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scoundrel [369]
1 year ago
6

If a satellite is orbiting the Earth in elliptical motion, then it will move _______________ (slowest, fastest) when its closest

to the Earth. While moving towards the Earth (along the path from D to A) there is a component of force in the __________________ (same, opposite) direction as the motion; this causes the satellite to ___________________ (slow down, speed up). While moving away from the Earth (along the path from A to D) there is a component of force in the _________________ (same, opposite) direction as the motion; this causes the satellite to ___________________ (slow
Physics
1 answer:
AVprozaik [17]1 year ago
8 0

Answer:fastest,same,slow down,opposite,slow

Explanation:

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A carousel that is 5.00 m in radius has a pair of 600-Hz sirens mounted on posts at opposite ends of a diameter. The carousel ro
Gelneren [198K]

Answer:

59cm

Explanation:

angular velocity = 0.8 rad/s

linear velocity = angular velocity * radius

                        =0.8rad/s * 5m

                        = 4 m/s

wavelength = (V + U)/F

where,

V is the velocity of the wave

U is the velocity of the source

F is the frequency of the source.

wavelength = (350 m/s + 4 m/s ) / 600 Hz

Wavelength = 0.59m or 59 cm

4 0
2 years ago
A bicyclist of mass 112 kg rides in a circle at a speed of 8.9 m/s. If the radius of the circle is 15.5 m, what is the centripet
kogti [31]
The centripetal force, Fc, is calculated through the equation, 
                                    Fc = mv²/r
where m is the mass,v is the velocity, and r is the radius. 
Substituting the known values,
                                     Fc = (112 kg)(8.9 m/s)² / (15.5 m)
                                         = 572.36 N
Therefore, the centripetal force of the bicyclist is approximately 572.36 N. 
4 0
2 years ago
Read 2 more answers
Three point charges are arranged on a line. Charge q3 = +5.00 nC and is at the origin. Charge q2 = -2.00 nC and is at x = 5.00 c
tatuchka [14]

Answer:

q₁= +0.5nC

Explanation:

Theory of electrical forces

Because the particle q3 is close to three other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

To solve this problem we apply Coulomb's law:

Two point charges (q1, q2) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:

o solve this problem we apply Coulomb's law:  

Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:  

F=K*q₁*q₂/d² Formula (1)  

F: Electric force in Newtons (N)

K : Coulomb constant in N*m²/C²

q₁,q₂:Charges in Coulombs (C)  

d: distance between the charges in meters

Data:

Equivalences

1nC= 10⁻⁹ C

1cm= 10⁻² m

Data

q₃=+5.00 nC =+5* 10⁻⁹ C

q₂= -2.00 nC =-2* 10⁻⁹ C

d₂= 5.00 cm= 5*10⁻² m

d₁= 2.50 cm=  2.5*10⁻² m

k = 8.99*10⁹ N*m²/C²

Calculation of magnitude and sign of q1

Fn₃=0 : net force on q3 equals zero

F₂₃:The force F₂₃ that exerts q₂ on q₃ is attractive because the charges have opposite signs,in direction +x.

F₁₃:The force F₂₃ that exerts q₂ on q₃ must go in the -x direction so that Fn₃ is zero, therefore q₁ must be positive and F₂₃ is repulsive.

We propose the algebraic sum of the forces on q₃

F₂₃ - F₁₃=0

\frac{k*q_{2} *q_{3} }{d_{2}^{2}  } -\frac{k*q_{1} *q_{3} }{d_{1}^{2}  }=0

We eliminate k*q₃ of the equation

\frac{q_{1} }{d_{1}^{2}  } = \frac{q_{2} }{d_{2}^{2}  }

q_{1} =\frac{q_{2} *d_{1} ^{2} }{d_{2}^{2}  }

q_{1} =\frac{2*10^{-9}*2.5^{2}*10^{-4}   }{5^{2}*10^{-4}  }

q₁= +0.5*10⁻⁹ C

q₁= +0.5nC

4 0
1 year ago
The difference in heights of the liquid in the two sides of the manometer is 43.4 cm when the atmospheric pressure is 755 mm hg.
Schach [20]

The atmospheric P is greater than the P in the flask, since the Hg level is lacking down lower on the side open to the atmosphere. 

43.4 cm x (10 mm / 1 cm) = 435 mm 

the density of Hg is 13.6 / 0.791 = 17.2 times better than the liquid in the manometer. This means that 1 mmHg = 17.2 mm of manometer liquid. 

435 mm manometer liquid x (1 mm Hg / 17.2 mm manometer liquid) = 25.3 mm Hg 

The pressure in the flask is 755 - 25.3 = 729.7 mmHg. 

729.7 mmHg x (1 atm / 760 mmHg ) = 0.960 atm.

4 0
1 year ago
Which is found farthest from the center of an atom?
xxTIMURxx [149]
Electron

Hope this helps
5 0
1 year ago
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