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pychu [463]
3 years ago
11

The specific gravity of a substance that has mass of 10 kg and occupies a volume of 0.02 m^3 is a) 0.5 b) 1.5 c) 2.5 d) 3.5 e) n

one of these
Engineering
1 answer:
zhuklara [117]3 years ago
5 0

Answer:

Specific gravity is 0.5 so option (a) is correct option

Explanation:

We have given mass of substance m =10 kg

Volume V=0.02m^3

Density of substance is given by Density\ d=\frac{mass}{volume}=\frac{10}{0.02}=500kg/m^3

Specific gravity is given by specific\ gravity=\frac{density\ of\ substance}{density\ f\ water}=\frac{500}{1000}=0.5

So option (a) is correct option

You might be interested in
The two boxcars A and B have a weight of 20000lb and 30000lb respectively. If they coast freely down the incline when the brakes
kolbaska11 [484]

Answer:

T=5.98 kips

Explanation:

First, introduce forces, acting on both cars:

on car A there are 4 forces acting: gravity force mA*g, normal reaction force, friction force and force T- it represents the interaction between cars A and B. On car B, there are three forces acting: gravity force, normal reaction force and force T. Note, that force T is acting on both cars, but it has opposite direction. Force T, acting on car A has direction, opposite to the friction force, whether the T, acting on B, is directed backwards- in the same direction with the friction force. Note, that both cars have the same acceleration, which is directed backwards.

Once the forces were established, we can write components of the Second Newtons Law on vertical and horizontal axes, considering that horizontal axis is directed backwards- in the same direction with the acceleration:

For car A on the vertical axis the equation is: -mAg+NA=0

For car A on the horizontal axis, the equation is: Ffr-T=mAa

For car B, on the vertical axis the equation is: -mBg+NB=0

For car B, on the horizontal axis, the equation is: T=mBa

We need to solve these equations to find force T, knowing that Ffr=μmAg, where

After the transformations, the equations for acceleration and force in the coupling will be:

a=(μmAg)/(mA+mB)=6.43 ft/s2- note, that the given answer is not correct for the given numerical values;

and force T: T=μmAmBg/(mA+mB)=6.0 kips- note, that the force answer is in line with the given numerical value

5 0
4 years ago
a horizontal jet of water (at 100c) that is 6 cm in diameter and has a velocity of 20 m/s is deflected by the vane as shown. if
Reptile [31]

The net resultant direct force and angle on the vane is created when the water jet exits the vane at position 2 with 92% of its initial velocity.

<h3>What is mean by velocity?</h3>
  • The speed at which a body or object is moving determines its direction of motion. A scalar quantity, speed is primarily. As a matter of fact, velocity is a vector quantity.
  • The rate at which distance changes is what it is. It measures the displacement's rate of change. A body's velocity is defined as its speed in a particular direction.
  • Velocity is a measure of how quickly a distance changes in relation to time. Having both magnitude and direction, velocity is a vector quantity.
  • The rate of change in a body's displacement with respect to time is referred to as velocity. In the SI, m/s is its unit.

Given,

External angle of Curved Vane = 158°

mean velocity at 1 = 12 m/s

Volumetric flow rate = 55 \mathrm{~m}^3 / \mathrm{h}=\frac{55}{3600} \mathrm{~m}^3 / \mathrm{s}$.

mean velocity at $2=12 \times 0.92=11.04 \mathrm{~m} / \mathrm{s}$

i) Force exerted in x - friction A C 1 =  Volume

F_{S_x} &=\rho A C_1\left[C_2 \cos \theta-C_1\right] \\

&=1000 \times \frac{55}{3600}\left[\left(11.04 \cos 158^{\circ}\right)-12\right]

i\rangle F_{\text {sc }}=\supseteq A c_1\left[C_2 \sin \theta\right] \\

&=1000 \times \frac{55}{3600} \times \text { TI. 04 } \sin (1589 \\

&F_{\text {syn }}=63.18 \mathrm{~N} \\

&\text { Angle } \Rightarrow \frac{F_{s y}}{F_{3 x}}-\tan \theta \\

&\tan \theta=\frac{63.18}{339.18}, \theta=160-10-5.3 \\

&\theta=\tan ^{-}\left(\frac{-63 \cdot 18}{339728}\right) \\

&\theta=-10.540^{\circ} \\

The complete question is:

A horizontal jet of water strikes a curved vane as shown in Figure C.1. The external angle of the curved vane is 158°.The mean velocity and volumetric flow rate of the water jet at position 1 are 12 m/s and 55 m³/h respectively. Due to friction, the water jet leaves the vane at position 2 with 92 % its original velocity.

(i) Direct force exerted by the water jet on the vane in the x - direction.

(ii) Direct force exerted by the water jet on the vane in the y - direction.

(ii) Net resultant direct force and angle on the vane.

