Answer:
Yes
Explanation:
The given parameters are;
The speed with which the fastball is hit, u = 49.1 m/s (109.9 mph)
The angle in which the fastball is hit, θ = 22°
The distance of the field = 96 m (315 ft)
The range of the projectile motion of the fastball is given by the following formula
Where;
g = The acceleration due to gravity = 9.81 m/s², we have;
Yes, given that the ball's range is larger than the extent of the field, the batter is able to safely reach home.
Answer:
(a) 1.16 s
(b)0.861 Hz
Explanation:
(a) Period : The period of a simple harmonic motion is the time in seconds, required for a object undergoing oscillation to complete one cycle.
From the question,
If 1550 cycles is completed in (30×60) seconds,
1 cycle is completed in x seconds
x = 30×60/1550
x = 1.16 s
Hence the period is 1.16 seconds.
(b) Frequency : This can be defined as the number of cycles that is completed in one seconds, by an oscillating body. The S.I unit of frequency is Hertz (Hz).
Mathematically, Frequency is given as
F = 1/T ........................... Equation 1
Where F = frequency, T = period.
Given: T = 1.16 s.
Substitute into equation 1
F = 1/1.16
F = 0.862 Hz
Hence thee frequency = 0.862 Hz
Answer:
0.11 kg
Explanation:
Ft = MV
Ft = momentum 5.22kg m/s
M = mass
V = velocity 48.3m/s
Therefore
5.22 = M x 48.3
Divide both sides by 48.3
5.22/48.3 = M x 48.3/48.3
0.11 = M
M = 0.11kg
Answer:
The volume of the balloon increases in the upper atmosphere.
Explanation:
p1= 1 atm
p2= 0.15 atm
V1= 15.6 L
V2= ?
p1*V1= p2 * V2
V2= (p1/p2)*V1
V2= 104 L
Complete Question
Planet D has a semi-major axis = 60 AU and an orbital period of 18.164 days. A piece of rocky debris in space has a semi major axis of 45.0 AU. What is its orbital period?
Answer:
The value is
Explanation:
From the question we are told that
The semi - major axis of the rocky debris
The semi - major axis of Planet D is
The orbital period of planet D is
Generally from Kepler third law
Here T is the orbital period while a is the semi major axis
So
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