All categories

3 years ago

A lead ball has a mass of 55.0 grams and a density of 11.4 g/cm3. what is the volume of the ball?

Physics

2 answers:

lesantik [10]3 years ago

6 0

Density=mass/volume therefore volume=mass/density; 55g/11.4g/cm^3= 4.82cm^3

densk [106]3 years ago

6 0

Answer:

Volume, $V=4.82\ cm^3$

Explanation:

Given that,

Mass of the lead ball, m = 55 grams

Density of lead ball, $d=11.4\ g/cm^3$

We need to find the volume of the ball. The formula of density is given by :

$d=\dfrac{m}{V}$

$V=\dfrac{m}{d}$

$V=\dfrac{55\ g}{11.4\ g/cm^3}$

$V=4.82\ cm^3$

So, the volume of the the lead ball is

$V=4.82\ cm^3$ . Hence, this is the required solution.

You might be interested in

Radio signals travel at a rate of 3 × 108 meters per second. how many seconds would it take for a radio signal to travel from a

nlexa [21]

<span>3.2x10^-2 seconds (0.032 seconds)
This is a simple matter of division. I also suspect it's an exercise in scientific notation, so here is how you divide in scientific notation:

9.6 x 10^6 m / 3x10^8 m/s

First, divide the significands like you would normally.
9.6 / 3 = 3.2

And subtract the exponent. So
6 - 8 = -2

So the answer is 3.2 x 10^-2
And since the significand is less than 10 and at least 1, we don't need to normalize it.

So it takes 3.2x10^-2 seconds for the radio signal to reach the satellite.</span>

6 0

3 years ago

A particle's position is given by z(t) = −(6.50 m/s2)t2k for t ≥ 0. (Express your answer in vector form.) a. Find the particle's

blondinia [14]

Answer:

a) $z'(t) =v(t) = -13t$

Now we can replace the velocity for t=1.75 s

$v(1.75s) = -13*1.75 =-22.75 \frac{m}{s}$

For t = 3.0 s we have:

$v(3.0s) = -13*3.0 =-39 \frac{m}{s}$

b) $v_{avg}= \frac{z_f - z_i}{t_f -t_i}$

And we can find the positions for the two times required like this:

$z_f = z(3.0s) = -(6.5 \frac{m}{s^2}) (3.0s)^2=-58.5m$

$z_i = z(1.75s) = -(6.5 \frac{m}{s^2}) (1.75s)^2=-19.906m$

And now we can replace and we got:

$V_{avg}= \frac{-58.5 -(-19.906) m}{3-1.75 s}= -30.875 \frac{m}{s}$

Explanation:

The particle position is given by:

$z(t) = -(6.5 \frac{m}{s^2}) t^2, t\geq 0$

Part a

In order to find the velocity we need to take the first derivate for the position function like this:

$z'(t) =v(t) = -13t$

Now we can replace the velocity for t=1.75 s

$v(1.75s) = -13*1.75 =-22.75 \frac{m}{s}$

For t = 3.0 s we have:

$v(3.0s) = -13*3.0 =-39 \frac{m}{s}$

Part b

For this case we can find the average velocity with the following formula:

$v_{avg}= \frac{z_f - z_i}{t_f -t_i}$

And we can find the positions for the two times required like this:

$z_f = z(3.0s) = -(6.5 \frac{m}{s^2}) (3.0s)^2=-58.5m$

$z_i = z(1.75s) = -(6.5 \frac{m}{s^2}) (1.75s)^2=-19.906m$

And now we can replace and we got:

$V_{avg}= \frac{-58.5 -(-19.906) m}{3-1.75 s}= -30.875 \frac{m}{s}$

8 0

3 years ago

The siren on an ambulance emits a sound of frequency 2.80×103Hz. If the ambulance is traveling at 26.0 m/s (93.6 km/h or 58.2 mi

Norma-Jean [14]

To solve this problem it is necessary to apply the concepts related to the described wavelength through frequency and speed. Mathematically it can be expressed as:

$\lambda = \frac{v}{f}$

Where,

$\lambda =$ Wavelength

f = Frequency

v = Velocity

Our values are given as,

$f = 2.8*10^3Hz$

$v = 340m/s \rightarrow$ Speed of sound

Keep in mind that we do not use the travel speed of the ambulance because we are in front of it. In case it approached or moved away we should use the concepts related to the Doppler effect:

Replacing we have,

$\lambda = \frac{340}{2.8*10^3}$

$\lambda = 0.1214m$

Therefore the frequency that you hear if you are standing in from of the ambulance is 0.1214m

5 0

3 years ago

If air temperature increased how would it effect precipitation

Oduvanchick [21]

Decrease Because Water Vapor Would Condense More Slowly

6 0

3 years ago

two punds of water vapor at 30 psia fill the 4ft3 left chmaber of a partitioned system. The right chmaber has twice the volume o

tamaranim1 [39]

Answer:

pressure of water will be 49.7 atm

Explanation:

given data

pressure = 30 psi = 2.04 atm

water = 2 pound = 907.18

mole of water vapor = 907.19 /2 = 50.4 mole

volume = 4 ft³ = 113.2 L

temperature = 40 F = 277.59 K

to find out

pressure of water

solution

we will apply here ideal gas condition

that is

PV = nRT .......................1

put here all value and here R = 0.0821 , T temperature and V volume and P pressure and n is no of mole

and we get here temperature

PV = nRT

2.04 × 113.2 = 50.4×0.0821×T

solve it and we get

T = 55.8 K

so we have given right chamber has twice the volume of the left chamber i.e

volume = twice of volume + volume

volume = 2(113.2) + 113.2

volume = 339.6 L

so from equation 1 pressure will be

PV = nRT

P(339.6) = 50.4 × ( 0.0821) × (277.59)

P = 3.3822 atm = 49.7 atm

so pressure of water will be 49.7 atm

7 0

3 years ago