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3 years ago

The line where the Earth and sky seem to meet forms the:

Physics

2 answers:

anzhelika [568]3 years ago

7 0

The answer is C, Horizon.

babunello [35]3 years ago

6 0

The ocean and the sky provide a clean, flat line where the Earth seems to meet the sky. If you're standing on the beach looking out at the sea, the part of the sea that "touches" the horizon is called the offing

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A small 24 kilogram canoe is floating downriver at a speed of 2 m/s. What is the canoe's kinetic energy?

USPshnik [31]

J can get answer on this way:
Ek=m*V*V/2= (24kg*2m/s*2m/s)/2=48 Ј

3 0

4 years ago

The information on a can of soda indicates that the can contains 355 mL. The mass of a full can of soda is 0.369 kg, while an em

Vilka [71]

Answer:

$\rho=995.50\ kg.m^{-3}$

$\bar w=9765.887\ N.m^{-3}$

$s=0.9955$

Explanation:

Given:

volume of liquid content in the can, $v_l=0.355\ L=3.55\times 10^{-4}\ L$

mass of filled can, $m_f=0.369\ kg$

weight of empty can, $w_c=0.153\ N$

So, mass of the empty can:

$m_c=\frac{w_c}{g}$

$m_c=\frac{0.153}{9.81}$

$m_c=0.015596\ kg$

Hence the mass of liquid(soda):

$m_l=m_f-m_c$

$m_l=0.369-0.015596$

$m_l=0.3534\ kg$

Therefore the density of liquid soda:

$\rho=\frac{m_l}{v_l}$ (as density is given as mass per unit volume of the substance)

$\rho=\frac{0.3534}{3.55\times 10^{-4}}$

$\rho=995.50\ kg.m^{-3}$

Specific weight of the liquid soda:

$\bar w=\frac{m_l.g}{v_l}=\rho.g$

$\bar w=995.5\times 9.81$

$\bar w=9765.887\ N.m^{-3}$

Specific gravity is the density of the substance to the density of water:

$s=\frac{\rho}{\rho_w}$

where:

$\rho_w=$ density of water

$s=\frac{995.5}{1000}$

$s=0.9955$

3 0

3 years ago

Read 2 more answers

A stone is thrown horizontally at 60.0 m/sm/s from the top of a very tall cliff. Calculate its horizontal position and vertical

svp [43]

Answer:

X-Positions: Y-Positions

x(0) = 0 y(0) = 0

x(2) = 120 m y(2) = 19.6 m

x(4) = 240 m y(4) = 78.4 m

x(6) = 360 m y(6) = 176.4 m

x(8) = 480 m y(8) = 313 m

x(10) = 600m y (10) = 490 m

Explanation:

X-Positions

First, we choose to take the horizontal direction as our x-axis, and the positive x-axis as positive.
After being thrown, in the horizontal direction, no external influence acts on the stone, so it will continue in the same direction at the same initial speed of 60. 0 m/s
So, in order to know the horizontal position at any time t, we can apply the definition of average velocity, rearranging terms, as follows:

$x = v_{ox} * t = 60.0 m/s * t(s)$

It can be seen that after 2 s, the displacement will be 120 m, and each 2 seconds, as the speed is constant, the displacement will increase in the same 120 m each time.

Y-Positions

We choose to take the vertical direction as our y-axis, taking the downward direction as our positive axis.
As both axes are perpendicular each other, both movements are independent each other also, so, in the vertical direction, the stone starts from rest.
At any moment, it is subject to the acceleration of gravity, g.
As the acceleration is constant, we can find the vertical displacement (taking the height of the cliff as the initial reference level), using the following kinematic equation:

$y = \frac{1}{2} * g* t^{2} = \frac{1}{2} * 9.8 m/s2 * t(s)^{2}$

Replacing by the values of t, we get the following vertical positions, from the height of the cliff as y = 0:
y(2) = 2* 9.8 m/s2 = 19.6 m
y(4) = 8* 9.8 m/s2 = 78.4 m
y(6) = 18*9.8 m/s2 = 176.4 m
y(8) = 32*9.8 m/s2 = 313.6 m
y(10)= 50 * 9.8 m/s2 = 490.0 m

5 0

3 years ago

Your friend from France came to visit you when she was packing she went on weather.com and found that the average temperature in

Marina CMI [18]

Answer:

Because there is not as much cold as it was in France.

Explanation:

The average temperature in France during January ranges from 2.7° to 7.2° celsius which makes it the coldest month of the year. But since she comes to know that average temperature in Annville ranges 31° celsius which implies that the temperature is normal there and therefore, she packs sleeveless tops and shorts. Coats would not be required in a hot weather and hence, she does not pack it.

8 0

3 years ago

A 42.2 kg sled is pulled forward

zaharov [31]

The net force on the sledge is 31.64N.

Frictional force = µkR

= 0.269 x 42.2 x 9.81 = 111.36

net force = 143N - 111.36N

= 31.64N

refer brainly.com/question/24557767

#SPJ2

7 0

2 years ago