All categories

3 years ago

Increased________will lower viscosity

Physics

2 answers:

MrRa [10]3 years ago

8 0

The answer is high temperatures

Tcecarenko [31]3 years ago

4 0

Increased temperatures will lower viscosity

You might be interested in

What will happen to plant height if the amount of available light is reduced due to global dimming? Which type of investigation

Lorico [155]

The plant will not grow. In fact it could have all the nutrients and all the water it needs, but without a sufficient amount of light, it could die because its leaves are meant for a certain minimum amount of light.

I'll come back and see if you have posted the question you wanted and edit my answer.

6 0

3 years ago

There is a 50 g sample of Ra-229. It has a half-life of 4 minutes.

Sloan [31]

Via half-life equation we have:

$A_{final}=A_{initial}(\frac{1}{2})^{\frac{t}{h} }$

Where the initial amount is 50 grams, half-life is 4 minutes, and time elapsed is 12 minutes. By plugging those values in we get:

$A_{final}=50(\frac{1}{2})^\frac{12}{4}=50(\frac{1}{2})^{3}=50(\frac{1}{8})=6.25g$

There is 6.25 grams left of Ra-229 after 12 minutes.

4 0

3 years ago

Puck B has twice the mass of puck A. Starting from rest, both pucks are pulled the same distance across frictionless ice by stri

gogolik [260]

Answer:

(a) 1 : 2

(b) same

Explanation:

Let the mass of puck A is m and the mass of puck B is 2 m.

initial speed for both the pucks is same as u and the distance is same for both is s.

let the tension is T for same.

The kinetic energy is given by

$K = 0.5 mv^2$

(a) As the speed is same, so the kinetic energy depends on the mass.

So, kinetic energy of A : Kinetic energy of B = m : 2m = 1 : 2

(b) A the distance s same so the final velocities are also same.

8 0

2 years ago

How can the frequency with which the direction of a current changes be regulated?

sesenic [268]

Answer:

In an inductive circuit, when frequency increases, the circuit current decreases and vice versa.

Explanation:

5 0

3 years ago

A box rests on top of a flat bed truck. The box has a mass of m = 16.0 kg. The coefficient of static friction between the box an

3241004551 [841]

Answer:

1) 1.31 m/s2

2) 20.92 N

3) 8.53 m/s2

4) 1.76 m/s2

5) -8.53 m/s2

Explanation:

1) As the box does not slide, the acceleration of the box (relative to ground) is the same as acceleration of the truck, which goes from 0 to 17m/s in 13 s

$a = \frac{\Delta v}{\Delta t} = \frac{17 - 0}{13} = 1.31 m/s2$

2)According to Newton 2nd law, the static frictional force that acting on the box (so it goes along with the truck), is the product of its mass and acceleration

$F_s = am = 1.31*16 = 20.92 N$

3) Let g = 9.81 m/s2. The maximum static friction that can hold the box is the product of its static coefficient and the normal force.

$F_{\mu_s} = \mu_sN = mg\mu_s = 16*9.81*0.87 = 136.6N$

So the maximum acceleration on the block is

$a_{max} = F_{\mu_s} / m = 136.6 / 16 = 8.53 m/s^2$

4)As the box slides, it is now subjected to kinetic friction, which is

$F_{\mu_s} = mg\mu_k = 16*9.81*0.69 = 108.3 N$

So if the acceleration of the truck it at the point where the box starts to slide, the force that acting on it must be at 136.6 N too. So the horizontal net force would be 136.6 - 108.3 = 28.25N. And the acceleration is

28.25 / 16 = 1.76 m/s2

5) Same as number 3), the maximum deceleration the truck can have without the box sliding is -8.53 m/s2

3 0

3 years ago