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Crank
3 years ago
11

If the earth shrank until its radius were only one-quarter its present size without changing its mass what would a 20 n object w

eigh st its surface?
Physics
1 answer:
Dahasolnce [82]3 years ago
5 0

Basing on the information given, we can calculate the new weight of the object by the following given:current weight = 20 Ng = 10m/s2

20N/4 = 5N

Thank you for your question. Please don't hesitate to ask in Brainly your queries. 
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During the spin cycle of a washing machine, the clothes stick to the outer wall of the barrel as it spins at a rate as high as 1
Darya [45]

To answer the two questions, we need to know two important equations involving centripetal movement:

v = ωr (ω represents angular velocity <u>in radians</u>)

a = \frac{v^{2}}{r}

Let's apply the first equation to question a:

v = ωr

v = ((1800*2π) / 60) * 0.26

Wait. 2π? 0.26? 60? Let's break down why these numbers are written differently. In order to use the equation v = ωr, it is important that the units of ω is in radians. Since one revolution is equivalent to 2π radians, we can easily do the conversion from revolutions to radians by multiplying it by 2π. As for 0.26, note that the question asks for the units to be m/s. Since we need meters, we simply convert 26 cm, our radius, into meters. The revolutions is also given in revs/min, and we need to convert it into revs/sec so that we can get our final units correct. As a result, we divide the rate by 60 to convert minutes into seconds.

Back to the equation:

v = ((1800*2π)/60) * 0.26

v = (1800*2(3.14)/60) * 0.26

v = (11304/60) * 0.26

v = 188.4 * 0.26

v = 48.984

v = 49 (m/s)

Now that we know the linear velocity, we can find the centripetal acceleration:

a = \frac{v^{2}}{r}

a = \frac{49^{2}}{0.26}

a = 9234.6 (m/s^{2})

Wow! That's fast!

<u>We now have our answers for a and b:</u>

a. 49 (m/s)

b. 9.2 * 10^{3} (m/s^{2})

If you have any questions on how I got to these answers, just ask!

- breezyツ

5 0
2 years ago
PLEASE HELP
Mashutka [201]

Answer:

It cannot be constant because if it does not change and each time it increases its strength and speed.

Explanation:

4 0
3 years ago
Why do the passengers in high-altitude jet planes feel the sensation of weight while passengers in an orbiting space vehicle, su
kykrilka [37]

Passengers in an aircraft are subject to the Normal and Gravity Force acting on them at a low 'orbit', so tiny that it can be many times compared to the same surface of the earth when speaking in general terms.

In a high orbit space vehicle or in the same space, said force decreases considerably or simply disappears, generating the sensation of weightlessness.

Remember that the Force of Gravity is given under the principle

F_g = \frac{GMm}{r^2}

Where,

G = Gravitational Universal constant

M = Mass of the planet

m = mass of the object

r = Distance from center of the planet

When the radius grows considerably the gravitational force begins to decrease.

7 0
2 years ago
Ivan drove to the mountains last weekend. There was heavy traffic on the way there, and the trip took 7 hours. When Ivan drove h
lesya692 [45]

Answer:252 miles

Explanation:

Given

During his way to mountain it took 7 hr to drive

and during his return trip it took 4 hr to return

Let x be the distance between home and mountain

average speed for return is  27 miles per hour faster than his former trip

let v be the speed on his way to mountain thus v+27 is his return speed

thus 7=\frac{x}{v}----1

for return trip

4=\frac{x}{v+27}-----2

divide  1 & 2

\frac{7}{4}=\frac{x\cdot (v+27)}{v\cdot x}

7v=4v+4\cdot 27

3v=4\cdot 27

v=36 mph

thus x=7\times 36=252\ miles

7 0
3 years ago
Can u help me. thank you​
katrin2010 [14]
I can give you a search engine that could help you with all ir hw its called socratic it uses everything on the internet to search for answers it’s literally a search engine
6 0
2 years ago
Read 2 more answers
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