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lawyer [7]
3 years ago
9

Calculat the sideral period of Mars given its synodic period is 780 days. 7. Calculate the synodic period of Venus given its sid

ereal period is 229 days.
Physics
1 answer:
Ad libitum [116K]3 years ago
4 0

Answer:

686 days for sideral period of Mars.

615 days for synodic period of Venus.

Explanation:

The equations we need to use is:

\frac{1}{P}=\frac{1}{E}\pm\frac{1}{S}

Where P is the sidereal period, S the synodic period and E the Earth's period (365 days) and where the + is used for inferior planes (Venus) and the - for superior ones (Mars).

For the sidereal period of Mars we then have:

P_M=\frac{1}{\frac{1}{E}-\frac{1}{S_M}}=\frac{1}{\frac{1}{365days}-\frac{1}{780days}}=686days

And for the synodic period of Venus we then have:

S_V=\frac{1}{\frac{1}{P_V}-\frac{1}{E}}=\frac{1}{\frac{1}{229days}-\frac{1}{365days}}=615days

(Commonly for Venus a sidereal period of 225 days is given, which changes this result to 587 days).

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A 4.00 µf capacitor is connected to a 12.0 v battery.
atroni [7]
(a) The capacitance of the capacitor is:
C=4 \mu F=4 \cdot 10^{-6}F
and the voltage applied across its plates is
V=12.0 V

The relationship between the charge Q on each plate of the capacitor, the capacitance and the voltage is:
C= \frac{Q}{V}
and re-arranging it we find the charge stored in the capacitor:
Q=CV=(4 \cdot 10^{-6} F)(12.0 V)=4.8 \cdot 10^{-5} C

(b) The electrical potential energy stored in a capacitor is given by
U= \frac{1}{2}CV^2
where C is the capacitance and V is the voltage. The new voltage is 
V=1.50 V
so the energy stored in the capacitor is
U= \frac{1}{2}(4 \cdot 10^{-6} F)(1.50 V)^2=4.5 \cdot 10^{-6} J
3 0
3 years ago
Read 2 more answers
An object is launched at a velocity of 20 m/s in a direction making an angle of 25° upward with the horizontal.
Hitman42 [59]

Answer:

(a) max. height = 3.641 m

(b) flight time = 1.723 s

(c) horizontal range = 31.235 m

(d) impact velocity = 20 m/s

Above values have been given to third decimal.  Adjust significant figures to suit accuracy required.

Explanation:

This problem requires the use of kinematics equations

v1^2-v0^2=2aS .............(1)

v1.t + at^2/2 = S ............(2)

where

v0=initial velocity

v1=final velocity

a=acceleration

S=distance travelled

SI units and degrees will be used throughout

Let

theta = angle of elevation = 25 degrees above horizontal

v=initial velocity at 25 degrees elevation in m/s

a = g = -9.81 = acceleration due to gravity (downwards)

(a) Maximum height

Consider vertical direction,

v0 = v sin(theta) = 8.452 m/s

To find maximum height, we find the distance travelled when vertical velocity = 0, i.e. v1=0,

solve for S in equation (1)

v1^2 - v0^2 = 2aS

S = (v1^2-v0^2)/2g = (0-8.452^2)/(2*(-9.81)) = 3.641 m/s

(b) total flight time

We solve for the time t when the vertical height of the object is AGAIN = 0.

Using equation (2) for vertical direction,

v0*t + at^2/2 = S    substitute values

8.452*t + (-9.81)t^2 = 3.641

Solve for t in the above quadratic equation to get t=0, or t=1.723 s.

