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nexus9112 [7]
2 years ago
12

Explain why viscosity is classified as a physical property

Physics
2 answers:
masha68 [24]2 years ago
6 0
The viscosity of a substance can be determined without changing the composition of the substance. ... It is a physical property because the composition of copper does not change as its shape changes.
dalvyx [7]2 years ago
3 0

Answer:

Copper's composition does not change as its shape changes, hence it is a physical feature.

Explanation:

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A hawk is flying horizontally at 18.0 m/s in a straight line, 230 m above the ground. A mouse it has been carrying struggles fre
Lisa [10]

Answer:

a) vd = 47.88 m/s

b) θ = 80.9°

c) t = 6.8 s

Explanation:

In the situation of the problem, you can assume that the trajectory of the hawk and the trajectory of the mouse form a rectangle triangle.

One side of the triangle is the horizontal trajectory of the hawk after 2.00s of flight, the other side of the triangle is the distance traveled by the mouse when it is falling down. And the hypotenuse is the trajectory of the hawk when it is trying to recover the mouse.

(a) In order to calculate the diving speed of the hawk, you first calculate the hypotenuse of the triangle.

One side of the triangle is c1 = (18.0m/s)(2.0s) = 36m

The other side of the triangle is c2 = 230m - 3m = 227 m

Then, the hypotenuse is:

h=\sqrt{(36m)^2+(227m)^2}=229.83m    (1)

Next, it is necessary to calculate the falling down time of the mouse, this can be done by using the following formula:

y=y_o+v_ot+\frac{1}{2}gt^2    (2)

yo: initial height = 230m

vo: initial vertical speed of the mouse = 0m/s

g: gravitational acceleration = -9.8m/s^2

y: final height of the mouse = 3 m

You replace the values of the parameters in (2) and solve for t:

3=230-4.9t^2\\\\t=\sqrt{\frac{227}{4.9}}=6.8s

The hawk traveled during 2.00 second in the horizontal trajectory, hence, the hawk needed 6.8s - 2.0s = 4.8 s to travel the distance equivalent to the hypotenuse to catch the mouse.

You use the value of h and 4.8s to find the diving speed of the hawk:

v_d=\frac{229.83m}{4.8s}=47.88\frac{m}{s}

The diving speed of the Hawk is 47.88m/s

(b) The angle is given by:

\theta=cos^{-1}(\frac{c_1}{h})=cos^{-1}(\frac{36m}{229.83m})=80.9 \°

Then angle between the horizontal and the trajectory of the Hawk when it is descending is 80.9°

(c) The mouse is falling down during 6.8 s

4 0
3 years ago
What is the resistance of a light bulb if a potential difference of 120 V will produce a current of 0.5 A in the bulb?
Yakvenalex [24]

Explanation:

Ohm's law:

V = IR

120 V = (0.5 A) R

R = 240 Ω

6 0
3 years ago
Read 2 more answers
A transparent oil with index of refraction 1.28 spills on the surface of water (index of refraction 1.33), producing a maximum o
Ad libitum [116K]

Answer:

The thickness of the oil slick is 1.95\times10^{-7}\ m

Explanation:

Given that,

Index of refraction = 1.28

Wave length = 500 nm

Order m = 1

We need to calculate the thickness of oil slick

Using formula of thickness

2nt= m\lambda

Where, n = Index of refraction

t = thickness

\lambda = wavelength

Put the value into the formula

2\times1.28 \times t=1\times\times500\times10^{-9}

t = \dfrac{1\times\times500\times10^{-9}}{2\times1.28 }

t=1.95\times10^{-7}\ m

Hence, The thickness of the oil slick is 1.95\times10^{-7}\ m

4 0
2 years ago
A student weighing 700 N climbs at constant speed to the top of an 8 m vertical rope in 10 s. The average power expended by the
Goshia [24]

To solve this problem we will use the concepts related to power, defined as the amount of energy applied over a period of time.

The energy in this case is the accumulated in the form of potential energy, over a period of time. Thus we will have that the mathematical expression of the power can be expressed as

P = \frac{E}{t}

Here,

E = Energy

t = time

As the energy is equal to the potential Energy we have tat

P = \frac{mgh}{t}

The weight (mg) of the man is 700N, the height (h) is 8m and the time is 10s, then:

P = \frac{700*8}{10}

P = 560W

Therefore the correct answer is A.

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3 years ago
Install webroot mac +1 (888)(210)(2883) install webroot macbook pro
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<h2><em><u>ookk</u></em></h2>

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