Answer:i dont remember
Explanation: ?>>> confused
Answer: An jack makes changing a tire easier because it lifts up the car to get the tire off of the ground.
Explanation:
Answer:
Explanation:
The equation for this is
where f is the frequency, v is the velocity, and lambda is the wavelength. Filling in:
and
which means that
the wavelength is 1.37 m, rounded to the correct number of significant digits.
Answer: c. 1.3 m/s^2
Explanation:
When he is at rest, is weight can be calculated as:
W = g*m
where:
m = mass of the man
g = gravitational acceleration = 9.8m/s^2
We know that at rest his weight is W = 824N, then we have:
824N = m*9.8m/s^2
824N/(9.8m/s^2) = m = 84.1 kg
Now, when the elevators moves up with an acceleration a, the acceleration that the man inside fells down is g + a.
Then the new weight is calculated as:
W = m*(g + a)
and we know that in this case:
W = 932N
g = 9.8m/s^2
m = 84.1 kg
Then we can find the value of a if we solve:
932N = 84.1kg*(9.8m/s^2 + a)
932N/84.1kg = 11.1 m/s^2 = 9.8m/s^2 + a
11.1 m/s^2 - 9.8m/s^2 = a = 1.3 m/s^2
The correct option is C
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