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WITCHER [35]
3 years ago
13

the density of a steel is 7.8.find the mass of steel cube of side 10 cm. find volume of steel if the mass is 8kg​

Physics
1 answer:
lukranit [14]3 years ago
3 0

Answer:

Mass of the steel cube =  7800 kg

Volume of the steel  = 1.025 cubic centimetre

Explanation:

Given:

The density of the steel   =  7.8

Side of the cube = 12 cm

<u>(1)The mass of steel cube :</u>

We know that,

Density  = \frac{Mass}{Volume}

We are given with density and sides of the cube

then volume of the cube

= (side)^3

= 10^3

= 1000 cubic centimetre

Now

7.8 = \frac{mass}{1000}

mass = 7.8 \times 1000

mass =  7800 kg

<u>(2)volume of steel:</u>

Given the mass  = 8 kg

Density  = \frac{Mass}{Volume}

Substituting the values

7.8 = \frac{8}{volume}

volume = \frac{8}{7.8}

volume = 1.025 cubic centimetre

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marishachu [46]
The sun is just one of the uncountable stars in the universe, as there are more than billions of galaxies containing billions of stars. 
3 0
3 years ago
f the CD rotates clockwise at 500 rpmrpm (revolutions per minute) while the last song is playing, and then spins down to zero an
vazorg [7]

Answer:

The magnitude of the angular acceleration is  a =  20.14 rad/s^2

Explanation:

From the question we are told that

   The angular speed of CD is  w_{CD} =  500 rpm =  \frac{500  rpm}{\frac{60 \ s }{1 \ min} } * \frac{2 \pi }{ 1 \ rev} = 52.37 rad/s

    time taken to decelerate is t_{CD} =  2.60\ s

    The final angular speed is  w_f= 0 \ rad/s

The angular acceleration is mathematically represented as

         a =  \frac{w_f - w_{CD}}{t}

substituting values

          a =  \frac{0 - 52.37}{2.60}

         a =  - 20.14 rad/s^2

The negative sign show that the CD is decelerating  but the magnitude is

       a =  20.14 rad/s^2

   

3 0
3 years ago
A permanent magnet is pushed into a wire, left there for a while, and then pulled out. During which time does a current run thou
lakkis [162]

Answer:

C. while the magnet is moving

Explanation:

Electromagnetic induction implies the production of electric current by mere movement of a magnet with respect to a coil or wire.

In the given question, current would be induced in the wire only when the magnet moves. That is either when the magnet is pushed into a wire, or when pulled out. But no current would flow through the wire when the magnet is left there for a while.

The current is induced because of the motion involved. Thus, the appropriate option is C.

8 0
3 years ago
A crate rests on a flatbed truck which is initially traveling at 17.9 m/s on a level road. The driver applies the brakes and the
Mamont248 [21]

Answer:

The minimum coefficient of friction required is 0.35.  

Explanation:

The minimum coefficient of friction required to keep the crate from sliding can be found as follows:

-F_{f} + F = 0      

-F_{f} + ma = 0      

\mu mg = ma

\mu = \frac{a}{g}

Where:

μ: is the coefficient of friction

m: is the mass of the crate

g: is the gravity

a: is the acceleration of the truck

The acceleration of the truck can be found by using the following equation:

v_{f}^{2} = v_{0}^{2} + 2ad

a = \frac{v_{f}^{2} - v_{0}^{2}}{2d}

Where:  

d: is the distance traveled = 46.1 m

v_{f}: is the final speed of the truck = 0 (it stops)      

v_{0}: is the initial speed of the truck = 17.9 m/s

a = \frac{-(17.9 m/s)^{2}}{2*46.1 m} = -3.48 m/s^{2}        

If we take the reference system on the crate, the force will be positive since the crate will feel the movement in the positive direction.  

\mu = \frac{a}{g}  

\mu = \frac{3.48 m/s^{2}}{9.81 m/s^{2}}

\mu = 0.35

Therefore, the minimum coefficient of friction required is 0.35.  

I hope it helps you!

4 0
3 years ago
Ksp for agbr is 5x10-13. what is the maximum concentration of silver ion that you can have in a 0.1 m solution of nabr?
liberstina [14]

Answer : The maximum concentration of silver ion is 5\times 10^{-12}m

Solution : Given,

K_{sp} for AgBr = 5\times 10^{-13}

Concentration of NaBr solution = 0.1 m

The equilibrium reaction for NaBr solution is,

NaBr(aq)\rightleftharpoons Na^++Br^-

The concentration of NaBr solution is 0.1 m that means,

[Na^+]=[Br^-]=0.1m

The equilibrium reaction for AgBr is,

                          AgBr\rightleftharpoons Ag^++Br^-

At equilibrium                     s       s

The expression for solubility product constant for AgBr is,

K_{sp}=[Ag^+][Br^-]

The concentration of Ag^+ = s

The concentration of Br^- = 0.1 + s

Now put all the given values in K_{sp} expression, we get

5\times 10^{-13}=(s)(0.1+s)

By rearranging the terms, we get the value of 's'

s=5\times 10^{-12}m

Therefore, the maximum concentration of silver ion is 5\times 10^{-12}m.

4 0
3 years ago
Read 2 more answers
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