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lana66690 [7]
3 years ago
10

The pressure exerted by a liquid column of depth 0.5 m on the base of its

Physics
1 answer:
Alik [6]3 years ago
4 0

Answer:

1020.4kg/m³

35m³

Explanation:

Given parameters:

Depth  = 0.5m

Pressure exerted  = 5000N/m²

Unknown = density of the liquid  = ?

Solution:

To find the density of this unknown liquid, we use the expression below:

       Pressure of liquid  = density x height x acceleration due to gravity

   5000 = density x 0.5 x 9.8

    Density  = 1020.4kg/m³

B.

Density of water  = 1000kg/m³

Mass of water  = 35000kg  

 Unknown:

Volume of water  = ?

Solution:

 The volume of water can be derived from the expression below:

   Volume  = \frac{mass}{density}  

  Volume  = \frac{35000}{1000}    = 35m³

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A ball is thrown upward from the ground with an initial speed of 25 m/s; at the same instant, another ball is dropped from a bui
Nimfa-mama [501]

Answer:

t = o.6 s

Explanation:

Let ball thrown from below be A and ball dropped from above be B.

A and B meet when they both are same level above the ground. Then let A moved up a distance d and B dropped a distance h. Then you know

d + h = 15 m  ---------------(1)

Now apply s = ut + \frac{1}{2}at²

To A upwards,

d = 25t - \frac{1}{2}gt²  -----------------(2)

To B downwards,

h = 0 + \frac{1}{2}gt² ----------------(3)

(1) = (2) + (3) ⇒ 15 = 25t

                         t = 0.6 s

5 0
3 years ago
B. The coefficient of friction between the tires and the road is 0.850 and the mass of the car is
Mrac [35]

Answer:

156.26N

Explanation:

The data needed are incomplete. Let the acceleration of the body be 3.5m/s²

Other given parameters

Mass = 1.35×10^1 = 13.5kg

coefficient of friction between the tires and the road = 0.850

Acceleration due to gravity = 9.8m/s²

According to Newton's second law:

Fnet = ma

Fnet = Fapp - Ff

Fapp is the applied force

Ff is the frictional force = umg

The equation becomes:

Fapp - Ff = ma

Fapp-umg = ma

Fapp - 0.85(13.5)(9.8) = 13.5(3.5)

Fapp - 109.0125 = 47.25

Fapp = 47.25+109.0125

Fapp = 156.2625N

Hence the applied force that caused the acceleration is 156.26N

Note that the acceleration of the car was assumed. Any value of acceleration can be used for the calculation.

8 0
3 years ago
6. What is the charge of an object that lost 1.20x1024 electrons?
Montano1993 [528]

Answer:

Hydrogen ions contains +1 charge while sulphate ions have 2 charge. ... No .of ions in one formula unit of MgC12 =3 Total no of ions in MgC12 =1.20x1024 x3=3.6x1024 ions.

Explanation:

8 0
3 years ago
Find the ratio of the diameter of aluminium to copper wire, if they have the same
kicyunya [14]

Answer:

1.24

Explanation:

The resistivity of copper\rho_1=2.65\times 10^{-8}\ \Omega-m

The resistivity of Aluminum,\rho_2=1.72\times 10^{-8}\ \Omega-m

The wires have same resistance per unit length.

The resistance of a wire is given by :

R=\rho \dfrac{l}{A}\\\\R=\rho \dfrac{l}{\pi (\dfrac{d}{2})^2}\\\\\dfrac{R}{l}=\rho \dfrac{1}{\pi (\dfrac{d}{2})^2}

According to given condition,

\rho_1 \dfrac{1}{\pi (\dfrac{d_1}{2})^2}=\rho_2 \dfrac{1}{\pi (\dfrac{d_2}{2})^2}\\\\\rho_1 \dfrac{1}{{d_1}^2}=\rho_2 \dfrac{1}{{d_2}^2}\\\\(\dfrac{d_2}{d_1})^2=\dfrac{\rho_1}{\rho_2}\\\\\dfrac{d_2}{d_1}=\sqrt{\dfrac{\rho_1}{\rho_2}}\\\\\dfrac{d_2}{d_1}=\sqrt{\dfrac{2.65\times 10^{-8}}{1.72\times 10^{-8}}}\\\\=1.24

So, the required ratio of the diameter of Aluminum to Copper wire is 1.24.

3 0
3 years ago
Hiran is standing beside the road when he hears a bird flying away from hip and chirping. The bird’s chirp has a frequency of 18
Murrr4er [49]

The frequency of bird chirping hear by hiran will be 1.77 kHz.

<u>Explanation:</u>

As per Doppler effect, the observer will feel a decrease in the frequency of the receiving signal if the source is moving away from the observer. So the shifted frequency is obtained using the below equation:

f'=\frac{c}{c+v_{s} }f

Here , c is the speed of sound, Vs is the velocity of source with which it is moving away. f is the original frequency of source and f' is the frequency shift heard by the observer.

As here, f = 1800 Hz, Vs= 6 m/s and c = 343 m/s, then

f'=\frac{343}{343+6} \times 1800=1.77\ kHz

So, the frequency of bird chirping hear by hiran will be 1.77 kHz.

4 0
3 years ago
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