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tino4ka555 [31]

3 years ago

12

At the surface of Jupiter's moon Io, the acceleration due to gravity is 1.81 m/s2 . A watermelon has a weight of 44.0 N at the s

urface of the earth. In this problem, use 9.81 m/s2 for the acceleration due to gravity on earth.

1 answer:

BabaBlast [244]3 years ago

6 0

Answer:

Weight at the surface of Jupiter's moon Io is 8.13 N .

Explanation:

Given :

Acceleration due to gravity at the surface of Jupiter's moon is $g_m=1.81\ m/s^2$ .

Weight of watermelon in earth , $W=44\ N$ .

Acceleration due to gravity at the surface of earth is $g=9.81\ m/s^2$ .

We know , weight is given by :

$W=mg\\m=\dfrac{W}{g}\\\\m=4.49\ kg$

Therefore , mass at the surface of Jupiter's moon Io is :

$W_m=mg_m\\\\W_m=4.49\times 1.81\\\\W_m=8.13 \ N$

Hence , this is the required solution .

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Whenever two Apollo astronauts were on the surface of the Moon, a third astronaut orbited the Moon. Assume the orbit to be circu

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Answer:

$v = 1,582 \ \frac{m}{s}$

Explanation:

We know that for circular motion the centripetal acceleration $a_c$ is:

$a_c = \frac{v^2}{r}$

where v is the speed and r is the radius.

The centripetal acceleration for the astronaut must be the gravitational acceleration due to the gravity, as there are no other force. So

$a_c = 1.27 \frac{m}{s^2}$ .

The radius of the orbit must be the radius of the Moon, plus the 270 km above the surface

$r = 1.7 * 10^6 \ m + 270 \ km$

$r = 1.7 * 10^6 \ m + 270 * 10^3 \ m$

$r = 1.7 * 10^6 \ m + 0.270 * 10^6 \ m$

$r = 1.97 * 10^6 \ m$

We can obtain the speed as:

$v^2 = a_c r$

$v = \sqrt{a_c r}$

$v = \sqrt{1.27 \frac{m}{s^2} * 1.97 * 10^6 \ m}$

$v = \sqrt{ 2.509 \ 10^6 \ \frac{m^2}{s^2}}$

$v = 1.582 \ 10^3 \ \frac{m}{s}$

$v = 1,582 \ \frac{m}{s}$

And this is the orbital speed.

7 0

3 years ago

A bicyclist, initially at rest, begins pedaling and gaining speed steadily for 4.90s during which she covers 25.0m.

emmasim [6.3K]

The bicyclist accelerates with magnitude <em>a</em> such that

25.0 m = 1/2 <em>a</em> (4.90 s)²

Solve for <em>a</em> :

<em>a</em> = (25.0 m) / (1/2 (4.90 s)²) ≈ 2.08 m/s²

Then her final speed is <em>v</em> such that

<em>v</em> ² - 0² = 2<em>a</em> (25.0 m)

Solve for <em>v</em> :

<em>v</em> = √(2 (2.08 m/s²) / (25.0 m)) ≈ 10.2 m/s

Convert to mph. If you know that 1 m ≈ 3.28 ft, then

(10.2 m/s) • (3.28 ft/m) • (1/5280 mi/ft) • (3600 s/h) ≈ 22.8 mi/h

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3 years ago

WILL NAME THE BRAINLIEST! An airplane undergoes the following displacements: It first flies 72 km in a direction of 30° East of

nekit [7.7K]

Answer:

82.1 km

Explanation:

We need to resolve each displacement along two perpendicular directions: the east-west direction (let's label it with x) and the north-south direction (y). Resolving each vector:

$A_x = (72) sin 30^{\circ} =36.0 km\\A_y = (72) cos 30^{\circ} = 62.4 km$

Vector B is 48 km south, so:

$B_x = 0\\B_y = -48$

Finally, vector C:

$C_x = -(100) cos 30^{\circ} =-86.6 km\\C_y = (100) sin 30^{\circ} = 50.0 km$

Now we add the components along each direction:

$R_x = A_x + B_x + C_x = 36.0 + 0 +(-86.6)=-50.6 km\\R_y = A_y+B_y+C_y = 62.4+(-48)+50.0=64.6 km$

So, the resultant (which is the distance in a straight line between the starting point and the final point of the motion) is

$R=\sqrt{R_x^2+R_y^2}=\sqrt{(-50.6)^2+(64.6)^2}=82.1 km$

4 0

3 years ago

A 2.20-kg object is attached to a spring and placed on frictionless, horizontal surface. A horizontal force of 29.0 N is require

g100num [7]

Answer:

a. 145 N/m b. 1.29 Hz c. 1.62 m/s d. 0 m e. 13.2 m/s² f. ± 0.2 m g. 2.9 J h. 0.54 m/s i. 4.39 m/s²

Explanation:

a. The force constant of the spring

The spring force F = kx and k = F/x where k is the spring constant. F = 29.0 N and x = 0.200 m

k = 29.0 N/0.200 m = 145 N/m

b. The frequency of oscillations, f

f = 1/2π√(k/m) m = mass = 2.20 kg

f = 1/2π√(145 N/m/2.20 kg) = 1.29 Hz

c. maximum speed of the object

The maximum elastic potential energy of the spring = maximum kinetic energy of the object

1/2kx² = 1/2mv²

v = (√k/m)x where v is the maximum speed of the object

v = (√145/2.2)0.2 = 1.62 m/s

d Where does the maximum speed occur?

The maximum speed occurs at 0 m

e. The maximum acceleration

a = kx/m = 145 × 0.2/2.2 = 13.2 m/s²

f. The maximum acceleration occurs at x = ± 0.2 m

g. The total energy of the system is the maximum elestic potential energy of the system

E = 1/2kx² = 1/2 × 145 × 0.2² = 2.9 J

h. When x = x₀/3

1/2k(x₀/3)² = 1/2mv²

kx₀²/9 = mv²

v = 1/3(√k/m)x₀ = 1/3(√145/2.2)0.2 = 0.54 m/s

i When x = x₀/3

a = kx₀/3m = 145 × 0.2/(2.2 × 3)= 4.39 m/s²

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3 years ago

If a reaction is exothermic and produces large quantities of heat reaching equilibrium, its equilibrium constant will be

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The equilibrium constant will be lowered and the equilibrium will shift to the left if the heat being produced is not removed.

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3 years ago

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