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3 years ago

A sample of copper has a volume of 23.4 cm3 if the density of copper is 8.9 gcm3 what is the coppers mass?

1 answer:

7 0

The answer is: " 208 g " .
_____________________________________________
Explanation:
__________________________________________
The formula/ equation for density is:
__________________________________________
D = m / V ; That is, "mass divided by volume" ;

Density is expressed as:
__________________________________________
   "mass per unit volume"; in which the "mass" is expressed in units of "g" ("grams") ; and the "unit volume" is expressed in units of:
"cm³ " or "mL";
_____________________________________________
{Note the exact equivalent: 1 cm³ = 1 mL }.
____________________________________________
→ The formula is: " D = m / V " ;
___________________________________________
in which:

"D" refers to the "density" (see above), which is: "8.9 g/cm³ " (given);

"m" refers to the "mass" , in units of "g" (grams), which is unknown; and we want to find this value;

"V" refers to the "volume", in units of "cm³ " ;
which is: "23.4 cm³ " (given);
_________________________________________________
We want to find the mass, "m" ; so we take the original equation/formula for the density:
_________________________________________________
D = m / V ;
_________________________________________________________
And we rearrange; to isolate "m" (mass) on ONE side of the equation; and then we plug in our known/given values;
to solve for "m" (mass); in units of "g" (grams) ;
___________________________________________________
Multiply each side of the equation by "V" ;
____________________________________________________
V * { D = m / V } ; to get:
____________________________________________________
V * D = m ;   ↔ m = V * D ;
___________________________________________________
Now, we plug in the given values for "V" (volume) and "D" (density) ; to solve for the mass, "m" ;
______________________________________________________
m = V * D ;

m = (23.4 cm³) * (8.9 g / 1 cm³) = (23.4 * 8.9) g = 208.26 g ;

→ Round to "208 g" (3 significant figures);
____________________________________
The answer is: " 208 g " .
_____________________________________________________

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When the atmosphere is not quite clear, one may sometimes see colored circles concentric with the Sun or the Moon. These are gen

stealth61 [152]

Answer:

D) diffraction

Explanation:

Corona is an optical phenomenon produced by the diffraction of sunlight or moonlight, as light moves through water droplets in the atmosphere.

This phenomenon produces one or more diffuse concentric rings of light around the Sun or Moon, usually seen as colored circles.

Therefore, the explanation for these phenomena of colored concentric circles, sometimes seen with the Sun or the Moon involves diffraction.

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3 years ago

Is a process that modifies light waves so they vibrate in a single plane

madam [21]

The process you're fishing for is "polarization", but that's a

misleading description.

Polarization doesn't do anything to change the light waves.

It simply filters out (absorbs, as with a polarizing filter) the

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The intensity of a light beam is always reduced after

polarizing it, because much (most) of the original light

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A laser light source may be thought of as an exception,

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8 0

4 years ago

A tugboat tows a ship with a constant force of magnitude F1. The increase in the ship's speed during a 10 s interval is 3.0 km/h

Yuki888 [10]

Answer:

The magnitude of F₁ is 3.7 times of F₂

Explanation:

Given that,

Time = 10 sec

Speed = 3.0 km/h

Speed of second tugboat = 11 km/h

We need to calculate the speed

$v_{1}=\dfrac{3.0\times10^{3}}{3600}$

$v_{1}=0.833\ m/s$

The force F₁is constant acceleration is also a constant.

$F_{1}=ma_{1}$

We need to calculate the acceleration

Using formula of acceleration

$a_{1}=\dfrac{v}{t}$

$a_{1}=\dfrac{0.833}{10}$

$a_{1}=0.083\ m/s^2$

Similarly,

$F_{2}=ma_{2}$

For total force,

$F_{3}=F_{2}+F_{1}$

$ma_{3}=ma_{2}+ma_{1}$

The speed of second tugboat is

$v=\dfrac{11\times10^{3}}{3600}$

$v=3.05\ m/s$

We need to calculate total acceleration

$a_{3}=\dfrac{v}{t}$

$a_{3}=\dfrac{3.05}{10}$

$a_{3}=0.305\ m/s^2$

We need to calculate the acceleration a₂

$0.305=a_{2}+0.083$

$a_{2}=0.305-0.083$

$a_{2}=0.222\ m/s^2$

We need to calculate the factor of F₁ and F₂

Dividing force F₁ by F₂

$\dfrac{F_{1}}{F_{2}}=\dfrac{m\times0.83}{m\times0.22}$

$\dfrac{F_{1}}{F_{2}}=3.7$

$F_{1}=3.7F_{2}$

Hence, The magnitude of F₁ is 3.7 times of F₂

3 0

3 years ago

A hydrogen atom in a galaxy moving with a speed of 6.65×106 m/???? away from the Earth emits light with a wavelength of 5.13×10−

Mumz [18]

Answer:

The observed wavelength on Earth from that hydrogen atom is $5.24\times 10^{-7}\ m$ .

Explanation:

Given that,

The actual wavelength of the hydrogen atom, $\lambda_a=5.13\times 10^{-7}\ m$

A hydrogen atom in a galaxy moving with a speed of, $v=6.65\times 10^6\ m/s$

We need to find the observed wavelength on Earth from that hydrogen atom. The speed of galaxy is given by :

$v=c\times \dfrac{\lambda_o-\lambda_a}{\lambda_a}$

$\lambda_o$ is the observed wavelength

$\lambda_o=\dfrac{v\lambda_a}{c}+\lambda_a\\\\\lambda_o=\dfrac{6.65\times 10^6\times 5.13\times 10^{-7}}{3\times 10^8}+5.13\times 10^{-7}\\\\\lambda_o=5.24\times 10^{-7}\ m$

So, the observed wavelength on Earth from that hydrogen atom is $5.24\times 10^{-7}\ m$ . Hence, this is the required solution.

8 0

3 years ago

A wheel has a constant angular acceleration of 4.5 rad/s2. during a certain 5.0 s interval, it turns through an angle of 128 rad

dalvyx [7]

The solution for this problem is through this formula:Ø = w1 t + 1/2 ã t^2
where:Ø - angular displacement w1 - initial angular velocity t - time ã - angular acceleration
128 = w1 x 4 + ½ x 4.5 x 5^2 128 = 4w1 + 56.254w1 = -128 + 56.25 4w1 = 71.75w1 = 71.75/4
w1 = 17.94 or 18 rad s^-1
w1 = wo + ãt
w1 - final angular velocity
wo - initial angular velocity
18 = 0 + 4.5t t = 4 s

3 0

4 years ago