Answer:
uh finish the question please lol.
<span>let the fsh jump with initial velocity (u) in direction (angle p) with horizontal
it can cross and reach top of trajectory if its top height h = 1.5m
and horizontal distance d = (1/2) Range
--------------------------------------...
let t be top height time
at top height, vertical component of its velocity =0
vy = 0 = u sin p - gt
t = u sin p/g
h = [u sin p]*t - 0.5 g[t[^2
1.5 = u^2 sin^2 p/g - u^2 sin^2 p/2g
u^2 sin^2 p/2g = 1.5
u^2 sin^2 p = 1.5*2*9.8 = 29.4
u sin p = 5.42 m/s >>>>>>>>>>>>>>> V-component
=====================
t = HALF the time of flight
d = (1/2) Range (R) = (1/2) [2 u^2 sin p cos p/g]
1 = u^2 sin p cos p/g
u sin p * u cos p = 9.8
5.42 * u cos p = 9.8
u cos p = 1.81 m/s >>>>>>>>>>>>> H-component
check>>
u = sqrt[u^2 cos^2 p + u^2 sin^2 p] = 5.71 m/s
u < less than fish's potential jump speed 6.26 m/s
so it will able to cross</span>
Answer:
Time = 80.91 seconds
Explanation:
Given the following data;
Velocity = 5.50 m/s.
Distance = 445 meters
To find the time;
Velocity can be defined as the rate of change in displacement (distance) with time. Velocity is a vector quantity and as such it has both magnitude and direction.
Mathematically, velocity is given by the equation;

Substituting into the formula, we have;
5.5 = 445/time
Time = 445/5.5
Time = 80.91 seconds
Answer:
The frequency of sound heard by the boy is 1181 Hz.
Explanation:
Given that,
Frequency of sound from alarm 
Speed = -8.25 m/s
Negative sign show the boy riding away from the car
Speed of sound = 343
We need to calculate the heard frequency
Using formula of frequency

Where,
= frequency of source
= speed of observer
= speed of source
= speed of sound
Put the value into the formula

here, source is at rest


Hence, The frequency of sound heard by the boy is 1181 Hz.
Answer:
7.16 m /s
Explanation:
The depth of the small pipe attached with the side wall of tank from the surface of water
h =( 3.1 - .48 )m
= 2.62 m
velocity of flow of water= √ 2 g h
= √ 2 x 9.8 x 2.62
= 7.16 m /s
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