PHYSICS COLLEGE

A weather balloon is designed to expand to a maximum radius of 24 m at its working altitude, where the air pressure is 0.030 atm and the temperature is 200 K. If the balloon is filled at atmospheric pressure and 349 K, what is its radius at liftoff

Answers

Answer 1

Answer:

Answer:

Radius at liftoff 8.98 m

Explanation:

At the working altitude;

maximum radius = 24 m

air pressure = 0.030 atm

air temperature = 200 K

At liftoff;

temperature = 349 K

pressure = 1 atm

radius = ?

First, we assume balloon is spherical in nature,

and that the working gas obeys the gas laws.

from the radius, we can find the volume of the balloon at working atmosphere.

Volume of a sphere = $(4)/(3) \pi r^(3)$

volume of balloon = $(4)/(3)$ x 3.142 x $24^(3)$ = 57913.34 m^3

using the gas equation,

$(P1V1)/(T1)$ = $(P2V2)/(T2)$

The subscript 1 indicates the properties of the gas at working altitude, and the subscript 2 indicates properties of the gas at liftoff.

imputing values, we have

$(0.03*57913.34)/(200)$ = $(1*V2)/(349)$

0.03 x 57913.34 x 349 = 200V2

V2 = 606352.67/200 = 3031.76 m^3 this is the volume occupied by the gas in the balloon at liftoff.

from the formula volume of a sphere,

V = $(4)/(3) \pi r^(3)$ = $(4)/(3)$ x 3.142 x $r^(3)$ = 3031.76

4.19 $r^(3)$ = 3031.76

$r^(3)$ = 3031.76/4.19

radius r of the balloon on liftoff = $\sqrt[3]{723.57}$ = 8.98 m

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A baseball player standing on a platform throws a baseball out over a level playing field. The ball is released from a point 3.50 m above the field with an initial speed of 14.3 m/s at an upward angle of 27.0 degrees above the horizontal. What horizontal distance will the ball travel before hitting the ground?

Answers

Answer:

22.1 m

Explanation:

$v_(o)$ = initial speed of ball = 14.3 m/s

$\theta$ = Angle of launch = 27°

Consider the motion of the ball along the vertical direction.

$v_(oy)$ = initial speed of ball = $v_(o) Sin\theta = 14.3 Sin27 = 6.5 ms^(-1)$

$a_(y)$ = acceleration due to gravity = - 9.8 ms⁻²

$t$ = time of travel

$y$ = vertical displacement = - 3.50 m

Using the kinematics equation that suits the above list of data, we have

$y = v_(oy) t + (0.5) a_(y) t^(2) \n- 3.50 = (6.5) t + (0.5) (- 9.8) t^(2)\n- 3.50 = (6.5) t - 4.9 t^(2)\nt = 1.74 s$

Consider the motion of the ball along the horizontal direction.

$v_(ox)$ = initial speed of ball = $v_(o) Cos\theta = 14.3 Cos27 = 12.7 ms^(-1)$

$X$ = Horizontal distance traveled

$t$ = time taken = 1.74 s

Since there is no acceleration along the horizontal direction, we have

$X = v_(ox) t\nX = (12.7)(1.74)\nX = 22.1 m$

The concept of photons applies to which regions of the electromagnetic spectrum?A. visible light only
B. infrared light, visible light, and UV light only
C. X-rays and gamma rays only
D. all regions of the spectrum

Answers

Answer:

D. all regions of the spectrum

Explanation:

I did some research ; )

6. A barber raises his customer's chair by applying a force of 150 N to thehydraulic piston of area 0.01 m2. If the chair is attached to a piston with an
area of 0.1 m², how much force is needed to raise the customer?
STEP 1: List the known
and unknown values F =
A=
A,
STEP 2: Write the
correct equation
STEP 3: Insert the
known values into the
equation to solve for
the unknown value

Answers

Answer:

15N

Explanation:

F¹=150N

A=0.01m2²

F2=?

A2=0.1m²

P=F/A

F1/A2=F2/A1

150/0.1=F2/0.01

22. What force is necessary to accelerate a 2500kg car from rest to 20m/s over 10s?(6 Points)
2N
250N
5000N
50000N

Answers

Answer:

50000N

Explanation:

Force = mass × acceleration

= 2500 × 20

= 50000N

A car with tires of radius 0.25 m come to a stop from 28.78 m/s (100 km/hr) in 50.0 m without any slipping of tires. Find: (a) the angular acceleration of the wheels; (b) number of revolutions made while coming to rest.

