Answer:
23.52 m/s
Explanation:
The following data were obtained from the question:
Time taken (t) to reach the maximum height = 2.4 s
Acceleration due to gravity (g) = 9.8 m/s²
Initial velocity (u) =..?
At the maximum height, the final velocity (v) is zero. Thus, we can obtain how fast the rock (i.e initial velocity)
was thrown as follow:
v = u – gt (since the rock is going against gravity)
0 = u – (9.8 × 2.4)
0 = u – 23.52
Collect like terms
0 + 23.52 = u
u = 23.52 m/s
Therefore, the rock was thrown at a velocity of 23.52 m/s.
Answer:
3.03e-19 J
Explanation:
Use the formula E = hc/λ
Where:
h (Planck's constant) = 6.626e-34 J*s
c (speed of light, constant) = 3.00e8 m/s
λ (wavelength) = 656e-9 m
E = energy (in Joules)
E = (6.626e-34 * 3.00e8) / 656e-9 = 3.03018293e-19 = 3.03e-19 J
V ( initial ) = 20 m/s
h = 2.30 m
h = v y * t + g t ² / 2
d = v x * t
1 ) At α = 18°:
v y = 20 * sin 18° = 6.18 m/s
v x = 20 * cos 18° = 19.02 m/ s
2.30 = 6.18 t + 4.9 t²
4.9 t² + 6.18 t - 2.30 = 0
After solving the quadratic equation ( a = 4.9, b = 6.18, c = - 2.3 ):
t 1/2 = (- 6.18 +/- √( 6.18² - 4 * 4.9 * (-2.3)) ) / ( 2 * 4.9 )
t = 0.3 s
d 1 = 19.02 m/s * 0.3 s = 5.706 m
2 ) At α = 8°:
v y = 20* sin 8° = 2.78 m/s
v x = 20* cos 8° = 19.81 m/s
2.3 = 2.78 t + 4.9 t²
4.9 t² + 2.78 t - 2.3 = 0
t = 0.46 s
d 2 = 19.81 * 0.46 = 9.113 m
The distance is:
d 2 - d 1 = 9.113 m - 5.706 m = 3.407 m
GOOD LUCK AND HOPE IT HELPS U
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Answer:
Current, I = 3.57A
Explanation:
A current of I amperes means that I Coulombs of charge flows through the conductor (heating coil) per second.
Therefore, in time t, the total charge (Q) passing through any point in which the current (C) flows will be given by the equation;
Q = It
Where; Q is the charge in coulombs; I is the current in amperes; t is the time in seconds.
From the question, we were given the following parameters;
Q = 25C, t= 7secs and I =?
From the equation, Q = It
We make current, I the subject of formula;
Thus, I = Q/t
Substituting into the equation;
I = 25/7
I = 3.57Amp.