1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
ira [324]
3 years ago
15

An object is lifted from the surface of aspherical planet to an altitude equal to the radius of the planet.As a result, what hap

pens to the mass and wieght of theobject?
massincreases; weight decreases
massdecreases; weight decreases
massincreases; weight increases
massincreases; weight remains the same
mass remainsthe same; weight decreases
mass remainsthe same; weight increases
mass remainsthe same; weight remains the same
Physics
1 answer:
serg [7]3 years ago
8 0

An object is lifted from the surface of a spherical planet to an altitude equal to the radius of the planet.  

As a result, the object's <em>mass remains the same</em>, and its <em>weight decreases</em> to 1/4 of whatever it is when the object is on the planet's surface.

You might be interested in
Earth is about 150 million kilometers from the Sun, and the apparent brightness of the Sun in our sky is about 1300 watts/m2. Us
nalin [4]

Answer:

13 W/m^2

Explanation:

The apparent brightness follows an inverse square law, therefore we can write:

I \propto \frac{1}{r^2}

where I is the apparent brightness and r is the distance from the Sun.

We can also rewrite the law as

\frac{I_2}{I_1}=\frac{r_1^2}{r_2^2} (1)

where in this problem, we have:

I_1 = 1300 W/m^2 apparent brightness at a distance r_1, where

r_1 = 150 million km

We want to estimate the apparent brightness at r_2, where r_2 is ten times r_1, so

r_2 = 10 r_1

Re-arranging eq.(1), we find I_2:

I_2 = \frac{r_1^2}{r_2^2}I_1 = \frac{r_1^2}{(10r_1)^2}(1300)=\frac{1}{100}(1300)=13 W/m^2

5 0
3 years ago
Lindsay is planning a flight from St. Catharines to Hamilton, which lies due west of St. Catharines. Her aircraft flies at a spe
olga nikolaevna [1]

Lindsay has to fly this plane towards this direction [W 12.5° S] to get to Hamilton.

From this question, the plane is still up in the air.

We have wind blowing in [W 60° N ]

To solve the problem we have to make use of the sine rule

\frac{SinA}{a}=\frac{SinB}{b} =\frac{SinC}{c}

We put the values in the equation, we have:

50/Sinθ = 200/sin60°

The next step is to cross multiply

50 x sin60° = 200Sinθ

50 x 0.8660 = 200sinθ

We make Sin θ the subject

Sine θ = 43.30/200

sine θ = 0.2165

we find the value of θ

θ = sine⁻¹(0.2165)

θ = 12.50

So Lindsay has to fly this plane towards this direction

[W 12.5° S]

Here is a similar question brainly.com/question/13338067?referrer=searchResults

7 0
2 years ago
Read 2 more answers
A seagull flies at a velocity of 9.00 m/s straight into the wind. (a) if it takes the bird 20.0 min to travel 6.00 km relative t
enot [183]

Here we will the speed of seagull which is v = 9 m/s

this is the speed of seagull when there is no effect of wind on it

now in part a)

if effect of wind is in opposite direction then it travels 6 km in 20 min

so the average speed is given by the ratio of total distance and total time

v_{avg} = \frac{6000}{20*60}

v_{avg} = 5m/s

now since effect of wind is in opposite direction then we can say

V_{net} = v_{bird} - v_{wind}

5 = 9 - v_{wind}

v_{wind}= 4 m/s

Part b)

now if bird travels in the same direction of wind then we will have

v_{net}= v_{bird} + v_{wind}

v_{net} = 9 + 4 = 13 m/s

now we can find the time to go back

time = \frac{distance}{speed}

time = \frac{6000}{13}

time = 7.7 minutes

Part c)

Total time of round trip when wind is present

T = t_1 + t_2

T = 20 + 7.7 = 27.7 min

now when there is no wind total time is given by

T = \frac{6000}{9} + \frac{6000}{9}

T = 22.22 min

So due to wind time will be more

4 0
3 years ago
What instrument measures air temperature?
zavuch27 [327]
Its a thermometer . i hope this helps you
5 0
3 years ago
A tennis ball is served horizontally from 2.4m above the ground at net is 12m away and point 0.9 high will be ball clear the net
Schach [20]

Explanation:

Let us first calculate  long does it take to go 12m at 30m/s( assumed speed)

12/30 = 0.4 seconds

horizontal distance the ball drop in that time

H= (0)(0.4)+1/2(-9.8)(0.4)2

H= -0.78m

negative sign shows that the height of the ball at the net from the top.

Height of the ball at the net and from the ground= H1-H=2.4-0.78=1.62m

As 1.62m>0.9m so the ball will clear the net.

H_1= V0y t’ + ½ g t’^2

-2.4= (0)t’ + ½ (-9.8) t’^2

t’= 0.69s

X’=V0x t’

X’=(30)(0.96)

X’= 20.7m

3 0
2 years ago
Other questions:
  • In the past, how did people use the systematic movement of celestial bodies? Select all that apply.
    7·2 answers
  • Approximately how old is the earth?
    13·1 answer
  • i need the answer asap plz A child rides her bike at a rate of 12.0 km/hr down the street. A squirrel suddenly runs in front of
    7·2 answers
  • If a running system has a total change in heat of 295 joules, and it's running at a temperature of 402 kelvin, what is the entro
    11·1 answer
  • What causes air to become less dense and rise?
    5·1 answer
  • The smallest shift you can reliably measure on the screen is about 0.2 grid units. This shift corresponds to the precision of po
    12·1 answer
  • A bicycle has a momentum of 12 kgm/s. What momentum would the bicycle have if it had: a) half the mass and twice the speed
    13·1 answer
  • Infer whether a circuit breaker should be connected in parallel to the circuit that it is protecting.
    13·1 answer
  • HEEEEEEELLLLLPPPPPPP 20 points and Brainliest
    6·1 answer
  • Eeeee<br> help pls<br> pls<br> pls
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!