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Sergeeva-Olga [200]
2 years ago
12

Julietta and Jackson are playing miniature golf. Julietta's ball rolls into a long. Straight upward incline with a speed of 2.95

m/s and accelerates at -0.876 m/s/s for 1.54 seconds until it reaches the top of the incline and then continues along an elevated section. Determine the length of the incline.
Physics
1 answer:
Verdich [7]2 years ago
6 0

Answer:

The length of the incline is 3.504 meters.

Explanation:

Let suppose that Julietta's ball decelerates uniformly, then we determine the length of the incline is determined by the following equation of motion:

\Delta s = v_{o}\cdot t +\frac{1}{2}\cdot a \cdot t^{2} (Eq. 1)

Where:

\Delta s - Length of the incline, measured in meters.

v_{o} - Initial speed of the ball, measured in meters per second.

a - Aceleration of the ball, measured in meters per square second.

t - Time, measured in second.

If we know that v_{o} = 2.95\,\frac{m}{s}, t = 1.54\,s and a = -0.876\,\frac{m}{s^{2}}, then the length of the incline is:

\Delta s = \left(2.95\,\frac{m}{s} \right)\cdot (1.54\,s)+\frac{1}{2}\cdot \left(-0.876\,\frac{m}{s^{2}} \right) \cdot (1.54\,s)^{2}

\Delta s = 3.504\,m

The length of the incline is 3.504 meters.

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Calculate the orbital period for Jupiter's moon Io, which orbits 4.22×10^5km from the planet's center (M=1.9×10^27kg) .
Verdich [7]

According to the <u>Third Kepler’s Law of Planetary motion</u> “<em>The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.</em>



In other words, this law states a relation between the orbital period T of a body (moon, planet, satellite) orbiting a greater body in space with the size a of its orbit.



This Law is originally expressed as follows:



<h2>T^{2} =\frac{4\pi^{2}}{GM}a^{3}    (1) </h2>

Where;


G is the Gravitational Constant and its value is 6.674(10^{-11})\frac{m^{3}}{kgs^{2}}



M=1.9(10^{27})kg is the mass of Jupiter


a=4.22(10^{5})km=4.22(10^{8})m  is the semimajor axis of the orbit Io describes around Jupiter (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)



If we want to find the period, we have to express equation (1) as written below and substitute all the values:



<h2>T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}    (2) </h2>

T=\sqrt{\frac{4\pi^{2}}{6.674(10^{-11})\frac{m^{3}}{kgs^{2}}1.9(10^{27})kg}(4.22(10^{8})m)^{3}}    



T=\sqrt{\frac{2.966(10^{27})m^{3}}{1.268(10^{17})m^{3}/s^{2}}}    



T=\sqrt{2.339(10^{10})s^{2}}    



Then:


<h2>T=152938.0934s    (3) </h2>

Which is the same as:



<h2>T=42.482h     </h2>

Therefore, the answer is:



The orbital period of Io is 42.482 h



7 0
3 years ago
1. A man walks round a park, first walking north for 80m, then turning right and walking
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Answer:

Total distance travelled = 210m

Explanation:

Distance travelled = 80m + 50m + 10m + 70m

                               = 210m

8 0
3 years ago
A closely wound search coil has an area of 3.21 cm2, 120 turns, and a resistance of 58.7 O. It is connected to a charge-measurin
Alexxx [7]

Answer:

The magnetic field in the System is 0.095T

Explanation:

To solve the exercise it is necessary to use the concepts related to Faraday's Law, magnetic flux and ohm's law.

By Faraday's law we know that

\epsilon = \frac{NBA}{t}

Where,

\epsilon  =electromotive force

N = Number of loops

B = Magnetic field

A = Area

t= Time

For Ohm's law we now that,

V = IR

Where,

I = Current

R = Resistance

V = Voltage (Same that the electromotive force at this case)

In this system we have that the resistance in series of coil and charge measuring device is given by,

R = R_c + R_d

And that the current can be expressed as function of charge and time, then

I = \frac{q}{t}

Equation Faraday's law and Ohm's law we have,

V = \epsilon

IR = \frac{NBA}{t}

(\frac{q}{t})(R_c+R_d) = \frac{NBA}{t}

Re-arrange for Magnetic Field B, we have

B = \frac{q(R_c+R_d)}{NA}

Our values are given as,

R_c = 58.7\Omega

R_d = 45.5\Omega

N = 120

q = 3.53*10^{-5}C

A = 3.21cm^2 = 3.21*10^{-4}m^2

Replacing,

B = \frac{(3.53*10^{-5})(58.7+45.5)}{120*3.21*10^{-4}}

B = 0.095T

Therefore the magnetic field in the System is 0.095T

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Assume that two of the electrons at the negative terminal have attached themselves to a nearby neutral atom. There is now a nega
devlian [24]

Answer: its negative

Explanation: becase it is

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A hummingbird can flutter its wings 4,800 times per minute.
Marianna [84]
80 flutters per second
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3 years ago
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