{\circ})=16.4 m/s$

The horizontal velocity remains constant, while the vertical velocity changes due to the acceleration along the vertical direction.

When the stone reaches the top of its parabolic path, the vertical velocity has became zero (because it is changing direction): so the speed of the stone is simply equal to the horizontal velocity, therefore

$v=18.9 m/s$

(b) 22.2 m/s

We can solve this part by analyzing the vertical motion only first. In fact, the vertical velocity at any height h during the motion is given by

$v_y^2 - u_y^2 = 2ah$ (1)

where

$u_y = 16.4 m/s$ is the initial vertical velocity

$v_y$ is the vertical velocity at height h

$a=g=-9.8 m/s^2$ is the acceleration due to gravity (negative because it is downward)

At the top of the parabolic path, $v_y = 0$ , so we can use the equation to find the maximum height

$h_{max} = \frac{-u_y^2}{2a}=\frac{-(16.4)^2}{2(-9.8)}=13.7 m$

So, at half of the maximum height,

$h = \frac{13.7}{2}=6.9 m$

And so we can use again eq(1) to find the vertical velocity at h = 6.9 m:

$v_y = \sqrt{u_y^2 + 2ah}=\sqrt{(16.4)^2+2(-9.8)(6.9)}=11.6 m/s$

And so, the speed of the stone at half of the maximum height is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{18.9^2+11.6^2}=22.2 m/s$

(c) 17.4% faster

We said that the speed at the top of the trajectory (part a) is

$v_1 = 18.9 m/s$

while the speed at half of the maximum height (part b) is

$v_2 = 22.2 m/s$

So the difference is

$\Delta v = v_2 - v_2 = 22.2 - 18.9 = 3.3 m/s$

And so, in percentage,

$\frac{\Delta v}{v_1} \cdot 100 = \frac{3.3}{18.9}\cdot 100=17.4\%$

So, the stone in part (b) is moving 17.4% faster than in part (a).

4 0

2 years ago

A series of waves with decreasing wavelength labeled radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, gamm

kotegsom [21]

Answer:

label A= radio waves, label C= infrared, Label D= visible Light, Label G= gamma rays.

Explanation:

hope it helped??

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6 0

2 years ago

Read 2 more answers

Two objects are placed in thermal contact and are allowed to come to equilibrium in isolation. the heat capacity of object a is

Harman [31]

Given:
Ca = 3Cb (1)
where
Ca = heat capacity of object A
Cb = heat capacity f object B

Also,
Ta = 2Tb (2)
where
Ta = initial temperature of object A
Tb = initial temperature of object B.

Let
Tf = final equilibrium temperature of both objects,
Ma = mass of object A,
Mb = mass of object B.

Assuming that all heat exchange occurs exclusively between the two objects, then energy balance requires that
Ma*Ca*(Ta - Tf) = Mb*Cb*(Tf - Tb) (3)

Substitute (1) and (2) into (3).
Ma*(3Cb)*(2Tb - Tf) = Mb*Cb*(Tf - Tb)
3(Ma/Mb)*(2Tb - Tf) = Tf - Tb

Define k = Ma/Mb, the ratio f the masses.
Then
3k(2Tb - Tf) = Tf - Tb
Tf(1+3k) = Tb(1+6k)
Tf = [(1+6k)/(1+3k)]*Tb

Answer:
$T_{f} =( \frac{1+6k}{1+3k} )T_{b}= \frac{1}{2}( \frac{1+6k}{1+3k})T_{a}$
where
$k= \frac{M_{a}}{M_{b}}$

7 0

2 years ago

Waves that move the particles of the medium parallel to the direction in which the waves are traveling are called

Nikolay [14]

1. a. longitudinal waves.

There are two types of waves:

- Transverse waves: in transverse waves, the oscillations of the wave occur in a direction perpendicular to the direction of propagation of the wave

- Longitudinal waves: in longitudinal waves, the oscillations of the waves occur parallel to the direction in which the waves are travelling.

So, these types of waves are called longitudinal waves.

2. d. a medium

There are two types of waves:

- Electromagnetic waves: these waves are produced by the oscillations of electric and magnetic field, and they can travel both in a medium and also in a vacuum (they do not need a medium to propagate)

- Mechanical waves: these waves are produced by the oscillations of the particles in a medium, so they need a medium to propagate - therefore, the correct choice is d. a medium

3. a. AM/FM radio

Analogue signals consist of continuous signals, which vary in a continuous range of values. On the contrary, digital signals consist of discrete signals, which can assume only some discrete values. For AM and FM radios, signals are transmitted by using analogue signals.

5 0

2 years ago

A satellite, orbiting the earth at the equator at an altitude of 400 km, has an antenna that can be modeled as a 1.76-m-long rod

ivann1987 [24]

Answer:

The inducerd emf is 1.08 V

Solution:

As per the question:

Altitude of the satellite, H = 400 km

Length of the antenna, l = 1.76 m

Magnetic field, B = $8.0\times 10^{- 5}\ T$