Answer:
681.6/ms
Explanation:
A reconnaissance plane flies 545 km away from its base at 568 m/s. then flies back to its base at 852 m/s.
What is its average speed?
Answer in its of m/s
Avg speed of the round trip is
2*568*852/(568+852)= 681.6/ms
Answer:
The minimum frequency of the coil is 7.1 Hz
Explanation:
Given;
number of turns, N = 200 turns
cross sectional area, A = 300 cm² = 300 x 10⁻⁴ m²
magnitude of magnetic field strength, B = 30 x 10⁻³ T
maximum value of the induced emf, E = 8 V
Maximum induced emf is given as;
E = NBAω
where
ω is angular velocity (ω = 2πf)
E = NBA2πf
where;
f is the minimum frequency, measured in hertz (Hz)
f = E / (NBA2π)
f = 8 / (200 x 30 x 10⁻³ x 300 x 10⁻⁴ x 2 x 3.142)
f = 7.073 Hz
f = 7.1 Hz
Therefore, the minimum frequency of the coil is 7.1 Hz
Answer: A) mass on earth surface = 5.91kg
B) mass on surface of jupiter = 5.91kg
C) weight on surface of jupiter = 10.697N
Explanation:
The relationship between weight (W), mass (m) and acceleration due gravity (g) is given below
W=mg
From the question, g= 9.8m/s² and weight on the surface on the earth is 58N
A) The mass of watermelon on earth is
m = 58/ 9.8 = 5.91kg
B) the mass of the watermelon on jupiter is 5.91kg.
You will notice this is the same as the mass of watermelon on earth and that is so because mass is a scalar quantity that does not depends on the distance away from the center of the earth (unlike weight which is a vector) thus making it constant all through any location.
C) mass of watermelon is 5.91kg, g=9.8m/s² weight of watermelon on jupiter is given below as
W = mg
W = 5.91 x 9.8
= 10.697N.
The frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.
To find the answer, we need to know about the time of flight and range of projectile motion.
<h3>What's the expression of range of a projectile motion?</h3>
- Range = U²× sin(2θ)/g
- U= initial velocity, θ= angle of projectile and g= acceleration due to gravity
- U=√{Range×g/sin(2θ)}
- Here, range= 2.20m, = 36.5°
- U= √{2.20×9.8/sin(73)}
U= √{2.20×9.8/sin(73)} = 22.5m/s
<h3>What's the expression of time of flight in projectile motion?</h3>
- Time of flight= (2×U×sinθ)/g
- So, T= (2×22.5×sin36.5°)/9.8
= 2.73 s
Thus, we can conclude that the frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.
Learn more about the range and time period of projectile motion here:
brainly.com/question/24136952
#SPJ1
A beta particle. Hoped I help. Sorry if it wrong.