Answers

Answer:

The answer is below

Explanation:

a) Using the formula:

$\omega^2=\omega_o^2+2\alpha \theta\n\n\omega=final\ angular\ velocity,\omega_o=initial\ anglular\ velocity,\alpha= angular\ acceleration,\n\theta=angular\ distance\n\nGiven\ that:\n\ninitial\ velocity(u)=28.78m/s,distance(s)=50\ m,radius(r)=0.25\ m,\nfinal/ velocity(v)=0(stop)\n\n\omega=v/r=(28.78m/s)/(0.25m) =115.12\ rad/s,\omega_o=0,\theta=s/r=(50\ m)/(0.25\ m)=200\ rad\n \n\omega^2=\omega_o^2+2\alpha \theta\n\n115.12^2=0^2+2\alpha(200)\n\n2\alpha(200)=13252.6144\n\n\alpha=33.13\ rad/s^2$

$\theta=200\ rad=200\ rad*(1\ rev)/(2\pi\ rad)=31.83\ rev$

A 75 kg man starts climbing a ladder that leans against a wall. If the weight of the ladder is negligible, determine how far up the ladder the man can climb before the ladders starts to slip. The coefficient of friction on both surfaces is μS=0.25

Answers

The man can climb $\bold { X (max) = 0.25* L* tan \alpha }$ , before the ladders starts to slip.

A - point at the top of the ladder

B - point at the bottom of the ladder

C - point where the man is positioned in the ladder

L- the length of the ladder

α - angle between ladder and ground

x - distance between C and B

The forces act on the ladder,

Horizontal reaction force (T) of the wall against the ladder

Vertical (upward) reaction force (R) of ground against the ladder.

Frictionalhorizontal ( to the left ) force (F)

Vertical( downwards) of the man,

mg = 75 Kg x 9.8 m/s² = 735 N

in static conditions,

∑Fx = T - F = 0 Since, T = F

∑Fy = mg - R = 0 Since, 735 - R = 0, R = 735

∑ Torques(b) = 0

In point B the torque produced by forces R and F is Zero

Then:

∑Torques(b) = 0

And the arm lever for each force,

mg = 735

Since, ∑Torques(b) = 0

$\bold {735* x* cos\alpha = F* L* sin\alpha }$ Since,T = F

$\bold {F = \frac {735* x* cos\apha }{L* sin\alpha }}$ $\bold { \frac {cos \alpha } { sin\alpha }= cot\alpha =\frac 1{tan\alpha}}$

$\bold {F = \frac {735* x* cos\apha }{L }}$

$\bold {F = 735* x* tan\alpha }}$

F < μR the ladder will starts slipping over the ground

μ(s) = 0.25

$\bold { X (max) = 0.25* L* tan \alpha }$

Therefore, the man can climb $\bold { X (max) = 0.25* L* tan \alpha }$ , before the ladders starts to slip. \

To know more about Torque,

brainly.com/question/6855614

Answer:

x (max) = 0,25*L*tanα

Explanation:

Letá call

A: point at the top of the ladder

B: the point at the bottom of the ladder

C: point where the man is up the ladder

L the length of the ladder

α angle between ladder and ground

"x" distance between C and B

Description

The following forces act on the ladder

Point A: horizontal (to the right) reaction (T) of the wall against the ladder

Point B : Vertical (upwards) reaction (R) of ground against the ladder

frictional horizontal ( to the left ) force (F)

Point C : Weight (vertical downwards)) of the man mg

mg = 75 Kg * 9,8 m/s² mg = 735 [N]

Then in static conditions:

∑Fx = T - F = 0 ⇒ T = F

∑Fy = mg - R = 0 ⇒ 735 - R = 0 ⇒ R = 735

∑Torques(b) = 0

Note: In point B the torque produced by forces R and F are equal to 0

Then:

∑Torques(b) = 0

And the arm lever for each force is:

mg = 735

d₁ for mg and d₂ for T

cos α = d₁/x then d₁ = x*cosα

sin α = d₂ / L then d₂ = L*sinα

Then:

∑Torques(b) = 0 ⇒ 735*x*cosα - T*L*sinα = 0

735*x*cosα = T*L*sinα

T = F then 735*x*cosα = F*L*sinα

F = (735)*x*cosα/L*sinα cos α / sinα = cotgα = 1/tanα

F = (735)*x*cotanα/L or F = (735)*x/L*tanα

When F < μ* R the ladder will stars slippering over the ground

μ(s) = 0,25 0,25*R = 735*x/L*tanα

x = 0,25*R*tanα*L/735

But R = mg = 735 then

0,25*L*tanα = x

Then x (max) = 0,25*L*tanα

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