Why couldn't nawaz hear the bell when the air had been pumped out of the jar?
1 answer:
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Answer:
y = 17 m
Explanation:
For this projectile launch exercise, let's write the equation of position
x = v₀ₓ t
y = t - ½ g t²
let's substitute
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
the maximum height the ball can reach where the vertical velocity is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
0 = v₀ sin θ - 9.8 t
Let's write our system of equations
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
0 = v₀ sin θ - 9.8 t
We have a system of three equations with three unknowns for which it can be solved.
Let's use the last two
v₀ sin θ = 9.8 t
we substitute
10 = (9.8 t) t - ½ 9.8 t2
10 = ½ 9.8 t2
10 = 4.9 t2
t = √ (10 / 4.9)
t = 1,429 s
Now let's use the first equation and the last one
45 = v₀ cos θ t
0 = v₀ sin θ - 9.8 t
9.8 t = v₀ sin θ
45 / t = v₀ cos θ
we divide
9.8t / (45 / t) = tan θ
tan θ = 9.8 t² / 45
θ = tan⁻¹ ( 9.8 t² / 45 )
θ = tan⁻¹ (0.4447)
θ = 24º
Now we can calculate the maximum height
v_y² = - 2 g y
vy = 0
y = v_{oy}^2 / 2g
y = (20 sin 24)²/2 9.8
y = 3,376 m
the other angle that gives the same result is
θ‘= 90 - θ
θ' = 90 -24
θ'= 66'
for this angle the maximum height is
y = v_{oy}^2 / 2g
y = (20 sin 66)²/2 9.8
y = 17 m
thisis the correct
5m/s
100m
Explanation:
Average speed is sum of distance distance traveled in a given time by a body.
Average speed=
Distance = 20m
time = 4s
Average speed = = 5m/s
For the spaceship;
Distance covered = speed x time
Speed = 50m/s
time = 2s
Distance covered = 50 x 2 = 100m
learn more:
Average speed brainly.com/question/5063905
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Answer:
The same as the escape velocity of asteorid A (50m/s)
Explanation:
The escape velocity is described as follows:
where is the universal gravitational constant,
is the mass of the asteroid and
is the radius
and since the scape velocity is 50m/s:
Now, if the astroid B has twice mass and twice the radius, we have that tha mass is:
and the radius is:
inserting these values into the formula for escape velocity:
and we have found that , so the two asteroids have the same escape velocity.
We found that the expression for escape velocity remains the same as for asteroid A, this because both quantities (radius and mass) doubled, so it does not affect the equation.
The answer is
Asteroid B would have an escape velocity the same as the escape velocity of asteroid A