Answer:
R = 4.24 x 10⁻⁴ m
Explanation:
given,
mass of the person = 60.3-kg
mass of the bullet = 10 gram = 0.01 Kg
velocity of bullet = 389 m/s
angle made with the horizontal = 45°
using conservation of momentum.
M v + m u = ( M + m ) V
60.3 x 0 + 0.01 x 389 = (60.3 + 0.01) V


V = 0.0645 m/s
for calculation of range


R = 4.24 x 10⁻⁴ m
the distance actor fall is R = 4.24 x 10⁻⁴ m
Answer:

Explanation:
We are asked to find the acceleration of an object.
According to Newton's Second Law of Motion, force is the product of mass and acceleration.

We know the object has a mass of 113 kilograms and it is exposed to a 110 Newton force.
Let's convert the units of force to make the problem easier later. 1 Newton is equal to 1 kilogram meter sper second squared. So, the force of 110 Newtons is equal to 110 kilogram meters per second squared.
Substitute the values into the formula.

We are solving for a, the acceleration, so we isolate this variable. It is being multiplied by 113 kilograms. The inverse operation of multiplication is division, so we divide both sides by 113 kg.


The units of kilograms (kg) cancel.


We are asked to round to one decimal place or the tenths place. The 7 in the hundredths place tells us to round the 9 up to a 0, but then we must also round the 0 in the ones place to a 1.

The acceleration of the object is approximately <u>1.0 meter per second squared.</u>
The change in frequency noticed by observers on sub b is : 2.66 Hz before the submarines pass each other.
<u>Given data:</u>
New speed of Sub b ( Vb ) = 12.4 m/s
F₁ = 1416.44 Hz
Frequency of sonar wave ( Fₐ ) = 1400 Hz
Speed of sound in water ( V ) = 1533 m/s
Speed of submarine A ( Va ) = 8.4 m/s
<h3>Determine the change in frequency observed on sub B </h3>
The change in frequency can be calculated using the formula below
F₁' = Fₐ ( V + Vb / V - Vₐ )
= 1400 ( 1533 + 12.4 / 1533 - 8.4 )
= 1400 ( 1545.4 / 1524.6 )
= 1419.1 Hz
Final step : <u>determine the change in frequency </u>
ΔF = F₁' - F₁
= 1419.1 - 1416.44
= 2.66 Hz.
Hence we can conclude that The change in frequency noticed by observers on sub b is : 2.66 Hz.
learn more about change in frequency : brainly.com/question/254161
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