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1 year ago

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A 40-cm-diameter, 300 g beach ball is dropped with a 4.0 mg ant riding on the top. The ball experiences air resistance, but the

ant does not. What is the magnitude of the normal force exerted on the ant when the ball's speed is 4.0 m/s?

1 answer:

mars1129 [50]1 year ago

5 0

Answer:

The normal force exerted on the ant is 0.75 N.

Explanation:

Given;

diameter of the ball, D = 40 cm = 0.4m

radius of the ball, r = 0.2m

mass of the beach ball, m₁ = 300 g = 0.3 kg

mass of the ant, m₂ = 4 x 10⁻⁶ kg

speed of the ball, v = 4 m/s

The area of the ball, assuming spherical ball is given by;

A = 4πr²

A = 4π(0.2)² = 0.5027 m²

The drag force (resistance) experienced by the spherical ball is given as;

$F_D = \frac{1}{2}C\rho Av^2$

where;

C is the drag coefficient of the spherical ball = 0.45

ρ is density of air = 1.21 kg/m³

$F_D = \frac{1}{2}C\rho Av^2\\\\F_D = \frac{1}{2}(0.45)(1.21) (0.5027)(4)^2\\\\F_D = 2.19 \ N$

The downward force of the ball due to its weight and that of the ant is given by;

$F_g = mg\\\\F_g =g(m_{ant} + m_{ball})\\\\F_g = g(4*10^{-6} \ kg\ + \ 0.3\ kg)\\\\F_g = g(0.300004 \ kg) \ \ \ (mass \ of \ the \ ant \ is \ insignificant)\\\\F_g = 9.8(0.3)\\\\F_g = 2.94 \ N$

The net downward force experienced by the ball is given by;

$F_{net} = F_g - F_D\\\\F_{net} = 2.94 \ N - 2.19 \ N\\\\F_{net} = 0.75 \ N$

This downward force experienced by the ball is equal to the normal reaction it exerts on the ant.

Thus, the normal force exerted on the ant is 0.75 N.

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a) Average power: 1425 W

b) Instantaneous power at 3.0 sec: 2850 W

Explanation:

a)

The motion of the object along the ramp is a uniformly accelerated motion (because the force applied is constant), so we can use the suvat equation

$s=ut+\frac{1}{2}at^2$

where

s = 18 m is the displacement along the ramp

u = 0 is the initial velocity

t = 3.0 s is the time taken

a is the acceleration of the object along the ramp

Solving for a,

$a=\frac{2s}{t^2}=\frac{2(18)}{(3.0)^2}=4 m/s^2$

Now we can apply Newton's second law to find the net force on the object:

$F=ma=(24 kg)(4 m/s^2)=96 N$

This net force is the resultant of the applied force forward ( $F_a$ ) and the component of the weight acting backward ( $mg sin \theta$ ), so we can find what is the applied force:

$F=F_a - mg sin \theta\\F_a = F+mg sin \theta = 96+(24)(9.8)(sin 37^{\circ})=237.5 N$

where

m = 24 kg is the mass of the object

$g=9.8 m/s^2$ is the acceleration of gravity

Now we can finally find what is the work done by the applied force, which is parallel to the ramp, therefore:

$W=F_a s = (237.6)(18)=4276 J$

where s = 18 m is the displacement.

Therefore the average power needed is:

$P=\frac{W}{t}=\frac{4276}{3}=1425 W$

b)

The instantaneous power at any point of the motion is given by

$P=F_av$

where

$F_a$ is the force applied

v is the velocity of the object

We already calculated the applied force:

$F_a=237.5 N$

While since this is a uniformly accelerated motion, we can find the velocity at the end of the 3.0 seconds using the suvat equation:

$v=u+at=0+(4)(3.0)=12.0 m/s$

And therefore, the instantaeous power at 3.0 sec is:

$P=Fv=(237.5)(12)=2850 W$

Learn more about power:

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#LearnwithBrainly

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2 years ago

When a 12.0-V battery causes 2.00 μC of charge to flow onto the plates of an air-filled capacitor, how much work did the battery

Mrac [35]

Answer:

Explanation:

Voltage, V = 12 V

Charge, q = 2 micro coulomb = 2 x 10^-6 C

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W = 0.5 x q x V

W = 0.5 x 2 x 10^-6 x 12

W = 12 x 10^-6 J

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A screwdriver is used to pry the lid off a paint can. The resistance arm is 0.5 cm long and the effort arm is 20 cm long. What i

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Answer:

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2 years ago

In a demonstration, a 4.00 cm2 square coil with 10 000 turns enters a larger square region with a uniform 1.50 T magnetic field

Kay [80]

Explanation:

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Speed, v = 100 m/s

An emf is induced in the coil which is given by :

$E=NBLv$

$E=10000\times 1.5\times 0.02\times 100$

E = 30000 V

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Let d is the gap between the two wires connected to the ends of the coil and still get a spark. So,

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2 years ago

If the 20-mm-diameter rod is made of A-36 steel and the stiffness of the spring is k = 61 MN/m , determine the displacement of e

ryzh [129]

Answer:

σ = 1.09 mm

Explanation:

Step 1: Identify the given parameters

rod diameter = 20 mm

stiffness constant (k) = 55 MN/m = 55X10⁶N/m

applied force (f) = 60 KN = 60 X 10³N

young modulus (E) = 200 Gpa = 200 X 10⁹pa

Step 2: calculate length of the rod, L

K = \frac{A*E}{L}K=

L

A∗E

L = \frac{A*E}{K}L=

K

A∗E

A=\frac{\pi d^{2}}{4}A=

4

πd

2

d = 20-mm = 0.02 m

A=\frac{\pi (0.02)^{2}}{4}A=

4

π(0.02)

2

A = 0.0003 m²

L = \frac{A*E}{K}L=

K

A∗E

L = \frac{(0.0003142)*(200X10^9)}{55X10^6}L=

55X10

6

(0.0003142)∗(200X10

9

)

L = 1.14 m

Step 3: calculate the displacement of the rod, σ

\sigma = \frac{F*L}{A*E}σ=

A∗E

F∗L

\sigma = \frac{(60X10^3)*(1.14)}{(0.0003142)*(200X10^9)}σ=

(0.0003142)∗(200X10

9

)

(60X10

3

)∗(1.14)

σ = 0.00109 m

σ = 1.09 mm

Therefore, the displacement at the end of A is 1.09 mm

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2 years ago