To learn more about velocity, refer to:

brainly.com/question/24681896

#SPJ4

4 0
2 years ago
Efficiency of___<br>motor is high​
Vsevolod [243]

Explanation: three phase induction

4 0
3 years ago
a) The reverse-saturation current of a pn junction diode is IS = 10−11 A. Determine the diode voltage to produce currents of (i)
creativ13 [48]

Answer:

a) i) Vd = 358 mV , Vd = 417.5 mV, Vd = 477.2 mV

   ii) Vd = -17.9 mV

b) Vd = 477.2 mV, Vd = 536.8 mV, Vd = 596.5 mV

   Vd = 161 mV

Explanation:

We will use the <u>ideal diode equation</u>:

I_{D} = I_{s}  (e^{\frac{qv_{d} }{nkT} }-1)

Where Id = Diode current

           Is = Reverse Saturation Current

           Vd = Diode Voltage

           q = charge of electron

           n = ideality factor

           k = Boltzmann constant

           T = Temperature in Kelvin

(a) We are given:

     Is = 10⁻¹¹ A

i) <u>Id = 10 μA</u>

Ideally, kT/q = 25.9mV and n=1. So,

q/nkT = 38.6

Id = Is e^(38.6Vd) -1

Id/Is + 1 = e^(38.6Vd)

ln(Id/Is + 1) = 38.6Vd

Vd = ln(Id/Is + 1)/38.6

     = ln [(10*10⁻⁶)/10⁻¹¹ + 1] / 38.6

     = 13.815/38.6

Vd = 0.3579

Vd = 358 mV

<u>Id = 100 µA</u>

Vd = ln(Id/Is + 1)/38.6

     = ln [(100*10⁻⁶)/10⁻¹¹ + 1] / 38.6

     = 16.118/38.6

Vd = 0.4175

Vd = 417.5 mV

<u>Id = 1 mA</u>

Vd = ln(Id/Is + 1)/38.6

     = ln [(1*10⁻³)/10⁻¹¹ + 1] / 38.6

    = 18.420/38.6

Vd = 0.4772

Vd = 477.2 mV

ii) Id = -5 x 10⁻¹² A

   Vd = ln(Id/Is + 1)/38.6

         = ln [(-5 x 10⁻¹²)/10⁻¹¹ + 1] / 38.6

         = -0.693/38.6

   Vd = -0.0179 V

   Vd = -17.9 mV

(b) Is = 10⁻¹³ A

i) <u>Id = 10 µA</u>

Vd = ln(Id/Is + 1)/38.6

     = ln [(10 x 10⁻⁶)/10⁻¹³ + 1] / 38.6

     = 18.42/38.6

Vd = 0.4772 V

Vd = 477.2 mV

<u>Id = 100 µA</u>

Vd = ln(Id/Is + 1)/38.6

     = ln [(100 x 10⁻⁶)/10⁻¹³ + 1] / 38.6

     = 20.723/38.6

     = 0.5368 V

Vd = 536.8 mV

<u>Id = 1 mA</u>

Vd = ln(Id/Is + 1)/38.6

     = ln [(1 x 10⁻³)/10⁻¹³ + 1] / 38.6

     = 23.025/38.6

     = 0.5965

Vd = 596.5 mV

ii) Id = -5 x 10⁻¹² A

   Vd = ln(Id/Is + 1)/38.6

         = ln [(-5 x 10⁻¹²)/10⁻¹³ + 1] / 38.6

         = ln (-49) / 38.6

     Not possible.

<u>Is = -10⁻¹⁴</u>

<u>Id = -5 x 10⁻¹² A</u>

Vd = ln(Id/Is + 1)/38.6

         = ln [(-5 x 10⁻¹²)/-10⁻¹⁴ + 1] / 38.6

         = 6.216/38.6

Vd = 0.161 V

Vd = 161 mV

4 0
4 years ago
A 21-mm-diameter steel bar is to be used as a torsion spring. If the torsional stress in the bar is not to exceed 109 MPa when o
Daniel [21]

Answer:

2m

Explanation:

The modulus of rigidity was omitted in the question and I'm using the modulus of rigidity of structural steel (G=79 GPa)

For the shaft, the maximum moment

T=π×rho×d³/16

Where π=3.14

Rho= torsional stress=109MPa

d=diameter of of shaft= 21mm

Inserting values of π, rho and d

T= 198,230.54Nmm

Polar moment of the shaft is given by

J= π D⁴/ 32

Also torsional moment (in radian) of the shaft is given as

α = L ×T / (J× G)

α = 32× L×T / (G× π× D⁴)

And in degrees,

αdegrees = 584× L×T / (G×D⁴)

Where L=length of shaft and G is modulus of rigidity

L=α×d⁴×G/584×T

Where α=15⁰, d=0.021m, G=79×10⁹Pa, T=198.23Nm

Substituting values for α, d, G and T

L=1.99m approximately 2m

5 0
3 years ago
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