Answer:
1.17 m
Explanation:
From the question,
s₁ = vt₁/2................ Equation 1
Where s₁ = distance of the reflecting object for the first echo, v = speed of the sound in air, t₁ = time to dectect the first echo.
Given: v = 343 m/s, t = 0.0115 s
Substitute into equation 1
s₁ = (343×0.0115)/2
s₁ = 1.97 m.
Similarly,
s₂ = vt₂/2.................. Equation 2
Where s₂ = distance of the reflecting object for the second echo, t₂ = Time taken to detect the second echo
Given: v = 343 m/s, t₂ = 0.0183 s
Substitute into equation 2
s₂ = (343×0.0183)/2
s₂ = 3.14 m
The distance moved by the reflecting object from s₁ to s₂ = s₂-s₁
s₂-s₁ = (3.14-1.97) m = 1.17 m
Answer:
282 m
Explanation:
Given:
v₀ = 20.1 m/s
v = 33.2 m/s
t = 10.6 s
Find: Δx
Δx = ½ (v + v₀) t
Δx = ½ (33.2 m/s + 20.1 m/s) (10.6 s)
Δx ≈ 282 m
Whenever an object is falling, its potential energy
is decreasing and its kinetic energy is increasing.
Olivia's potential energy is decreasing and her kinetic energy
is increasing as she moves toward the right side of the picture,
all the way from W, through X, to the bottom of the arc.
Answer:
E = 0.18 J
Explanation:
given,
Potential of the battery,V = 9 V
Charge on the circuit, Q = 20 m C
= 20 x 10⁻³ C
energy delivered in the circuit
E = Q V
E = 20 x 10⁻³ x 9
E = 180 x 10⁻³
E = 0.18 J
Energy delivered in the circuit is equal to E = 0.18 J
Answer:
Explanation:
We are asked to find the final velocity of the boat.
We are given the initial velocity, acceleration, and time. Therefore, we will use the following kinematic equation.
The initial velocity is 2.7 meters per second. The acceleration is 0.15 meters per second squared. The time is 12 seconds.
= 2.7 m/s - a= 0.15 m/s²
- t= 12 s
Substitute the values into the formula.
Multiply the numbers in parentheses.
Add.
The final velocity of the boat is <u>4.5 meters per second in the positive direction.</u>
