Answer:
Explanation:
You can approach an expression for the instantaneous velocity at any point on the path by taking the limit as the time interval gets smaller and smaller. Such a limiting process is called a derivative and the instantaneous velocity can be defined as.#3
For the special case of straight line motion in the x direction, the average velocity takes the form: If the beginning and ending velocities for this motion are known, and the acceleration is constant, the average velocity can also be expressed as For this special case, these expressions give the same result. Example for non-constant acceleration#1
Not if both speeds are in the same units.
However, if the 254 is 'centimeters per time' and the 100 is 'inches per time',
then the speeds are equal.
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
Answer:
55000 W/m²
Explanation:
Parameters given:
Surface temperature, T = 1000°C
Hear transfer coefficient, h = 55 W/m²C
Convection heat transfer coefficient is given as:
h = Heat flux/Temperature
Hence, Heat Flux, q, is given as:
q = h * T
q = 55 * 1000 = 55000 W/m²C
Answer:
a) T ’= 0.999 s
, b) t = 3596.4 s
Explanation:
The angular velocity of a simple pendulum is
w = √g / L
The angular velocity, frequency and period are related
w = 2π f = 2π / T
2π / T = √ g / L
T = 2π √ L / g
L = T² g / 4π²
L = 1² 9.8 / 4π²
L = 0.248 m
To know the effect of the temperature change let's use the thermal expansion ratios
ΔL = α L ΔT
ΔL = 24 10⁻⁶ 0.248 (-4 - 20)
ΔL = 142.8 10⁻⁶ m
Lf - L = -142. 8 10⁻⁶
Lf = 142.8 10⁻⁶ + 0.248
Lf = 0.2479 m
Let's calculate new period
T ’= 2π √ L / g
T ’= 2π √ (0.2479 / 9.8)
T ’= 0.999 s
We can see that the value of the period is reduced so that the clock is delayed
b) change of time in 1 hour
When the clock is at 20 ° C in one hour it performs 3600 oscillations, for the new period the time of this number of oscillations is
t = 3600 0.999
t = 3596.4 s
Therefore the clock is delayed almost 4 s
