Answer:
20.96 m/s
Explanation:
Using the equations of motion
y = uᵧt + gt²/2
Since the puck slides off horizontally,
uᵧ = vertical component of the initial velocity of the puck = 0 m/s
y = vertical height of the platform = 2 m
g = 9.8 m/s²
t = time of flight of the puck = ?
2 = (0)(t) + 9.8 t²/2
4.9t² = 2
t = 0.639 s
For the horizontal component of the motion
x = uₓt + gt²/2
x = horizontal distance covered by the puck
uₓ = horizontal component of the initial velocity = 20 m/s
g = 0 m/s² as there's no acceleration component in the x-direction
t = 0.639 s
x = (20 × 0.639) + (0 × 0.639²/2) = 12.78 m
For the final velocity, we'll calculate the horizontal and vertical components
vₓ² = uₓ² + 2gx
g = 0 m/s²
vₓ = uₓ = 20 m/s
Vertical component
vᵧ² = uᵧ² + 2gy
vᵧ² = 0 + 2×9.8×2
vᵧ = 6.26 m/s
vₓ = 20 m/s, vᵧ = 6.26 m/s
Magnitude of the velocity = √(20² + 6.26²) = 20.96 m/s
Answer:
21 m
Explanation:
The motion of the frog is a uniform motion (constant speed), therefore we can find the distance travelled by using

where
d is the distance covered
v is the speed
t is the time
The frog in this problem has a speed of
v = 2.1 m/s
and therefore, after t = 10 s, the distance it covered is

Answer:
F=(-4.8*10^22,0,0) N
Explanation:
<u>Given :</u>
We are given the magnitude of the momentum of the planet and let us call this momentum (p_now) and it is given by p_now = 2.60 × 10^29 kg·m/s. Also, we are given the force exerted on the planet F = 8.5 × 10^22 N. and the angle between the planet and the star is Ф = 138°
Solution :
We are asked to find the parallel component of the force F The momentum here is not constant, where the planet moving along a curving path with varying speed where the rate change in momentum and the force may be varying in magnitude and direction. We divide the force here into two parts: a parallel force F to the momentum and a perpendicular force F' to the momentum.
The parallel force exerted to the momentum will speed or reduce the velocity of the planet and does not change its moving line. Let us apply the direction cosines, we could obtain the parallel force as next
F=|F|cosФp (1)
Where the parallel force F is in the opposite direction of p as the angle between them is larger than 90°. Now we can plug our values for 0 and I F I into equation (1) to get the parallel force to the planet
F=|F|cosФp
=-4.8*10^22 N*p
<em>As this force is in one direction, we could get its vector as next </em>
F=(-4.8*10^22,0,0) N
F=(0,-4.8*10^22,0) N
F=(0,0-4.8*10^22) N
The cosine of 138°, the angle between F and p is, is a negative number, so F is opposite to p. The magnitude of the planet's momentum will decrease.
A. iodine is located in period 5, group 17 and is classified
as a nonmetal
Group is the column, period is the raw.
1 astronomical unit = 149597870700m
Enrico should divide distance in meters with this number.