So time for the flight = 1.723 s

(c) Horiontal range

We know the horizontal velocity is constant (neglect air resistance) at

vh = v*cos(theta) = 25*cos(25) = 18.126 m/s

Time of flight = 1.723 s

Horizontal range = 18.126 m/s * 1.723 s = 31.235 m

(d) Magnitude of object on hitting ground, Vfinal

By symmetry of the trajectory, Vfinal = v = 20, or

Vfinal = sqrt(v0^2+vh^2) = sqrt(8.452^2+18.126^2) = 20 m/s

7 0
4 years ago
How are distance and time related when describing motion?
Bingel [31]
an object moves along a straight line, the distance travelled can be represented by a distance-time graph. In a distance-time graph, the gradient of the line is equal to the speed of the object. The greater the gradient (and the steeper the line) the faster the object is moving
3 0
3 years ago
A car is traveling at a constant speed of 12m/s , when the driver accelerates the car reaches a speed of 26 m/s in 6s what is th
DaniilM [7]

Acceleration = (change in speed) / (time for the change)

Change in speed = (end speed) - (start speed)

Change in speed = (26 m/s) - (12 m/s) = 14 m/s

Time for the change = 6 s

Acceleration = (14 m/s) / (6 s)

Acceleration = (14/6) (m/s²)

<em>Acceleration = 2.33 m/s²</em>

7 0
3 years ago
In each case the momentum before the collision is: (2.00 kg) (2.00 m/s) = 4.00 kg * m/s
Ivan

Answer:

Check Explanation.

Explanation:

Momentum before collision = (2)(2) + (2)(0) = 4 kgm/s

a) Scenario A

After collision, Mass A sticks to Mass B and they move off with a velocity of 1 m/s

Momentum after collision = (sum of the masses) × (common velocity) = (2+2) × (1) = 4 kgm/s

Which is equal to the momentum before collision, hence, momentum is conserved.

Scenario B

They bounce off of each other and move off in the same direction, mass A moves with a speed of 0.5 m/s and mass B moves with a speed of 1.5 m/s

Momentum after collision = (2)(0.5) + (2)(1.5) = 1 + 3 = 4.0 kgm/s

This is equal to the momentum before collision too, hence, momentum is conserved.

Scenario C

Mass A comes to rest after collision and mass B moves off with a speed of 2 m/s

Momentum after collision = (2)(0) + (2)(2) = 0 + 4 = 4.0 kgm/s

This is equal to the momentum before collision, hence, momentum is conserved.

b) Kinetic energy is normally conserved in a perfectly elastic collision, if the two bodies do not stick together after collision and kinetic energy isn't still conserved, then the collision is termed partially inelastic.

Kinetic energy before collision = (1/2)(2.00)(2.00²) + (1/2)(2)(0²) = 4.00 J.

Scenario A

After collision, Mass A sticks to Mass B and they move off with a velocity of 1 m/s

Kinetic energy after collision = (1/2)(2+2)(1²) = 2.0 J

Kinetic energy lost = (kinetic energy before collision) - (kinetic energy after collision) = 4 - 2 = 2.00 J

Kinetic energy after collision isn't equal to kinetic energy before collision. This collision is evidently totally inelastic.

Scenario B

They bounce off of each other and move off in the same direction, mass A moves with a speed of 0.5 m/s and mass B moves with a speed of 1.5 m/s

Kinetic energy after collision = (1/2)(2)(0.5²) + (1/2)(2)(1.5²) = 0.25 + 3.75 = 4.0 J

Kinetic energy lost = 4 - 4 = 0 J

Kinetic energy after collision is equal to kinetic energy before collision. Hence, this collision is evidently elastic.

Scenario C

Mass A comes to rest after collision and mass B moves off with a speed of 2 m/s

Kinetic energy after collision = (1/2)(2)(0²) + (1/2)(2)(2²) = 4.0 J

Kinetic energy lost = 4 - 4 = 0 J

Kinetic energy after collision is equal to kinetic energy before collision. Hence, this collision is evidently elastic.

c) An impossible outcome of such a collision is that A stocks to B and they both move off together at 1.414 m/s.

In this scenario,

Kinetic energy after collision = (1/2)(2+2)(1.414²) = 4.0 J

This kinetic energy after collision is equal to the kinetic energy before collision and this satisfies the conservation of kinetic energy.

But the collision isn't possible because, the momentum after collision isn't equal to the momentum before collision.

Momentum after collision = (2+2)(1.414) = 5.656 kgm/s

which is not equal to the 4.0 kgm/s obtained before collision.

This is an impossible result because in all types of collision or explosion, the second law explains that first of all, the momentum is always conserved. And this evidently violates the rule. Hence, it is not possible.

6 0
3 years ago
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