Answer: 60m/s
Explanation:
From the diagram:
Θ = 30°
Vertical resolution (y-axis) :
Voy = VoSinΘ
g in the upward direction = negative (-) = - g
Vfinal = 0
Distance (H) traveled along y =
Time taken to reach maximum height :
From v = u + at
0 = usinΘ - gt
gt = usinΘ
t = usinΘ / g
Horizontal resolution:
S = ut + 1/2at^2
Substituting t = usinΘ / g ; Voy = usinΘ
S = (usinΘ × usinΘ / g) - 1/2 g × (usinΘ /g)^2
S = (u^2sin^2Θ / g) - (u^2sin^2Θ / 2g)
S = (u^2sin^2Θ) / 2g
Now if S = maximum height = 45m
Then,
45 = [Vo^2sin^2(30°)] / 2(10)
45 =[ Vo^2 * (0.5)^2] / 20
45 =( Vo^2 * 0.25) / 20
20 * 45 = Vo^2 * 0.25
900 / 0.25 = Vo^2
3600 = Vo^2
Vo = sqrt(3600)
Vo = 60m/s
Answer:
t = 23.255 s, x = 2298.98 m, v_y = - 227.90 m / s
Explanation:
After reading your extensive writing, we are going to solve the approach.
The initial speed of the plane is 250 miles / h and it is at an altitude of 2650 m; In general, planes fly horizontally for launch, therefore this is the initial horizontal speed.
As there is a mixture of units in different systems we are going to reduce everything to the SI system.
v₀ₓ = 250 miles h (1609.34 m / 1 mile) (1 h / 3600 s) = 111.76 m / s
y₀ = 2650 m
Let's set a reference system with the x-axis parallel to the ground, the y-axis is vertical. As time is a scalar it is the same for vertical and horizontal movement
Y axis
y = y₀ + v₀ t - ½ g t²
the initial vertical velocity when the cargo is dropped is zero and when it reaches the floor the height is zero
0 = y₀ + 0 - ½ g t²
t =
t = √(2 2650/ 9.8)
t = 23.255 s
Therefore, for the cargo to reach the desired point, it must be launched from a distance of
x = v₀ₓ t
x = 111.76 23.255
x = 2298.98 m
at the point and arrival the speed is
vₓ = v₀ₓ = 111.76
vertical speed is
v_y = v_{oy} - gt
v_y = 0 - gt
v_y = - 9.8 23.25 555
v_y = - 227.90 m / s
the negative sign indicates that the speed is down
in the attachment we have a diagram of the movement
F = qE + qV × B
where force F, electric field E, velocity V, and magnetic field B are vectors and the × operator is the vector cross product. If the electron remains undeflected, then F = 0 and E = -V × B
which means that |V| = |E| / |B| and the vectors must have the proper geometrical relationship. I therefore get
|V| = 8.8e3 / 3.7e-3
= 2.4e6 m/sec
Acceleration a = V²/r, where r is the radius of curvature.
a = F/m, where m is the mass of an electron,
so qVB/m = V²/r.
Solving for r yields
r = mV/qB
= 9.11e-31 kg * 2.37e6 m/sec / (1.60e-19 coul * 3.7e-3 T)
= 3.65e-3 m
Answer:
B. 7.5 m/s^2
Explanation:
To find acceleration you need to subtract the final velocity by the starting velocity then divide that by the time
a= v-v/t
a= 60-0/8
a= 60/8
a=7.5 m/s^2
Answer:
1 casparian strips are present in the root of endodermis.
2 the endarch condition is the character fature of stem.