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1 year ago

A FBD of a rocket launching into space should include:

Physics

1 answer:

Vladimir [108]1 year ago

3 0

Answer:

Explanation:

the force of the rocket engine pushing it up, the force of gravity pulling it down, maybe some force of air resistance as the rocket goes fast, hmmm Free Body Diagrams (FBD) should have any and all forces on the model, unless they are negligible . or so slight they really make little difference in the total outcome.

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What sentence best supports the statement that hormones are involved in the regulation of homeostasis? A. The hormone cortisol s

juin [17]

B. The hormone erythropoeitin increases the production of red blood cells when oxygen levels are low.

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1 year ago

The initial velocity of a 4.0-kg box is 11 m/s, due west. After the box slides 4.0 m horizontally, its speed is 1.5 m/s. Determi

ankoles [38]

Answer:

F = - 59.375 N

Explanation:

GIVEN DATA:

Initial velocity = 11 m/s

final velocity = 1.5 m/s

let force be F

work done = mass* F = 4*F

we know that

Change in kinetic energy = work done

kinetic energy = $= \frac{1}{2}*m*(v_{2}^{2}-v_{1}^{2})$

kinetic energy = $= \frac{1}{2}*4*(1.5^{2}-11^{2})$ = -237.5 kg m/s2

-237.5 = 4*F

F = - 59.375 N

7 0

2 years ago

You throw a baseball at an angle of 30.0∘∘ above the horizontal. It reaches the highest point of its trajectory 1.05 ss later. A

garri49 [273]

Answer:

The speed with which the baseball leaves the hand = 20.58 m/s

Explanation:

The time take to reach highest height during a projectile's flight is given by

t = (u sin θ)/g

u = initial velocity of the baseball = ?

θ = angle of throw above the horizontal

g = acceleration due to gravity = 9.8 m/s²

1.05 = (u sin 30)/9.8

u = (1.05 × 9.8)/0.5

u = 20.58 m/s

7 0

1 year ago

A proton is released from rest at the positive plate of a parallel-plate capacitor. It crosses the capacitor and reaches the neg

WINSTONCH [101]

Answer:

$=2,012,319.36 \ m/s$

Explanation:

-The only relevant force is the electrostatic force

-The formula for the electrostatic force is:

$F = Eq$

E is the electric field and q is the magnitude of the charge.

#Since the electric field is the same in both cases, and the charge of the protons and electrons have the same magnitude, you can state that the magnitude of the electric forces acting in both proton and electron are the same.

$F_e = F_p\\\\F_e= Force \ on \ electron\\F_p = Force \ on \ proton$

-Applying Newton's 2nd Law:

$F=ma$

$F_e=M_ea_e$

$F_p=M_pa_p$

#equate the two forces:

$F_e = F_p\\\\M_ea_e=M_pa_p\\\\a_e=\frac{M_pa_p}{M_e}$

#The equations for velocity in uniform acceleration:

$V_f^2=V_o^2+2ad\\\\V_o^2=0\\\\\therefore V_f^2=2ad$

#For the proton:

$V_f^2=2a_pd\\\\a_p=\frac{V_f^2}{2d}\\\\a_p=\frac{47000m/s)^2}{2d}$

#For the electron:

$V_f^2=2{a_e}^2\times 2d\\\\A_e=M_p\times A_p/M_e\\\\V_f^2=M_p\times (47000m/s)^2/2d\times2d/M_e\\\\V_f^2=M_p\times (47000m/s)^2/M_e\\\\V_f=47000m/s\times\sqrt{\frac{M_p}{M_e}}$

The mass values of the proton and electron are:

$M_p=1.67\times 10^{-27} kg\\\\M_e=9.11\times10^{-31}kg$

The speed of the ion is therefore calculated as:

$V_f=47000m/s\times\sqrt{\frac{M_p}{M_e}}\\\\=47000m/s\times\sqrt{\frac{1.67\times10^{-27}}{9.11\times10^{-31}}\\\\=2,012,319.36 \ m/s$

Hence, the ion's speed at the negative plate is $=2,012,319.36 \ m/s$

7 0

2 years ago

A copper wire has radius 0.800 mm and carries current I at 20.0°C. A silver wire with radius 0.500 mm carries the same current a

IgorC [24]

Answer:

Ecu/Eag = 0.46

Explanation:

E = PI/A

Ecu = Pcu × I/A

Pcu = 1.72×10^-8 ohm-meter

r = 0.8 mm = 0.8/1000 = 8×10^-4 m

A = πr^2 = π×(8×10^-4)^2 = 6.4×10^-7π

Ecu = 1.72×10^-8I/6.4×10^-7π = 0.026875I/1

Eag = Pag × I/A

Pag = 1.47×10^-8 ohm-meter

r = 0.5 mm = 0.5/1000 = 5×10^-4 m

A = πr^2 = π × (5×10^-4)^2 = 2.5×10^-7π

Eag = 1.47×10^-8I/2.5×10^-7π = 0.0588I/π

Ecu/Eag = 0.026875I/π × π/0.0588I = 0.46

7 0

2 years ago