Q = mass water x specific heat water x delta T.
<span>714,000 = mass water x specific heat water x 30.
Substitute specific heat water and solve for mass water.</span>
The maximum speed of the donkey is 10.72m/s
The question is based on the principle of motion in one dimension and hence formulas of motion in one dimension can be applied.
It is given that donkey attains an acceleration of 1.6 m/s^2
The time taken to accelerate to given speed is 6.7 seconds
We use the formula v=u + at to find the fastest speed
v is the final or maximum speed
u is the initial speed which in this case is 0 as the donkey is at rest
a is the acceleration of the donkey
t is the time taken in seconds
v = u + at
v= 0 + 1.6 x 6.7
= 10.72 m/s
Hence the donkey obtains the speed of 10.72 m/s
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I would assume air resistance is negligible and so the acceleration of the package would be approximately 9.81 m/s².
Taking downwards as positive, use v²=u²+2as.
v²=(-2)²+2(9.81)(14)
v=16.7 m/s
Answer:
option B
Explanation:
given,
diameter of the rotating space = 2 Km
Force exerted at the edge of the space = 1 g
force experienced at the half way = ?
As the object is rotating in the circular part
Force is equal to centripetal acceleration.
at the edge
g = ω² r
ω is the angular velocity of the particle
r is the radius.
now, acceleration at the half way
g' = ω² r'
People at the halfway experience g/2
hence, the correct answer is option B
Answer:
(a) 0.613 m
(b) 0.385 m
(c) vₓ = 1.10 m/s, vᵧ = 3.50 m/s
v = 3.68 m/s², θ = 72.6° below the horizontal
Explanation:
(a) Take down to be positive.
Given in the y direction:
v₀ = 0 m/s
a = 10 m/s²
t = 0.350 s
Find: Δy
Δy = v₀ t + ½ at²
Δy = (0 m/s) (0.350 s) + ½ (10 m/s²) (0.350 s)²
Δy = 0.613 m
(b) Given in the x direction:
v₀ = 1.10 m/s
a = 0 m/s²
t = 0.350 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (1.10 m/s) (0.350 s) + ½ (0 m/s²) (0.350 s)²
Δx = 0.385 m
(c) Find: vₓ and vᵧ
vₓ = aₓt + v₀ₓ
vₓ = (0 m/s²) (0.350 s) + 1.10 m/s
vₓ = 1.10 m/s
vᵧ = aᵧt + v₀ᵧ
vᵧ = (10 m/s²) (0.350 s) + 0 m/s
vᵧ = 3.50 m/s
The magnitude is:
v² = vₓ² + vᵧ²
v = 3.68 m/s²
The direction is:
θ = atan(vᵧ / vₓ)
θ = 72.6° below the horizontal
