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aleksandrvk [35]
3 years ago
14

If the jumping frog can give itself the same initial speed, regardless its direction. how is the maximum vertical height related

to its maximum horizontal range?
Physics
1 answer:
DaniilM [7]3 years ago
4 0
A projectile motion is characterized by a motion whose path is alike to that of an arc. It already has its derived equations. For the maximum height H and the maximum range R, the equations are:

H = v₀²sin²θ/2g       ;       R = v₀²sin(2θ)/g

As you can observe, they are almost alike. Since they both contain v₀²/g, let's isolate this term for both equations:

v₀²/g = 2H/sin²θ
v₀²/g = R/sin (2θ)

Equating both equations:
2H/sin²θ = R/sin (2θ)
H = Rsin²θ/2sin(2θ)

This is the equation that relates H to R.
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A railroad car of mass 2.50*10^4 kg is moving with a speed of 4.00 m/s. It collides and couples with three other coupled railroa
NemiM [27]

Answer:

(a) 2.5 m/s

(b) 37.5 KJ

Explanation:

(a)

From the law of conservation of momentum, Initial momentum=Final momentum

mV_1+3mV_2=(m+3m)V_f=4mV_f

V_1+3V_2=4V_f and making V_f the subject then

V_f=\frac {V_1+3V_2}{4} and since V_1 is initial velocity of car, value given as 4 m/s, V_2 is the initial velocity of the three cars stuck together, value given as 2 m/s and v_f is the final velocity which is unknown. By substitution

V_f=\frac {4+(3\times2)}{4}=2.5 m/s

(b)

Initial kinetic energy is given by

\frac {mV_1^{2}}{2}+\frac {3mV_2^{2}}{2}=\frac {m(V_1^{2}+3V_2^{2}}{2}=\frac {2.5\times 10^{4}(4^{2}+3(2^{2}))}{2}=350\times10^{3} J= 350 KJ

Final kinetic energy is given by

\frac {4mV_f^{2}}{2}=\frac {4\times 2.5\times 10^{4}\times 2.5^{2}}{2}=312.5\times 10^{3} J=312.5 KJ

The energy lost is given by subtracting the final kinetic energy from the initial kinetic energy hence

Energy lost=350-312.5=37.5 KJ

6 0
3 years ago
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andrew-mc [135]

Answer:

59.18 kg

Explanation:

use f=ma

f= 580 N

a = 9.8 m/s 2

weigh(m) doesn't change only force(F) changes

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