The muzzle speed is 144 m/s.
Let x = the angle of elevation.
Then
u = 144 cos(x), the horizontal velocity
v = 144 sin(x), the vertical launch velocity
Assume no air resistance, and g = 9.8 m/s².
To hit the target 780 m away, the time of flight is
t = (780 m)/[144 cos(x)] = 5.4167 sec(x) s
In terms of the vertical velocity,
(144 sin(x) m/s)*(5.4167 sec(x) s) + (1/2)*(-9.8 m/s²)*(5.4167 sec(x) m/s)² = (780 m)
780 tan(x) - 143.769 sec²x = 780
tan(x) - 0.1843 (1 + tan²x) = 1
0.1843 tan²(x) - tan(x) + 1.1843 = 0
tan²x - 5.4259 tan(x) + 6.4259 = 0
Solve with the quadratic formula.
tan(x) = 0.5[5.4259 +/- √(3.737)] = 3.6795 or 1.7464
x = tan⁻¹ 3.6795 = 74.8°
or
x = tan⁻¹ 1.7464 = 60.2°
Answer: (74.8°, 60.2°)
Answer:
speed of golf ball is 1.15 × m/s
and % of uncertainty in speed = 2.07 × %
Explanation:
given data
mass = 45.9 gram = 0.0459 kg
speed = 200 km/hr = 55.5 m/s
uncertainty position Δx = 1 mm = m
to find out
speed of the golf ball and % of speed of the golf ball
solution
we will apply here heisenberg uncertainty principle that is
uncertainty position ×uncertainty momentum ≥ ......1
Δx × ΔPx ≥
here uncertainty momentum ΔPx = mΔVx
and uncertainty velocity = ΔVx
and h = 6.626 × Js
so put here all these value in equation 1
× 0.0459 × ΔVx =
ΔVx = 1.15 × m/s
and
so % of uncertainty in speed = ΔV / m
% of uncertainty in speed = 1.15 × / 55.5
% of uncertainty in speed = 2.07 × %
Answer:
If the buoyant force is greater than the object's weight, the object will rise to the surface and float. If the buoyant force is less than the object's weight, the object will sink. If the buoyant force equals the object's weight, the object will remain suspended at that depth.
Explanation:
Not much explaining to do here!
Answer:
- Here we use the conservation of momentum theorem.
- m stands for mass, and v stands for velocity. The numbers refer to the respective objects.
- m1v1 + m2v2 = m1vf1 + m2vf2
- Since the equation is perfectly inelastic, the final velocity of both masses is the same. Let’s account for this in our formula.
- m1v1 + m2v2 = vf(m1 + m2)
<u>Let’s substitute in our givens.</u>
(0.002 kg)(700 m/s) + (5 kg)(0 m/s) = vf(0.002 kg + 5 kg)
I assume you are proficient in algebra I, so I will not include the steps to simplify this equation.
Note that I have considered the bullet’s velocity to be in the positive direction,
The answer is vf = 0.280 m/s