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2 years ago

Can someone help me figure out how to do a function formula for fx and fy

Physics

1 answer:

masha68 [24]2 years ago

4 0

Answer: How to solve for FX and FY?

to find fx(x, y): keeping y constant, take x derivative; • to find fy(x, y): keeping x constant, take y derivative. f(x1,...,xi−1,xi + h, xi+1,...,xn) − f(x) h . ∂y2 (x, y) ≡ ∂ ∂y ( ∂f ∂y ) ≡ (fy)y ≡ f22. similar notation for functions with > 2 variables.

Explanation:

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How to get stud multipliers in lego star wars the skywalker saga?.

Anna11 [10]

You can see the Stud Multipliers right away in your Holoprojector menu under the Extras tab.

7 0

1 year ago

A 126- kg astronaut (including space suit) acquires a speed of 2.70 m/s by pushing off with her legs from a 1800-kg space capsul

jeka94

The change in the speed of the space capsule will be -0.189 m/s.

The average force exerted by each on the other will be 567 N.

The kinetic energy of each after the push for the astronaut and the capsule are 459.27 J and 32.14 J.

<h3>Given:</h3>

Mass of the astronaut, $m_a$ = 126 kg

Speed he acquires, $v_{a}$ = 2.70 m/s

Mass of the space capsule, $m_{c}$ = 1800kg

The initial momentum of the astronaut-capsule system is zero due to rest.

$P_f = m_av_a + m_cv_c$

$P_I$ = 0

$m_av_a + m_cv_c = 0$

$v_c =\frac{- m_a v_a}{m_c}}\\\\$

$= \frac{126* 2.70}{1800}$

$= - 0.189$ m/s

Therefore,

According, to the impulse-momentum theorem;

FΔt = ΔP

ΔP = m Δv

ΔP = 126×2.70

= 340.2 kgm/sec

t is time interval = 0.600s

F = ΔP/Δt

F = 340.2/0.600

= 567 N

Therefore, the average force exerted by each on the other will be 567 N.

The Kinetic Energy of the astronaut;

K.E = $\frac{1}{2} m v^2$

$= \frac{1}{2}$ × 126 × $(2.70) ^2$

= 459.27 J

The Kinetic Energy of the capsule;

K.E = $\frac{1}{2} m v^2$

= $\frac{1}{2}$ ×1800× $(0.189) ^2$

= 32.14 J

Therefore, the kinetic energy of each after the push for the astronaut and the capsule are 459.27 J and 32.14 J.

Learn more about kinetic energy here:

brainly.com/question/26520543

#SPJ1

3 0

1 year ago

While standing at the edge of the roof of a building, a man throws a stone upward with an initial speed of 6.95 6.95 m/s. The st

qwelly [4]

Answer:

18.1347 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² = a

$v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-6.95^2}{2\times -9.81}\\\Rightarrow s=2.4619\ m$

Total height the ball falls is 2.4619+14.3 = 16.7619 m

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 16.7619+0^2}\\\Rightarrow v=18.1347\ m/s$

The speed at which the stone reaches the ground is 18.1347 m/s

8 0

2 years ago

The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hou

Inessa05 [86]

Answer:

a) 4.40 s

b) 2.20 s

Explanation:

Given parameters are:

At constant power ,

initial speed of the car, $v_0=0$

final speed of the car, $v=32$ mph

At full power,

initial speed of the car, $v_0=0$

final speed of the car, $v=64$ mph

At constant power, $KE = \frac{1}{2} mv^2$

At full power, $KE = \frac{1}{2} m(2v)^2$

So $KE_f = 4KE_i$

So, time to reach 64 mph speed is 4 times more than the initial time

$t = 4*1.10 =4.40$ s

$v=v_0+at\\a=\frac{v-v_0}{t}=\frac{32-0}{1.1/3600}=104727.27$ $miles/hours^2$

For final 64 mph speed,

$v=v_0+at\\t=\frac{v-v_0}{a}=\frac{64-0}{104727.27} = 6.111*10^{-4}$ $hours$ = $6.111*10^{-4}*3600=2.20$ s

7 0

2 years ago

A proton moving with a velocity of 4.0 × 104 m/s enters a magnetic field of 0.20 t. if the angle between the velocity of the pro

svlad2 [7]

Answer:

Magnitude of the force on proton = F = 1.1085 × 10^-15 N

Explanation:

Charge on proton = q = 1.60 × 10^-19 C

Velocity of proton = V = 4.0 × 10^4 m/s

Magnetic field = B = 0.20 T

Angle between V and B = θ = 60

We know that,

F = qVBsin θ = (1.60 × 10^-19)( 4.0 × 10^4)( 0.20)sin(60)

F = 1.1085 × 10^-15 N

8 0

2 years ago

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