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3 years ago

What effect does a surfactant have on the surface tension of water

Physics

1 answer:

SpyIntel [72]3 years ago

6 0

Answer:

It weakens its hydrogen bonds causing it to spread out over a surface.

Explanation:

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1. What are the 2 main categories of nonmetals?

Anettt [7]

the 2 main categories of nonmetals are REACTIVE NONMETALS an NOBLE GAS.

hope it helps

7 0

3 years ago

The acceleration due to gravity on the moon is about 1/6 of the acceleration due to gravity on the earth. A net force F acts hor

uysha [10]

Answer:

c. $a_m$

Explanation:

We are given that

Acceleration due to gravity on the moon= $a_m$

Acceleration due to gravity on the earth= $a_e$

$g_m=\frac{1}{6}g_e$

Net force due to am on an object on moon= $F_{net}=ma_m$

There is no friction and no drag force and there is no gravity involved

Then, the force acting on an object on earth= $F=ma_e$

$F=F_{net}$ (given)

$ma_m=ma_e$

$a_e=a_m$

Hence, option c is true.

3 0

3 years ago

A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum

s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle= $q=-2.74\times 10^{-6} C$

Kinetic energy of particle= $K_E=6.65\times 10^{-10} J$

Initial time= $t_1=6.36 s$

Final potential difference= $V_2=0.351 V$

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

$qV=K.E$

Using the formula

$2.74\times 10^{-6}V_1=6.65\times 10^{-10} J$

$V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V$

Initial voltage= $V_1=2.43\times 10^{-4} V$

$\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2$

Using the formula

$\frac{V_1}{V_2}=(\frac{6.36}{t})^2$

$\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}$

$t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}$

$t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}$

$t=241.7 s$

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0

3 years ago

A truck on a straight road starts from rest, accelerating at 2.00m/s^2 until it reaches a speed of 20.0m/s. Then the truck trave

Dennis_Churaev [7]

Answer:

Explanation:

a )

Time to reach the speed of 20 m/s with an acceleration of 2 m/s² can be calculated as follows .

v = u + a t

20 = 0 + 2 t

t = 20 /2 = 10 s .

Total time = 10 s + 20 s + 5 s = 35 s .

b) Average velocity = Total distance travelled / total time

Distance travelled in first 10 s

S₁ = ut + 1/2 a t²

= 0 + .5 x 2 x 10²

= 100 m

Distance travelled in next 20 s

S₂= 20s x 20 m/s = 400 m

Distance travelled in last 5 s .

deceleration in last 5 s

v = u + at

0 = 20 m/s + a x 5

a = - 4 m/s²

v² = u² - 2 a s

0 = (20 m/s)² - 2 x 4 m/s² x s

s = 50 m

S₃ = 50 m

Total distance = S₁ + S₂ + S₃

= 100 m + 400 m + 50 m

= 550 m .

Average velocity = 550 m / 35 s

= 15.71 m /s .

3 0

3 years ago

A pilot is flying a small airplane at 40 m/s in a circular path with a radius of 200 m. The pilot has a mass of 80.5 kg

g100num [7]

Try This Method(substitute your values)
Centripetal force = Mass X Velocity^2/ Radius
F= mv^2/r 
so 80.5 X 40^2/200 
=644N

4 0

3 years ago