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pochemuha
3 years ago
10

What is cubical expansivity of liquid while freezing

Physics
2 answers:
Elena L [17]3 years ago
7 0

Answer:

"the ratio of increase in the volume of a solid per degree rise of temperature to its initial volume" -web

Explanation:

tbh up above ✅

kolezko [41]3 years ago
7 0

Answer:

cubic meter

Explanation:

Increase in volume of a body on heating is referred to as volumetric expansion or cubical expansion

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When object goes under acceleration
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when object goes under acceleration

c).its velocity always increases

<h3><u>Additional</u><u> </u><u>information</u><u>:</u><u>-</u></h3>

★ Acceleration: Rate of increase in velocity.

★ Velocity: Distance travelled by a body per unit time in given direction is called velocity .

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3 years ago
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What term was used to describe the final digit
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Significant digits (also called significant figures or “sig figs” for short) indicate the precision of a measurement. A number with more significant digits is more precise. For example, 8.00 cm is more precise than 8.0 cm.

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3 years ago
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

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Name the 2 types of tissue that form your skin?
QveST [7]

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Epithelial tissue and Muscle tissue

Explanation:

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The sum of kinetic energy and potential energy within a system is called
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