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Naya [18.7K]
3 years ago
15

Water inside a tub rises from 15 mL to 19 mL when an object is submerged. If the mass of the object is 249, what's the density o

f
the object?
O A.6 g/ml
O B. 0.8 g/mL
O C.4g/ml
O D. 0.12 g/mL
Physics
1 answer:
Alex_Xolod [135]3 years ago
6 0

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume}  \\

From the question

mass = 249 g

volume = final volume of water - initial volume of water

volume = 19 - 15 = 4 mL

We have

density =  \frac{249}{4}  \\

We have the final answer as

<h2>62.25 g/mL</h2>

Hope this helps you

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4. Derive<br>the relation, P= hd g​
Sergeu [11.5K]

p=F/A

or,P=d×V×G/A (m=d×V)

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or,P=d×H×G

4 0
3 years ago
Two electrons with a charge of magnitude 1.6×10-19 C in an atom are separated by 1.5×10-10 m, the typical size of an atom. What
vesna_86 [32]

Answer:

1.02\cdot 10^{-8} N, repulsive

Explanation:

The magnitude of the electric force between two charged particles is given by Coulomb's law:

F=k\frac{q_1 q_2}{r^2}

where:

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q_1, q_2 are the two charges of the two particles

r is the separation between the two charges

The force is:

- repulsive if the two charges have  same  sign

- Attractive if the two charges have opposite signs

In this problem, we have two electrons, so:

q_1=q_2=1.6\cdot 10^{-19}C is the magnitude of the two electrons

r=1.5\cdot 10^{-10} m is their separation

Substituting into the formula, we find the electric force between them:

F=(8.99\cdot 10^9)\frac{(1.6\cdot 10^{-19})^2}{(1.5\cdot 10^{-10})^2}=1.02\cdot 10^{-8} N

And the force is repulsive, since the two electrons have same sign charge.

4 0
3 years ago
Water at room temperature is discharged from a pipe at a rate of 1000 gallons per minute (gpm). Express this flow rate in cubic
marshall27 [118]

Answer

given,

discharge rate from pipe = 1000 gallons/minutes

now,

flow rate in  cubic meters per second

1 gallon = 0.00378541 m³

1 min = 60 s

Q = 1000\times \dfrac{0.00378541\ m^3}{1\ gallon}\times \dfrac{1\ min}{60\ s}

Q = 0.063 m³/s

flow rate in  liters per minute

1 gallon = 3.78541 L

 Q = 1000\times \dfrac{3.78541\ m^3}{1\ gallon}

 Q = 3785.41 m³/min

flow rate in cubic feet per second

 1 gallon = 0.133681 ft³

 1 min = 60 s

Q = 1000\times \dfrac{0.133681\ ft^3}{1\ gallon}\times \dfrac{1\ min}{60\ s}

Q = 2.23 ft³/s

4 0
3 years ago
A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.05 104 m/s. What will its speed be wh
Elanso [62]

Answer:

The value is  v  =  2.3359 *10^{4} \ m/s

Explanation:

From the question we are told that

  The  initial speed is u =  2.05 *10^{4} \  m/s

 Generally the total energy possessed by the space probe when on earth is mathematically represented as

             T__{E}} =  KE__{i}} +  KE__{e}}

Here  KE_i is the kinetic energy of the space probe due to its initial speed which is mathematically represented as

          KE_i =   \frac{1}{2}  *  m  *  u^2

=>       KE_i =   \frac{1}{2}  *  m  *  (2.05 *10^{4})^2

=>       KE_i =  2.101 *10^{8} \ \ m \ \ J

And  KE_e is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

       KE_e =  \frac{1}{2}  *  m *  v_e^2

Here v_e is the escape velocity from earth which has a value v_e =  11.2 *10^{3} \  m/s

=>    KE_e =  \frac{1}{2}  *  m *  (11.3 *10^{3})^2

=>    KE_e =  6.272 *10^{7} \  \  m  \ \   J

Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as

        KE_p =  \frac{1}{2}  *  m *  v^2

Generally from the law energy conservation we have that

        T__{E}} =  KE_p

So

       2.101 *10^{8}  m  +  6.272 *10^{7}  m  =   \frac{1}{2}  *  m *  v^2

=>     5.4564 *10^{8} =   v^2

=>     v =  \sqrt{5.4564 *10^{8}}

=>     v  =  2.3359 *10^{4} \ m/s

4 0
3 years ago
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