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Ulleksa [173]
3 years ago
12

Satellite in a circular orbit of radius r around planet x has an orbital period t. if planet x had one-fourth as much mass, the

orbital period of this satellite in an orbit of the same radius would be
Physics
1 answer:
postnew [5]3 years ago
7 0
<span>Satellite in a circular orbit of radius r around planet x has an orbital period t. if planet x had one-fourth as much mass, the orbital period of this satellite in an orbit of the same radius would be 2t</span>
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What is the % error in using g = 10.0 m/s^2? (Take the value ofg as 9.8 m/s^2)
Alenkasestr [34]

Answer:

So percentage error will be 2 %

Explanation:

We have given initial value of acceleration due to gravity g=10m/sec^2

And final value of acceleration due to gravity g=9.8m/sec^2

We have to find the percentage error

We know that percentage error is given by percentage\ error=\frac{initial\ value-final\ value}{initial\ value}\times 100

So percentage\ error=\frac{10-9.8}{10}\times 100=2 %

4 0
3 years ago
What would you predict for this elephant's stride frequency? That is, how many steps per minute will the elephant take?
Sonja [21]
<span>` You can consider T to be in units of seconds/step. Frequency is the inverse of period, so 
1/T = frequency and has units of steps per second. There will be 60 times as many steps in a minute.</span>
8 0
3 years ago
Read 2 more answers
The speed of light (c) is:
Radda [10]

Answer:

A

Explanation:

Speed of light is 299 792 458 m / s. So option A is answer.

7 0
2 years ago
Thomas and John are carrying a 43kg cylinder head on a 510cm X 510mm board. The cylinder head with dimensions of 43cm X 250mm li
tatyana61 [14]

John carry the heaviest load.

<h3>How to find out who is carrying the heavy load?</h3>

Write down given data from questions:

Board=510cm X 510mm.

Cylinder head with dimensions=43cm X 250mm.

Cylinder lies across the board 210cm from john.

Find out: Who is carry the heaviest load?​

Calculation:

We assume that mass of cylinder head = x kg

Then weight=x x 9*81

                 W=9.81x  Newton.

Weight per unit length= Weight/Total leanth

Weight per unit length= 9.81x/43

(w/l)=0.23x N/cm

From equation contition: (F_{J} +F_{T} =9.81x n)

F_{T} (510)=9.81x  (210+21.5)

F_{T} (510)=9.81 x (231.5)

F_{T} =4.452x N.

F_{J} =9.81x-4.452x

F_{J} =5.358 xN

Therefore F_{J} \geq F_{T}

To learn more about mass per unit length, refer to:

brainly.com/question/24180692

#SPJ9

4 0
2 years ago
A (B + 25.0) g mass is hung on a spring. As a result, the spring stretches (8.50 A) cm. If the object is then pulled an addition
Morgarella [4.7K]

Answer:

Time period of the osculation will be 2.1371 sec

Explanation:

We have given mass m = (B+25)

And the spring is stretched by (8.5 A )

Here A = 13 and B = 427

So mass m = 427+25 = 452 gram = 0.452 kg

Spring stretched x= 8.5×13 = 110.5 cm

As there is additional streching of spring by 3 cm

So new x = 110.5+3 = 113.5 = 1.135 m

Now we know that force is given by F = mg

And we also know that F = Kx

So mg=Kx

K=\frac{mg}{x}=\frac{0.452\times 9.8}{1.135}=3.90N/m

Now we know that \omega =\sqrt{\frac{K}{m}}

So \frac{2\pi }{T} =\sqrt{\frac{K}{m}}

\frac{2\times 3.14 }{T} =\sqrt{\frac{3.90}{0.452}}

T=2.1371sec

8 0
3 years ago
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