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hodyreva [135]
3 years ago
14

In concept simulation 10.2 you can explore the concepts that are important in this problem. astronauts on a distant planet set u

p a simple pendulum of length 1.2 m. the pendulum executes simple harmonic motion and makes 100 complete oscillations in 360 s. what is the magnitude of the acceleration due to gravity on this planet
Physics
1 answer:
Margarita [4]3 years ago
5 0
Period of a simple pendulum = 2π √(L/G)

(360s/100) = 2π √(1.2m/G)

1.8s / π = √1.2m / √G

√G · (1.8s/π) = √1.2m

√G = (π · √1.2m) / 1.8s

Square each side:

G = π² · 1.2m / 3.24 s²

G = (1.2 · π² / 3.24)  m/s²

G = 3.66 m/s²

So I just went and looked up Mars gravity. Floogle says it's 3.711 m/s² there.
That seems awfully close ... only 1.4% greater than our astronauts measured. 
You don't suppose . . . . .
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What is the distance to a star whose parallex is 0.1 sec?
Arte-miy333 [17]

Answer:

30.86\times 10^{13} km

Explanation:

Given the parallex of the star is 0.1 sec.

The distance is inversely related with the parallex of the star. Mathematically,

d=\frac{1}{P}

Here, d is the distance to a star which is measured in parsecs, and P is the parallex which is measured in arc seconds.

Now,

d=\frac{1}{0.1}\\d=10 parsec

And also know that,

1 parsec=3.086\times 10^{13} km

Therefore the distance of the star  is 30.86\times 10^{13} km away.

6 0
3 years ago
A 50.0-kg box rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.300 and the
Dimas [21]

Answer:

98N and 147N

Explanation:

We have the following information:

m=50kg\\\mu_s =0.4\\\mu_k = 0.2\\F=140N

We can find the static fricton force as follow,

F=\mu_s * N

Where N is the normal force (mg)

F=0.3*50*9.8\\F=147N

Static friction force at 147N is greater than the force applied hence body does not move.

F=\mu_k N = 0.2*50*9.8= 98N

3 0
4 years ago
Saturn is 890,700,000 miles from the Sun. What is the distance in meters?
Maslowich

Answer:

1.4936 trillion meters

Explanation:

From an average distance of 886 million miles Saturn is 9.5 astronomical units away from the sun

6 0
3 years ago
Read 2 more answers
The body falls in a free fall 11s.Calculate: a) from what height the body of the paddle) what path did it fall during the last s
JulsSmile [24]

Answer:

a) h = 593.50 m

b) h₁₁ = 103 m

c) vf = 107.91 m/s

Explanation:

a)

We will use second equation of motion to find the height:

h = v_{i}t + (\frac{1}{2})gt^2

where,

h = height = ?

vi = initial speed = 0 m/s

t = time taken = 11 s

g = 9.81 /s²

Therefore,

h = (0\ m/s)(11\ s) + (\frac{1}{2})(9.8\ m/s^2)(11\ s)^2\\\\

<u>h = 593.50 m</u>

b)

For the distance travelled in last second, we first need to find velocity at 10th second by using first equation of motion:

v_{f} = v_{i} + gt

where,

vf = final velocity at tenth second = v₁₀ = ?

t = 10 s

vi = 0 m/s

Therefore,

v_{10} = 0\ m/s + (9.81\ m/s^2)(10\ s)\\\\v_{10} = 98.1\ m/s

Now, we use the 2nd equation of motion between 10 and 11 seconds to find the height covered during last second:

h = v_{i}t + (\frac{1}{2})gt^2

where,

h = height covered during last second = h₁₁ =  ?

vi = v₁₀ = 98.1 m/s

t = 1 s

Therefore,

h_{11} = (98.1\ m/s)(1\ s) + (\frac{1}{2})(9.8\ m/s^2)(1\ s)^2\\\\

<u>h₁₁ = 103 m</u>

c)

Now, we use first equation of motion for complete motion:

v_{f} = v_{i} + gt

where,

vf = final velocity at tenth second = ?

t = 11 s

vi = 0 m/s

Therefore,

v_{f} = 0\ m/s + (9.81\ m/s^2)(11\ s)

<u>vf = 107.91 m/s</u>

8 0
4 years ago
A force of 10 N causes a spring to extend by 20 mm. Find: a) the spring constant of the spring in N/m​
Mars2501 [29]

Answer:

formula used K=F/∆l

∆l is the elongation of the spring

  1. F=10N
  2. ∆l=20mm===> 0.02m
  3. K=10N divided 0.02m= 500N/m
6 0
3 years ago
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