For the first one it’s the guh sound
And the second one it’s the kuh sound
Projectile motion is characterized by an arc-shaped direction of motion. It is acted upon by two vector forces: horizontal component and vertical component. The horizontal component is in constant velocity motion, while the vertical component is in constant acceleration motion. These two motions are independent of each other.
Now, the total velocity of the space probe at the end of the projectile motion is determined through this equation:
V = √(Vx² + Vy²)
where Vx is the velocity in the horizontal direction and Vy is the velocity in the vertical direction.
Let's find Vx first. Assuming that the space probe was launched at an angle horizontal the Earth's surface, the launching angle is 0°. Thus, the initial velocity is 2.44×10⁴ m/s.
For Vy, the free falling motion is
Vy = √(2gh), where g is 9.81 m/s² and h is the distance traveled vertically by the space probe which represents the radius of the Earth equal to 6.37×10⁶ meters. Therefore,
V = √{(2.44×10⁴)² + [√(2×9.81×6.37×10⁶ )]²}
V = 26,839.14 m/s
Answer:
85.8 m/s
Explanation:
We know that the length of the circular path, L the plane travels is
L = rθ where r = radius of path and θ = angle covered
Now,its speed , v = dL/dt = drθ/dt = rdθ/dt + θdr/dt
where dθ/dt = ω = angular speed = v'/r where v' = maximum speed of plane and r = radius of circular path
Now, from θ = θ₀ + ωt where θ₀ = 0 rad, ω = angular speed and t = time,
θ = θ₀ + ωt = 0 + ωt = ωt
So, v = rdθ/dt + θdr/dt
v = rω + ωtdr/dt
v = (r + tdr/dt)ω
v = (r + tdr/dt)v'/r
v = v' + tv'/r(dr/dt)
v = v'[1 + t(dr/dt)/r]
Given that v' = 110 m/s, t = 33.0s, r = 120 m and dr/dt = rate at which line is shortened = -0.80 m/s (negative since it is decreasing)
So, v = 110 m/s[1 + 33.0 s(-0.80 m/s)/120 m]
v = 110 m/s[1 + 11.0 s(-0.80 m/s)/40 m]
v = 110 m/s[1 + 11.0 s(-0.02/s)]
v = 110 m/s[1 - 0.22]
v = 110 m/s(0.78)
v = 85.8 m/s
Answer:
T = 15.03°C
Explanation:
given data:
copper specific heat = Sc = 0.385 J/g °C
iron specific iron = Si = 0.450 J/g °C
specific heat of ethanol = Se = 2.46 J/g °C
net heat loss is equal to zero
T = 15.03°C
KE = 1/2 mv^2 is the relationship betwee mass and kinetic energy
