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forsale [732]
3 years ago
6

not all objects have a volume that is measured easily. If you were to determine the mass, volume, and density of your textbook,

a container of milk, and an air-filled balloon, how would you do it
Physics
2 answers:
Maru [420]3 years ago
8 0

Answer:

Explanation:

Volume is defined as the space occupied by any object. As there are three forms of substance found in nature which is solid, liquid and gas , different methods are used to calculate there volumes.

Mass can be calculated by using various measuring tools. For volume we need different techniques as each object has different nature.

1: Volume of book - As a book is in shape of a cuboid, we can measure the length, breath and height of the book. As volume of cuboid is equal to the product of these three values, we will get the volume of book.

2: Volume of a container of milk : Try to pour the milk into any container that has a specific shape and is filled up to the rim. In this vase the volume of the container will be the volume of milk.

3 : Volume of air - filled balloon : As we know density of air is 28.97 kg/m^3, we can simply divide the mass of air balloon with the density.

Density of book and milk can be calculated by dividing mass of these objects with there respective volume.

Maslowich3 years ago
3 0

Text book: We can measure the mass of the text book easily by weighing machine, to measure the volume we need to measure the length, width, and height of the text book by the ruler, by multiplying these dimension we can get the volume of the text book, and by dividing the mass of the book with its volume we can get the density of the book.

Milk Container: We can measure the mass of the milk container easily by weighing machine, now (assuming the milk container is cylindrical in shape) we need to measure its height, and and diameter and by the formula (π*r^2*h) we can measure its volume, and and by dividing the mass  with its volume we can get the density of the milk container.

Air filled balloon: we can measure the mass of the air filled balloon by weighing it weight machine, we know that the density of air is 28.97 kg/m^3, by dividing the mass of the balloon with the denisty of air we can get the volume of the balloon.

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lbvjy [14]

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3 0
3 years ago
A Sound wave has a frequency of 192 HZ and travels across a football field (91.4m) in 2 seconds. what is the wavelength of the s
igor_vitrenko [27]

The wavelength of the sound wave is 0.24 m

Explanation:

First of all, we calculate the  speed of the wave, which is the ratio between the distance covered by the wave and the time taken:

v=\frac{d}{t}

where, for the wave in this problem

d = 91.4 m

t = 2 s

Substituting,

v=\frac{91.4}{2}=45.7 m/s

Now we can find the wavelength of the wave by using the wave equation:

v=f \lambda

where

v is the speed of the wave

f is its frequency

\lambda is the wavelength

For the wave here,

f = 192 Hz

v = 45.7 m/s

Solving for the wavelength,

\lambda = \frac{v}{f}=\frac{45.7}{192}=0.24 m

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3 0
3 years ago
You launch a water balloon from the ground with a speed of 8.3 m/s at an angle of 27°. a. What is the horizontal component of th
solmaris [256]

a) The horizontal component of the velocity is 7.4 m/s

b) The vertical component of the velocity is 3.8 m/s

c) The balloon reaches the highest point after 0.39 s

d) The maximum height is 0.74 m

e) The total time of flight is 0.78 s

f) The range of the balloon is 5.77 m

Explanation:

a)

The motion of the balloon is the motion of a projectile, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

The horizontal component of the velocity (which is constant) is given by

v_x = u cos \theta

where

u = 8.3 m/s is the initial velocity of the balloon

\theta=27^{\circ} is the angle of projection

Substituting,

v_x = (8.3)(cos 27^{\circ})=7.4 m/s

b)

The vertical component of the initial velocity of a projectile is given by

u_y = u sin \theta

where

u is the initial velocity

\theta is the angle of projection

Here we have

u = 8.3 m/s

\theta=27^{\circ}

Substituting,

u_y = (8.3)(sin 27^{\circ})=3.8 m/s

c)

The vertical component of the velocity of the balloon follows the suvat equation

v_y = u_y - gt

where

v_y is the vertical velocity at time t

u_y = 3.8 m/s is the initial vertical velocity

g=9.8 m/s^2 is the acceleration of gravity

The balloon reaches the maximum height when the vertical velocity becomes zero:

v_y = 0

So we get:

0=u_y -gt\\t=\frac{u_y}{g}=\frac{3.8}{9.8}=0.39 s

d)

The maximum height of the balloon can be calculated using the suvat equation:

s=u_y t - \frac{1}{2}gt^2

where

u_y = 3.8 m/s is the initial vertical velocity

g=9.8 m/s^2 is the acceleration of gravity

t = 0.39 s is the time at which the highest point is reached

Substituting,

s=(3.8)(0.39)-\frac{1}{2}(9.8)(0.39)^2=0.74 m

e)

The total time of flight of a projectile is twice the time needed to reach the maximum height, and it is given by

t=\frac{2u_y}{g}

where

u_y is the initial vertical velocity

g is the acceleration of gravity

Here we have

u_y = 3.8 m/s

g=9.8 m/s^2

Substituting,

t=\frac{2(3.8)}{9.8}=0.78 s

f)

The range of a projectile is the horizontal distance covered by the projectile, so it can be found by multiplying its horizontal velocity (which is constant) by the time of flight:

d=v_x t

where

v_x is the horizontal velocity

t is the time of flight

Here we have

v_x = 7.4 m/s

t = 0.78 s

Substituting,

d=(7.4)(0.78)=5.77 m

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7 0
3 years ago
Neon has 3 naturally occurring isotopes, Neon-20, Neon-21, and Neon-22. Neon's average atomic mass on the periodic table is 20.1
Dmitry [639]
Neon - 20 is the most abundant. Abundance matters as the average atomic mass has to take into account the amount of one isotope. The more abundance the more mass, the closed the average atomic mass is to the mass of that one isotope
6 0
3 years ago
A nonconducting sphere has radius R = 2.81 cm and uniformly distributed charge q = +2.35 fC. Take the electric potential at the
Sladkaya [172]

Answer:

(a). The electric potential at 1.650 cm is -1.219\times10^{-4}\ V.

(b). The electric potential at 2.81 cm is -3.759\times10^{-4}\ V.

Explanation:

Given that,

Radius of sphere R=2.81 cm

Charge = +2.35 fC

Potential at center of sphere

V = 0

(a). We need to calculate the potential at a distance r = 1.60 cm

Using formula of potential difference

V_(r)-V_(0)=-\int_{0}^{r}{E(r)}dr

V_{r}-0=-\int_{0}^{r}{\dfrac{qr}{4\pi\epsilon_{0}R^3}}dr

V_{r}=-(\dfrac{qr^2}{8\pi\epsilon_{0}R^3})_{0}^{1.60\times10^{-2}}

V_{r}=-(\dfrac{2.35\times10^{-15}\times(1.60\times10^{-2})^2}{8\times\pi\times8.85\times10^{-12}\times(2.81\times10^{-2})^3})

V_{r}=-0.00012190\ V

V_{r}=-1.219\times10^{-4}\ V

The electric potential at 1.650 cm is -1.219\times10^{-4}\ V.

(b). We need to calculate the potential at a distance r = R

Using formula of  potential difference

V_{R}=-\dfrac{2.35\times10^{-15}}{8\pi\times8.85\times10^{-12}\times2.81\times10^{-2}}

V_{R}=-0.0003759\ V

V_{R}=-3.759\times10^{-4}\ V

The electric potential at 2.81 cm is -3.759\times10^{-4}\ V.

Hence, This is the required solution.

5 0
3 years ago
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