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GuDViN [60]
3 years ago
14

A pendulum with a period of 1 s on Earth, where the acceleration due to gravity is g, is taken to another planet, where its peri

od is 2s. The acceleration due to gravity on the other planet is most nearly
1. ag = g
2. ag =g/4
3. ag = 4 g
4. ag =g/2
5. ag = 2 g
Physics
1 answer:
Vsevolod [243]3 years ago
4 0

Answer:

g ’= ¼ g     correct answer is 2

Explanation:

The expression for the angular velocity of a simple pendulum is

          w = √ g / L

The angular velocity and the period are related

          w = 2π / T

           T = 2π √L / g

Let's look for the length of the pendulum with the data on Earth

            L = T² g / 4π²

            L = 1² g / 4π²

            L = 2.533 10⁻² 9.8

            L = 0.2482 m

Now let's look for the gravity of the planet

             g’=  4π² L / T’²

             g’= 4π² 0.2483 / 2²

             g’= 2.45 m / s²

The relationship between this value and the earth's gravity is

           g ’/ g = 2.45 / 9.80

           g ’= 0.25 g

           g ’= ¼ g

correct anwers is 2

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The weather conditions of an area may be expected to change over a(n) _________ period of time.
jenyasd209 [6]

Answer:

Your answer should be 2. Short

Explanation:

4 0
3 years ago
In a fireworks display, a rocket is launched from the ground with a speed of 18.0 m/s and a direction of 51.0° above the horizon
Masja [62]

Answer:38.66 m

Explanation:

Given

launch angle \theta =51^{\circ}

launch velocity u=18 m/s

center of mass continue to travel its original Path so it center of mass will be at a distance of

R=\frac{u^2\sin 2\theta }{g}

R=\frac{18^2\sin 102}{9.8}

R=32.33 m

Center of mass will be at x=32.33 m

(b)if one of the piece will be at x=26  m then other will be at

x_{com}=\frac{m_1x_1+m_2x_2}{m_1+m_2}

x_{com}=\frac{\frac{m}{2}\cdot 26+\frac{m}{2}\cdot x_0}{\frac{m}{2}+\frac{m}{2}}

32.33=\frac{26+x_0}{2}

x_0=38.66 m

8 0
3 years ago
Two power lines run parallel for a distance of 222 m and are separated by a distance of 40.0 cm. if the current in each of the t
earnstyle [38]
1) Magnitude of the force:

The magnetic field generated by a current-carrying wire is
B= \frac{\mu_0I}{2 \pi r}
where
\mu_0 is the vacuum permeability
I is the current in the wire
r is the distance at which the field is calculated

Using I=135 A, the current flowing in each wire, we can calculate the magnetic field generated by each wire at distance 
r=40.0 cm=0.40 m, 
which is the distance at which the other wire is located:
B= \frac{\mu_0 I}{2 \pi r}= \frac{(4 \pi \cdot 10^{-7} N/A^2)(135 A) }{2 \pi (0.40 m)}=6.75 \cdot 10^{-5} T

Then we can calculate the magnitude of the force exerted on each wire by this magnetic field, which is given by:
F=ILB=(135 A)(222 m)(6.75 \cdot 10^{-5}T)=2.03 N

2) direction of the force: 
The two currents run in opposite direction: this means that the force between them is repulsive. This can be determined by using the right hand rule. Let's apply it to one of the two wires, assuming they are in the horizontal plane, and assuming that the current in the wire on the right is directed northwards:
- the magnetic field produced by the wire on the left at the location of the wire on the right is directed upward (the thumb of the right hand is directed as the current, due south, and the other fingers give the direction of the magnetic field, upward)

Now let's apply the right-hand rule to the wire on the right:
- index finger: current --> northward
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A similar procedure can be used on the wire on the left, finding that the force exerted on it is directed westwards, so the force between the two wires is repulsive.
6 0
3 years ago
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REY [17]
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100,000 watts  =  100,000 joules/sec
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1 gigawatt  =  1 gigajoule/sec
1 petawatt  =  1 petajoule/sec

We don't care what frequency the transmission is using,
or who their morning DJ is.

5 0
3 years ago
What is the result of a mutation during replication
Sonbull [250]

The correct answer is A. The strands are different. Hope this helps! :)

8 0
3 years ago
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