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labwork [276]
3 years ago
14

3. Take sugar, oil, corn syrup, a glass and water. Pour the water in the glass and then add each of the above the substances one

after the other. Observe and note your observations. And give reason for your observation.
Physics
1 answer:
hoa [83]3 years ago
7 0

Here are the observations

<u>S</u><u>u</u><u>g</u><u>a</u><u>r</u><u>:</u><u>-</u>

  • Sugar is soluble in water
  • so It will dissolve in water .

<u>C</u><u>o</u><u>r</u><u>n</u><u> </u><u>s</u><u>y</u><u>r</u><u>u</u><u>p</u><u>:</u><u>-</u>

  • Corn syrup is also basically a sugar.
  • It will dissolve in water too .
  • If we shake the mixture in glass then corn syrup will be dissolved.

<u>O</u><u>i</u><u>l</u><u>:</u><u>-</u>

  • Oil is not soluble in water
  • Hence it won't dissolve in water.
  • It will float over water and make two layers
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ListenA bicycle and its rider have a combined mass of 80. kilograms and a speed of 6.0 meters per second. What is the magnitude
Setler [38]

Answer:

a) 1.2\times 10^2\ N

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

a = Acceleration

v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{0-6}{4}\\\Rightarrow a=-1.5\ m/s^2

The acceleration of the bicycle and rider is -1.5 m/s²

Force

F=ma\\\Rightarrow F=80\times -1.5\\\Rightarrow F=-120\ N=-1.2\times 10^2\ N

The magnitude of the average force needed to bring the bicycle and its rider to a stop is 1.2\times 10^2\ N

3 0
3 years ago
________ is a measure of how many waves pass by in one second.
Sergio [31]
C. Frequency is a measure of how many waves pass by in one second.
4 0
3 years ago
2) Two railway tracks are parallel to west east direction. Along one track, train A moves with a speed of 45 m/s from
Natalija [7]

Answer:

105 m/s

Explanation:

Given that the speed of train A, V_A = 45 m/s from west to east.

Speed of train B, V_B = 60 m/s from east to west.

Train B is moving in the opposite direction with respect to the speed of train A. Assuming that the speed from east to west direction is positive.

So, the speed of train A from east to west= - 45 m/s

The speed of train B w.r.t train A = V_B - V_A=60-(-45)=60+45=105 m/s

Hence, the speed of train B w.r.t train A is 105 m/s from east to west.

5 0
3 years ago
A gray kangaroo can bound across level ground with each jump carrying it 9.1 m from the takeoff point. Typically the kangaroo le
Alona [7]

Answer:

u = 10.63 m/s

h = 1.10 m

Explanation:

For Take-off speed ..

by using the standard range equation we have

R = u² sin2θ/g

R = 9.1 m

θ = 26º,

Initial velocity = u

solving for u

u² = \frac{Rg}{sin2\theta}

u^2 = \frac{9.1 x 9.80}{sin26}

u^2 = 113.17

u = 10.63 m/s

for Max height

using the standard h(max) equation ..

v^2 = (v_osin\theta)^2 -2gh

h =\frac{(v_o^2sin\theta)^2}{2g}

h  =  \frac{(113.17)(sin26)^2}{(2 x 9.80)}}

h = 1.10 m

7 0
3 years ago
g The electric power needs of a community are to be met by windmills with 40-m-diameter rotors. The windmills are to be located
Ksenya-84 [330]

Answer:

Explanation:

Given Data

The diameter of the wind mills is d = 40m

Velocity of the air is V = 6 m / s

Required power output is:  P ₀ = 2100 k W

Expression to calculate the exergy of the air is

E = V ² / 2

Substitute the value in above expression

E = ( 6 m / s ) ² / 2

E = 18 m ² / s ² x (1kJ/kg / 1000m²/s²)

E = 0.018 k J / k g

Expression to calculate the density of the air is

P v =m R T

m /v = P  /RT ⋯ ⋯( I )

Here  

m  is the mass of the air,  

v  is the volume of the air,  

P  is the atmospheric pressure,  

T  is the standard temperature at the atmospheric pressure and  

R  is the gas constant

As the density is

ρ = m /V

Substitute the value in expression (I)

ρ = 101  kP a /( 0.287 k J / k g ⋅ K ) ( 298 K )

ρ = 1.180 k g / m ²

Expression to calculate the mass flow rate is

m = ρ A V ⋯ ⋯ ( I I )

Here  A  is the area of the windmill

Expression to calculate the  A  is

A = π /4  d ²

Substitute the value in above expression

A = π /4 ( 40 m ²)

A = 1256.63 m ²

Substitute the value in expression (II)

m = ( 1.180 k g / m ³) ( 1256.63 m ²) ( 6 m / s )

m = 8896.94  k g / s

Expression to calculate the maximum power available to the windmill is

P w = m ( V ² /2 )

Substitute the value in above expression

P w = 8896.94  k g / s ( (6m/s)²/2 )

P w = 160144.92 W  × ( 1 W /1000 k W )

P w = 160.144 k W

Expression to calculate the number of windmills required is

n = P o /P w

Substitute the value in above expression

n=2100kw/160.144kw

n=13.11

8 0
3 years ago
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