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melamori03 [73]

3 years ago

10

The freezing point of water is 0ÁC, and the boiling point is 100ÁC. This range of temperatures can be found naturally on the Ear

th, which means that....
A. water makes up less than 30% of the Earth's surface.
B. water cannot move through the roots of plants.
C. water can exist in all three states on Earth.
D. water has a very low specific heat.

1 answer:

viva [34]3 years ago

8 0

C.
The range of temperatures on Earth allows water to exist in all of its states.

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A turtle and a rabbit are in a 150 meter race. The rabbit decides to give the turtle a 1 minute head start. The turtle moves at

yan [13]

Answer:

a) $s_{T} = 30\,m$ , b) $t = 5\,min$ , c) $\Delta t = 6.667\,s$ , d) $\Delta s_{R} = 33.333\,m$ , e) $t' = 11.667\,s$ , f) The rabbit won the race.

Explanation:

a) As turtle moves at constant speed, its position is determined by the following formula:

$s_{T} = v_{T}\cdot t$

Where:

$t$ - Time, measured in seconds.

$v_{T}$ - Velocity of the turtle, measured in meters per second.

$s_{T}$ - Position of the turtle, measured in meters.

Then, the position of the turtle when the rabbit starts to run is:

$s_{T} = \left(0.5\,\frac{m}{s} \right)\cdot (60\,s)$

$s_{T} = 30\,m$

The position of the turtle when the rabbit starts to run is 30 meters.

b) The time needed for the turtle to finish the race is:

$t = \frac{s_{T}}{v_{T}}$

$t = \frac{150\,m}{0.5\,\frac{m}{s} }$

$t = 300\,s$

$t = 5\,min$

The time needed for the turtle to finish the race is 5 minutes.

c) As rabbit experiments a constant acceleration until maximum velocity is reached and moves at constant speed afterwards, the time required to reach such speed is:

$v_{R} = v_{o,R} + a_{R}\cdot \Delta t$

Where:

$v_{R}$ - Final velocity of the rabbit, measured in meters per second.

$v_{o,R}$ - Initial velocity of the rabbit, measured in meters per second.

$a_{R}$ - Acceleration of the rabbit, measured in $\frac{m}{s^{2}}$ .

$\Delta t$ - Running time, measured in second.

$\Delta t = \frac{v_{R}-v_{o,R}}{a_{R}}$

$\Delta t = \frac{10\,\frac{m}{s}-0\,\frac{m}{s}}{1.50\,\frac{m}{s^{2}} }$

$\Delta t = 6.667\,s$

The time taken by the rabbit to reach maximum speed is 6.667 s.

d) On the other hand, the position reached by the rabbit when maximum speed is reached is determined by the following equation of motion:

$v_{R}^{2} = v_{o,R}^{2} + 2\cdot a_{R}\cdot \Delta s_{R}$

$\Delta s_{R} = \frac{v_{R}^{2}-v_{o,R}^{2}}{2\cdot a_{R}}$

$\Delta s_{R} = \frac{v_{R}^{2}-v_{o,R}^{2}}{2\cdot a_{R}}$

Where $\Delta s_{R}$ is the travelled distance of the rabbit from rest to maximum speed.

$\Delta s_{R} = \frac{\left(10\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot \left(1.50\,\frac{m}{s^{2}} \right)}$

$\Delta s_{R} = 33.333\,m$

The distance travelled by the rabbit from rest to maximum speed is 33.333 meters.

e) The time required for the rabbit to finish the race can be determined by the following expression:

$t' = \frac{\Delta s_{R}}{v_{R}}$

$t' = \frac{150\,m-33.333\,m}{10\,\frac{m}{s} }$

$t' = 11.667\,s$

The time required for the rabbit from rest to maximum speed is 11.667 seconds.

f) The animal with the lowest time wins the race. Now, each running time is determined:

Turtle:

$t_{T} = 300\,s$

Rabbit:

$t_{R} = 60\,s + 6.667\,s + 11.667\,s$

$t_{R} = 78.334\,s$

The rabbit won the race as $t_{R} < t_{T}$ .

7 0

3 years ago

A truck with 0.410 m radius tires travels at 25.0 m/s. What is the angular velocity of the rotating tires in radians per second?

Eduardwww [97]

relation between linear velocity and angular velocity is given as

$v = R\omega$

here

v = linear speed

R = radius

$\omega$ = angular speed

now plug in all data in the equation

$25.0 = 0.410 \omega$

$\omega = \frac{25}{0.410}$

$\omega = 60.9 rad/s$

so rotating speed is 60.9 rad/s

6 0

2 years ago

What makes a model not useful?

lina2011 [118]

Answer is b hope this helps

8 0

2 years ago

Give THREE examples of objects that have potential energy

Tasya [4]

A ball kept on 3rd floor of a building.

A pendulum bob kept at 3m height

A stone thrown vertically upward.

A pressed spring.

A squashed spunge ball.

4 0

3 years ago

Starting from rest, a 2-m-long pendulum swings from an angleof

Andrews [41]

Answer:

D.) 1m/s

Explanation:

Assume the initial angle of the swing is 12.8 degree with respect to the vertical. We can calculate the vertical distance from this initial point to the lowest point by first calculate the vertical distance from this point the the pivot point:

$L_1 = L*cos(12.8) = 2*0.975 = 1.95 m$

where L is the pendulum length

The vertical distance from the lowest point to the pivot point $L_2$ is the pendulum length 2m

this means the vertical distance from this initial point to the lowest point is simply:

$L_3 = L_2 - L_1 = 2 - 1.95 = 0.05 m$

As the pendulum travel (vertically) from the initial point to the bottom point, its potential energy is converted to kinetic energy:

$E_p = E_k$

$mgh = mv^2/2$

where m is the mass of the pendulum, g = 10 m/s2 is the constant gravitational acceleration, h = 0.05 is the vertical it travels, v is the pendulum velocity at the bottom, which we are trying to solve for.

The m on both sides of the equation cancel out

$v^2 = 2gh = 2*10*0.05 = 1$

$v = \sqrt{1} = 1 m/s$

so D is the correct answer

5 0

3 years ago