No, it couldn't be.
On that scale, Neptune would be almost 1,740 MILES from the sun.
ON THAT SCALE !
Answer:
xf = 5.68 × 10³ m
yf = 8.57 × 10³ m
Explanation:
given data
vi = 290 m/s
θ = 57.0°
t = 36.0 s
solution
firsa we get here origin (0,0) to where the shell is launched
xi = 0 yi = 0
xf = ? yf = ?
vxi = vicosθ vyi = visinθ
ax = 0 ay = −9.8 m/s
now we solve x motion: that is
xf = xi + vxi × t + 0.5 × ax × t² ............1
simplfy it we get
xf = 0 + vicosθ × t + 0
put here value and we get
xf = 0 + (290 m/s) cos(57) (36.0 s)
xf = 5.68 × 10³ m
and
now we solve for y motion: that is
yf = yi + vyi × t + 0.5 × ay × t
² ............2
put here value and we get
yf = 0 + (290 m/s) × sin(57) × (36.0 s) + 0.5 × (−9.8 m/s2) × (36.0 s) ²
yf = 8.57 × 10³ m
Answer:
The potential energy of the ball is 784 joules. And the kinetic energy of it is 392 while falling halfway down.
Explanation:
PE = mass (2kg) * Gravitational acceleration (9.8 m/s^2)* height (40 meters)
KE = 1/2 mass (1 kg) * velocity^2 (19.8)
Answer:
The speed will be "1.06 m/s".
Explanation:
The given values are:
Momentum,
m1 = 244 g
m2 = 45.2 g
On applying momentum conservation
,
Let v2 become the final golf's speed.
From Momentum Conservation
⇒ 
⇒ 
On putting the estimated values, we get
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
