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Lelu [443]
3 years ago
9

A granite monument has a volume of 25,365.4 cm. The density of granite is 2.7 g/cms. Use this information to calculate the mass

of the monument to the nearest tenth. The mass of the granite monument is​
Physics
1 answer:
olchik [2.2K]3 years ago
8 0

Answer:

m = 68,486,6 g is the answer.

Explanation:

To calculate mass you use formula:

m= V*r

To avoid remembering this formula you can see the type of unit on each given variable. We can see that we have g/cm^3 and cm^3. If we multiply them, we negate cm^3 and cm^3 and we are left with g which is unit for mass.

Hope I helped! ☺

You might be interested in
Two cars have identical horns, each emitting a frequency of fs = 406 Hz. One of the cars is moving with a speed of 10.5 m/s towa
Archy [21]

Answer:

f_{B}=12.8 Hz

Explanation:

Let's start finding the frequency heard by the bystander due to the moving car. We need to use Doppler effect here:

f_{obs}=f_{s}\left(\frac{v}{v-v_{car}}\right)    

v is the speed of sound (v = 343 m/s)  

So, we have:

f_{obs}=406\left(\frac{343}{343-10.5}\right)=418.8 Hz                  

Now, the beat frequency heard by the bystander is the combine of the frequencies, it means the difference between them. Therefore the equation is given by:

f_{B}=f_{obs}-f_{s}=418.8-406=12.8 Hz    

I hope it helps you!

4 0
3 years ago
a boy pulls 23 kg box with a 100 N force at 39 above a horizontal surface if the coefficient of kinetic friciton between the box
tatuchka [14]

Answer:

Total work done is 2606.08 J.

Explanation:

Given :

Mass of box , m = 23 kg .

Force applied , F = 100 N .

Angle from horizon , \theta=39^o.

Coefficient of kinetic friction , \mu=0.16.

Distance travelled by box , d = 34 m .

Now ,

Total work done = work done by boy + work done by friction.

W=Fdcos\theta+f_sdcos\theta=Fd-\mu(mg) \\\\W=100\times cos39^o\times 34-0.16\times 23 \times 9.8 \\\\W=77.71\times 34-36.06=2606.08\ J.

Hence , this is the required solution.

6 0
3 years ago
A car's bumper is designed to withstand a 5.04-km/h (1.4-m/s) collision with an immovable object without damage to the body of t
aleksklad [387]

Answer:

3420.39 N

Explanation:

Applying,

Fd = 1/2(mv²-mu²)................. Equation 1

Where F = force on the bumber, d = distance, m = mass of the car, v = final velocity, u = initial velocity.

make F the subject of the equation

F = (mv²-mu²)/2d............... Equation 2

From the question,

Given: m = 890 kg, v = 0 m/s (to rest), u = 1.4 m/s, d = 0.255 m

Substitute these values into equation 2

F = [(890×0²)-(890×1.4²)]/(2×0.255)

F = -1744.4/0.51

F = -3420.39 N

The negative sign denotes that the force in opposite direction to the motion of the car.

5 0
3 years ago
Pls help me… I missed the whole lesson and I have no clue what to do-
galben [10]

Answer:

1) 50 facing towards the right

2) 150 facing right

3) 200 facing right

4) 0- no direction

5) 50- facing left

6) 50 facing right

Explanation:

forces in opposite directions and equal magnitudes counteract each other. in number 2 they face the same direction so they would just be added. in number 4 they oppose each other so would be subtracted

8 0
3 years ago
If the specific heat of a metal is 0.850 J/g °C, what is its atomic weight?
Liula [17]

As we know that with respect to oxygen atom taken as reference the product of atomic mass and specific heat of a metal will remain constant.

this product is equal to 0.38

so here we will say that let atomic mass of the metal is M

so with respect to oxygen atom its mass is given as

m = \frac{M}{16}

now we will have

\frac{M}{16} \times 0.850 = 0.38

now we will have

M = 7

so atomic mass of the metal is 7 g/mol

8 0
4 years ago
Read 2 more answers
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