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
Answer:
44.13015
Explanation:
use the 9.8067 newtons to 1 kg conversion
<h3>Hello there!</h3>
Here, you are looking for the amount of heat put in for water, at a mass of 187 grams, to change by 80 degrees.
The equation commonly accepted to find the answer to questions like these is the specific heat formula.
The equation is Q = mc∆T, where Q is the amount of energy put in to raise the temperature by a certain amount, m is the mass, c is the specific heat capacity, and ΔT is the amount of temperature change.
The information given:
m = 187 grams
c = specific heat capacity of water, or in this case 1 calorie, or 4.184 joules (which is what we will be using)
ΔT = 80 degrees
Now just plug everything in to solve.
Q = 187 * 4.184 * 80
Q = 62592.64
So you have your answer: 62592.64 joules.
Hope this helped!
Answer:
5 m/s2
Explanation:
The total acceleration of the circular motion is made of 2 components: centripetal acceleration and linear acceleration of 4 m/s2. They are perpendicular to each other.
The centripetal acceleration is the ratio of instant velocity squared and the radius of the circle
So the magnitude of the total acceleration is
Answer:
5.571 sec
Explanation:
angular frequency = √ (k/m) = √ (49.3 / 5) = 3.14 rad/s
Period To = 2π / angular frequency
Period To = 2π/3.14 = 2 × 3.14 / 3.142 = 2.00 sec which you got
T measured by the observer = To / (√ (1 - (v²/c²))) = 2 / √( 1 - 0.871111) = 2 / 0.35901 = 5.571 sec
t=2.00/(1-√((2.80*10^8)^2/(3.00*10^8)^2))= should have been ( To / (√ (1 - (v²/c²))). where To = 2.00 sec
