All categories

3 years ago

Why are some liquids more viscous than others?

1 answer:

7 0

The force that opposes the movement of an object through water is called drag. This is a type of frictional force. This force normally depends on the density and the viscosity of the fluid in question. The liquid which has more density and more viscosity or stickiness will produce a greater amount of drag force on an object than a fluid that is less dense and less viscous in nature. River water normally has less drag than that of sea water.

You might be interested in

Why is there convection in the outer core and what is the result of this?

mars1129 [50]

Answer:

re believed to influence the Earth's magnetic field. ... As heat is transferred outward toward the mantle, the net trend is for the inner boundary of the liquid region to freeze, causing the solid inner core to grow

Explanation:

6 0

3 years ago

Two trains travel at right angles to each other after leaving the same train station at the same time. Two hours later they are

Vikentia [17]

Answer:

Speed of faster train equals 29 mph

Explanation:

Let the speed of slower train be 'x' miles per hour and the speed of faster train be 'y' miles per hour.

Distance that slower train covers in 2 hours= $2\times x miles$

Distance that faster train covers in 2 hours= $2\times y miles$

Since they move at right angles the distance between them can be found by Pythagoras formula as

$d^{2}=(2x)^{2}+(2y)^{2}\\\\d^{2}=4(x^{2}+y^{2})\\(70.46)^{2}=4(x^{2}+y^{2})\\\\\therefore (x^{2}+y^{2})=\frac{(70.46)^{2}}{4}\\\\(x^{2}+y^{2})=\frac{4964.6}{4}\\\\(x^{2}+y^{2})=1241.15$

It is given that $y=x+9$

Using this in the above equation we get

$(x^{2}+y^{2})=1241.15\\\\x^{2}+(x+8)^{2}=1241.15\\2x^{2}+16x+64=1241.15\\2x^{2}+16x-1177.15=0$

This is a quadratic equation in 'x'

Comparing with standard quadratic equation $ax^{2}+bx+c$ we get value of x as

$(x^{2}+y^{2})=1241.15\\\\x^{2}+(x+8)^{2}=1241.15\\2x^{2}+16x+64=1241.15\\2x^{2}+16x-1177.15=0\\\\x=\frac{-16\pm \sqrt{(16)^{2}-4\times 2\times -1177.15}}{2\times 2}\\\\x=\frac{-16\pm 98.35}{4}\\\\x=20.58mph(\because speed\neq$

Thus speed of faster train = 28.58 mph

Speed of faster train = 19 mph

3 0

4 years ago

Read 2 more answers

A 5.00-kg object is attached to one end of a horizontal spring that has a negligible mass and a spring constant of 280 N/m. The

lina2011 [118]

Answer:

1) v = 0.45 m/s

2) v = 0.65 m/s

3) v = 0.75 m/s

Explanation:

1) We can find the speed of the object by conservation of energy:

$E_{i} = E_{f}$

$\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}$

Where:

k: is the spring constant = 280 N/m

v: is the speed of the object =?

m: is the mass of the object = 5.00 kg

x: is the displacement of the spring

$\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.08 m)^{2} + \frac{1}{2}5.00 kgv^{2}$

$v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.08 m)^{2}}{5.00 kg}} = 0.45 m/s$

2) When the object is 5.00 cm (0.050 m) from equilibrium, the speed of the object is:

$\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}$

$\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.05 m)^{2} + \frac{1}{2}5.00 kgv^{2}$

$v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.05 m)^{2}}{5.00 kg}} = 0.65 m/s$

3) When the object is at the equilibrium position, the speed of the object is:

$\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}$

$\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0 m)^{2} + \frac{1}{2}5.00 kgv^{2}$

$v = \sqrt{\frac{280N/m(0.10 m)^{2}}{5.00 kg}} = 0.75 m/s$

I hope it helps you!

8 0

3 years ago

Choose the correct statement about the periodic table's organization from the choices below.

Ksivusya [100]

We need to be able to the see the choices to help you answer the question:)

7 0

3 years ago

How far east will the plane travel in 1 hour?

MaRussiya [10]

Let Vpa be the velocity of plane with respect to air and Vp and Va be the velocities of plane and air respectively with respect to ground.
Vpa = Vp - Va 
Vp = Vpa + Va [all these quantities are vectors] 
so by parallelogram law of vector addition 
Vp = (Va^2 + Vpa^2 + 2*Va*Vpa*cos120)^(1/2) 
Vp^2 = 33^2 + 140^2 - 33*140 
Vp^2 = 16096 
Vp = 126.763m/s 
(2) tan b = (33*sin120)/(140+33/2) 
b = 10.34 degrees North of East 
(3) component of velocity of plane in east = Vp cos(10.34) = 126.763*1 = 126.7m/s 
distance travelled by plane in east = 126.7 * 60*60=456120 m

3 0

3 years ago