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sweet [91]
3 years ago
8

SOMEONE HELP ME PLEASE!!!!!

Physics
1 answer:
Allushta [10]3 years ago
4 0

Explanation:

HEY PLS DON'T JOIN THE ZOOM CALL OF A PERSON WHO'S ID IS 825 338 1513 (I'M NOT SAYING THE PASSWORD) HE IS A CHILD PREDATOR AND A PERV. HE HAS LOTS OF ACCOUNTS ON BRAINLY BUT HIS ZOOM NAME IS MYSTERIOUS MEN.. HE ASKS FOR GIRLS TO SHOW THEIR BODIES AND -------- PLEASE REPORT HIM IF YOU SEE A QUESTION LIKE THAT. WE NEED TO TAKE HIM DOWN!!! PLS COPY AND PASTE THIS TO OTHER COMMENT SECTIONS!!

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12 g of coal contains approximately 6 × 1023 atoms of carbon.
Phantasy [73]

Answer:

5*10^25 atoms of carbon

Explanation:

<u>Check from the periodic table:</u>

M(carbon) = 12.01 g/mol

<u>Convert kg into g:</u>

1 kg = 1000 g

<u>Calculate the number of moles of carbon:</u>

n(carbon) = (1000 g) ÷ (12.01 g/mol) = 83.26 mol

<u>Convert moles into number of particles:</u>

#(carbon) = n × NA = 83.26 mol × 6*10^23

= 5*10^25 particles

<em>N</em><em>o</em><em>t</em><em>e</em><em>:</em><em> </em><em>NA represents the Avogadro's Number, which has the value 6.02*10^23 particles/mol</em><em>.</em>

3 0
4 years ago
The distance covered by a car moving with a speed of 36 km/h in 20 minutes is
Hunter-Best [27]

Answer:

D=12000m

Explanation:

Distance

Speed=distance/time

Where, speed is given in m/s

Time is given in s

Distance is given in m

Data

S=36km/h

T=20minutes

D=?

D=S*T

D=(36*1000m)/3600s*(20*60s)

D=36000m/3600s*1200s

D=10m/s*1200s

D=12000m

5 0
3 years ago
1A.)A railroad car of mass 21700 kg moving with a speed of 3.36 m/s collides
Pepsi [2]

Answer:

(a) 2.03 m/s

(b) 2.15 m/s

Explanation:

(a)

From the law of conservation of momentum, the sum of initial momentum equals the sum of final momentum

Momentum, p=mv where m is the mass and v is the velocity

m_1v_1+m_2v_2=(m_1+m_2)v_c where v_c is the common velocity, v_1 and v_2 are velocities of the first railroad car  and the second railroad car respectively, m_1 and m_2 are masses of the first railroad car  and the second railroad car respectively

Substituting the given figures then

21700\times 3.36+41700\times 1.34=(21700+41700)v_c\\63400v_c=128790\\v_c=\frac {128790}{63400}=2.031388013\approx 2.03 m/s

(b)

Momentum, p=mv where m is the mass and v is the velocity

m_1v_1+m_2v_2=m_1v_3+m_2v_4 where v_1 and v_2 are initial velocities of the ball moving towards west and the other ball moving in opposite direction respectively, m_1 and m_2 are masses of the first railroad car  and the second railroad car respectively

Taking west as positive then the opposite direction will be negative hence

0.455\times 7.36+(1.01\times -12.4)= 0.455\times -15.4 + (1.01\times v_4}

Note that when the 0.455 rebounds, it moves towards East and since we took West as positive that is why we give 15.4 as a negative value.

-2.1682=1.01v_4\\v_4=\frac {-2.1682}{1.01}=-2.146732673\approx 2.15 m/s towards East

8 0
4 years ago
Bro i need help omg.
sergey [27]

Answer:

25.

Explanation:

8 0
3 years ago
Read 2 more answers
A large, 68.0-kg cubical block of wood with uniform density is floating in a freshwater lake with 20.0% of its volume above the
LenaWriter [7]

Answer:

a) V = 0.085 m^3

b) m = 17 kg

Explanation:

1) Data given

mb = 68 kg (mass for the block)

20% of the block volume is floating

100-20= 80% of the block volume is submerged

2) Notation

mb= mass of the block

Vw= volume submerged

mw = mass water displaced

V= total volume for the block

3) Forces involved (part a)

For this case we have two forces the buoyant force (B), defined as the weight of water displaced acting upward and the weight acting downward (W)

Since we have an equilibrium system we can set the forces equal. By definition the buoyant force is given by :

B = (mass water displaced) g = (mw) g   (1)

The definition of density is :

\rho_w = \frac{m_w}{V_w}

If we solve for mw we got m_w = \rho_w V_w  (2)

Replacing equation (2) into equation (1) we got:

B = \rho_w V_w g (3)

On this case Vw represent the volume of water displaced = 0.8 V

If we replace the values into equation (3) we have

0.8 ρ_w V g = mg  (4)

And solving for V we have

 V =  (mg)/(0.8 ρ_w g )

We cancel the g in the numerator and the denominator we got

V = (m)/(0.8 ρ_w)

V = 68kg /(0.8 x 1000 kg/m^3) = 0.085 m^3

4) Forces involved (part b)

For this case we have bricks above the block, and we want the maximum mass for the bricks without causing  it to sink below the water surface.

We can begin finding the weight of the water displaced when the block is just about to sink (W1)

W1 = ρ_w V g

W1 = 1000 kg/m^3 x 0.085 m^3 x 9.8 m/s^2 = 833 N

After this we can calculate the weight of water displaced before putting the bricks above (W2)

W2 = 0.8 x 833 N = 666.4 N

So the difference between W1 and W2 would represent the weight that can be added with the bricks (W3)

W3 = W1 -W2 = 833-666.4 N = 166.6 N

And finding the mass fro the definition of weight we have

m3 = (166.6 N)/(9.8 m/s^2) = 17 Kg

8 0
3 years ago
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