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3 years ago

Draw an accurate, labelled diagram of a thermometer

Physics

1 answer:

olga nikolaevna [1]3 years ago

7 0

You wear very specific on what type of thermometer he wanted and what specific information you wanted so I just made one like this, hope it helps anyway! -Sam

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If you are walking to the elevator on a cruise ship , what are you probably using your frame of reference

omeli [17]

Answer:

The Elevator

Explanation:

The frame of reference is the immovable item and remains at the same place and can be used for guiding some one. For example I am telling you an address by saying that you will cross the mega store and then there will a Ferrari and on the next street is my home. So you kept searching my home and then you called me again. I checked from my window before receiving call and saw that the ferrari was gone. So it took more time which means that the frame of reference choosen was movable item. The frame of reference choosen is always immovable. So here I want to move towards elevator which is immovable so choosed it. Furthermore, ship is movable item and the ocean doesn't specify right direction thatswhy stars were used in the ancient times.

7 0

3 years ago

Assume the earth to be a nonrotating sphere with mass MEME and radius RERE. If an astronaut weights WW on the ground, what is hi

Solnce55 [7]

Answer:

The weight at a distance 2 RE from surface of earth is <em>W/9</em>

Explanation:

For the value of acceleration due to gravity (g), we have a formula, that is:

g = (G)(ME)/(RE)² ----- equation (1)

where,

G = Gravitational Constant

ME = Mass of Earth

RE = Radius of Earth

g = Acceleration due to gravity on surface of earth = 9.8 ms²

When the person goes 2RE, distance above earth's surface. Then the total distance from center of earth becomes: 2RE + RE = 3RE.

Therefore, equation (1) becomes:

gh = (G)(ME)/(3RE)²

where,

gh = acceleration due to gravity at height

gh = (G)(ME)/(RE)²9

using equation (1), we get:

gh = g/9

Now, he weight is given by formula:

W = mg ------- equation (2)

At height 2RE

Wh = (m)(gh)

where,

Wh = Weight at height = ?

m = mass of astronaut

Therefore, using vale of gh, we get:

Wh = mg/9

Using equation (2), we get:

6 0

3 years ago

An experimental rocket designed to land upright falls freely from a height of 2.59 102 m, starting at rest. At a height of 86.9

aleksandr82 [10.1K]

Answer:

The acceleration required by the rocket in order to have a zero speed on touchdown is 19.96m/s²

The rocket's motion for analysis sake is divided into two phases.

Phase 1: the free fall motion of the rocket from the height 2.59*102m to a height 86.9m

Phase 2: the motion of the rocket due to the acceleration of the rocket also from the height 86.9m to the point of touchdown y = 0m.

Explanation:

The initial velocity of the rocket is 0m/s when it started falling from rest under free fall. g = 9.8m/s² t1 is the time taken for phase 1 and t2 is the time taken for phase2.

The final velocity under free fall becomes the initial velocity for the accelerated motion of the rocket in phase 2 and the final velocity or speed in phase 2 is equal to zero.

The detailed step by step solution to the problems can be found in the attachment below.

Thank you and I hope this solution is helpful to you. Good luck.

5 0

3 years ago

A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle

xz_007 [3.2K]

Answer:

a) $\Delta{t} = 5.39s$

b) the motorcycle travels 155 m

Explanation:

Let $t_2-t_1 = \Delta{t}$ , then consider the equation of motion for the motorcycle (accelerated) and for the car (non accelerated):

$v_{m2}=v_0+a\Delta{t}\\x+d=(\frac{v_0+v_{m2}}{2} )\Delta{t}\\v_c = v_0 = \frac{x}{\Delta{t}}$

where:

$v_{m2}$ is the speed of the motorcycle at time 2

$v_{c}$ is the velocity of the car (constant)

$v_{0}$ is the velocity of the car and the motorcycle at time 1

d is the distance between the car and the motorcycle at time 1

x is the distance traveled by the car between time 1 and time 2

Solving the system of equations:

$\left[\begin{array}{cc}car&motorcycle\\x=v_0\Delta{t}&x+d=(\frac{v_0+v_{m2}}{2}}) \Delta{t}\end{array}\right]$

$v_0\Delta{t}=\frac{v_0+v_{m2}}{2}\Delta{t}-d \\\frac{v_0+v_{m2}}{2}\Delta{t}-v_0\Delta{t}=d\\(v_0+v_{m2})\Delta{t}-2v_0\Delta{t}=2d\\(v_0+v_0+a\Delta{t})\Delta{t}-2v_0\Delta{t}=2d\\(2v_0+a\Delta{t})\Delta{t}-2v_0\Delta{t}=2d\\a\Delta{t}^2=2d\\\Delta{t}=\sqrt{\frac{2d}{a}}=\sqrt{\frac{2*58}{4}}=\sqrt{29}=5.385s$

For the second part, we need to calculate x+d, so you can use the equation of the car to calculate x:

$x = v_0\Delta{t}= 18\sqrt{29}=96.933m\\then:\\x+d = 154.933$

3 0

3 years ago

What is the fate of solar radiation that reaches the earth?

expeople1 [14]

When solar radiation reaches the Earth it quickly dissipates as most of the radiation and UV rays are blocked by ozone layer, but more radiation and UV rays are able to get through because of global warming.

6 0

3 years ago