This is a projectile motion problem, so, we use the formula for trajectory:
y =xtanα + gx^2/2v^2(cosα)^2
where
y is the vertical distance (y = 50 m)
x is the horizontal distance (x=90 m)
α is the angle of trajectory; since it levels of HORIZONTALLY, α = 0°
v is the initial velocity
g is the acceleration due to gravity which is 9.81 m/s^2
Substituting to the formula,
50 =90tan(0°) + (9.81)(90)^2/2v^2(cos0°)^2
v = 28.2 m/s
Answer:
Explanation:
Conservation of momentum
Initial momentum is zero as both gun and bullet are motionless in the frame of reference before firing.
0 = 0.012(1500.0) + 4.0v
v = - 4.5 m/s
Must be a rail gun as the fastest powder driven bullets are moving at roughly half the speed described.
Formula is calculated by
GDP deflator =Nominal GDP/Real GDP ×(100/1)
Answer:
4.56×10¯⁷¹ N
Explanation:
From the question given above, the following data were obtained:
Distance apart (r) = 1.10 m
Force (F) =?
NOTE:
Gravitational constant (G) = 6.67×10¯¹¹ Nm² /Kg²
Mass of electron = 9.1×10¯³¹ Kg
Mass of the two elections = M₁ = M₂ = 9.1×10¯³¹ Kg
Thus, we can obtain the force of attraction between the two elections as illustrated below:
F = GM₁M₂ / r²
F = 6.67×10¯¹¹ × (9.1×10¯³¹)² / (1.1)²
F = 4.56×10¯⁷¹ N
Thus, the force of attraction between the two elections is 4.56×10¯⁷¹ N