Answer: 1,224 km/h
Explanation:
To do this, we pick the first unit and convert
Picking m first and converting to km:
Since we're converting from a non-prefix to a prefix, we divide the value by the prefix were taking it to. In this case, kilo = 10³ which means we're going to divide our value by 1000 to convert it from m to km
340 m/s ÷ 1000 = 0.34 km/s
Now, let's convert our seconds to hour:
We'll need to calculate how many hours is equivalent to one second first;
1 hr = 60×60 seconds
X hr = 1 second
*Cross multiply*
1 × 1 = X × 60 × 60
1 = 3,600 X
X = 1 / 3,600
X = 2.778×10⁻⁴ hour
So, in the place of "1 Second", we're going to be inserting 2.778×10⁻⁴ hour instead
0.34 km / s = 0.34 km / 2.778×10⁻⁴ hour
(0.34 / 2.778×10⁻⁴) km/hour
1,224 km/h.
340 m/s = 1,224 km/h
Cocking your head would be most useful for detecting the LOCATION of a sound.
Answer:
<h3>1.43m/s²</h3>
Explanation:
According to newtons second law.
F = mass * acceleration
If the doll has a mass of 0.2 kg, and the robot has a mass of 0.5 kg, the resulting mass will be 0.7kg
Force applied = 1N
acceleration = Force/mass
Substitute the values and get acceleration
acceleration = 1/0.7
acceleration = 1.43m/s²
Hence the magnitude of the acceleration of the robot is 1.43m/s²
Answer:
Questions that cannot be answered through scientific investigation are those that relate to personal preference, moral values, the supernatural, or unmeasurable phenomena.
Answer:
a) p₀ = 1.2 kg m / s, b) p_f = 1.2 kg m / s, c) θ = 12.36, d) v_{2f} = 1.278 m/s
Explanation:
a system formed by the two balls, which are isolated and the forces during the collision are internal, therefore the moment is conserved
a) the initial impulse is
p₀ = m v₁₀ + 0
p₀ = 0.6 2
p₀ = 1.2 kg m / s
b) as the system is isolated, the moment is conserved so
p_f = 1.2 kg m / s
we define a reference system where the x-axis coincides with the initial movement of the cue ball
we write the final moment for each axis
X axis
p₀ₓ = 1.2 kg m / s
p_{fx} = m v1f cos 20 + m v2f cos θ
p₀ = p_f
1.2 = 0.6 (-0.8) cos 20+ 0.6 v_{2f} cos θ
1.2482 = v_{2f} cos θ
Y axis
p_{oy} = 0
p_{fy} = m v_{1f} sin 20 + m v_{2f} cos θ
0 = 0.6 (-0.8) sin 20 + 0.6 v_{2f} sin θ
0.2736 = v_{2f} sin θ
we write our system of equations
0.2736 = v_{2f} sin θ
1.2482 = v_{2f} cos θ
divide to solve
0.219 = tan θ
θ = tan⁻¹ 0.21919
θ = 12.36
let's look for speed
0.2736 = v_{2f} sin θ
v_{2f} = 0.2736 / sin 12.36
v_{2f} = 1.278 m / s
