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madreJ [45]
3 years ago
12

One of the asteroids, Ida, looks like an elongated potato. Surprisingly it has a tiny (compared to Ida) spherical moon! This moo

n called Dactyl has a mass of 4.20 × 10^16 kg, and a radius of 1.57 × 10^4 meters, according to Wikipedia. Ida has a radius of 3.14 x 10^4 meters.
Find the acceleration of gravity on the surface of this little moon.
Physics
1 answer:
Mashcka [7]3 years ago
8 0

Answer:

g = 0.0114 m/s²

Explanation:

The value of acceleration due to gravity on the surface of the moon can be given by the following formula:

g = \frac{Gm}{r^2}

where,

g = acceleration due to gravity on the surface of moon = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

m = mass of moon = 4.2 x 10¹⁶ kg

r = radius of moon = 1.57 x 10⁴ m

Therefore,

g= \frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(4.2\ x\ 10^{16}\ kg)}{(1.57\ x\ 10^4\ m)^2}

<u>g = 0.0114 m/s²</u>

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Two people start biking from the same point. One bikes east at 9 mph, the other south at 28 mph. What is the rate at which the d
Irina-Kira [14]

Answer:

(a) 31.75 mph

(b) 31.77 mph

Explanation:

(a) For t = 10 min = 10 / 60 = 1 / 6 hour

According to the question,

dx /dt = 15 mph

dy / dt = 28 mph

After 10 minutes

x = 1/6 x 15 = 2.5 miles

y = 1/6 x 28 = 4.67 miles

d = \sqrt{x^{2}+y^{2}}=\sqrt{2.5^{2}+4.67^{2}} = 5.3 miles

According to diagram

D^{2} = {x^{2}+y^{2}}

Differentiate both sides with respect to t.

2D dD/dt = 2 x dx/dt + 2y dy/dt

D dD/dt = x dx/dt + y dy/dt

5.3 dD/dt = 2.5 (15) + 4.67 (28)

dD/dt = 31.75 miles/hour

(b) For t = 55 minutes = 55 / 60 hours

After 55 / 60 hours

x = 55 / 60 (15) = 13.75 miles

y = 55 / 60 (28) = 25.67 miles

d = \sqrt{x^{2}+y^{2}}=\sqrt{13.75^{2}+25.67^{2}} = 29.12 miles

According to diagram

D^{2} = {x^{2}+y^{2}}

Differentiate both sides with respect to t.

2D dD/dt = 2 x dx/dt + 2y dy/dt

D dD/dt = x dx/dt + y dy/dt

29.12 dD/dt = 13.75 (15) + 25.67 (28)

dD/dt = 31.77 miles/hour

8 0
4 years ago
The Sun exerts a gravitational force of 29.9 N on a rock that's located in a river bed here on Earth. How much gravitational for
11Alexandr11 [23.1K]

Answer:

F = 29.9 N

Explanation:

It is given that, The Sun exerts a gravitational force of 29.9 N on a rock that's located in a river bed here on Earth. We need to find the gravitational force the rock exert on the Sun. It is a case of Newton's third law of motion which states that the force acting from one object to another is equal to the force acing from second object to the first object and the two forces must be in opposite direction. Hence, the gravitational force the rock exert on the Sun is same i.e. 29.9 N.

4 0
4 years ago
3.) A 20cm long tube has an inner diameter of 0.85cm and an outer
IceJOKER [234]

Answer:

The answer is below

Explanation:

The resistance of a conductor (R) is given by:

R=\frac{\rho L}{A}\\\\where\ L \ is\ the\ conductor\ length\ and \ A\ is\ the\ cross\ sectional\ area and ρ is resistivity

The length = 20cm = 0.2 m

The tube has an outer diameter of 1.1 cm (0.011 m), hence the outer area is:

A₁ = π * diameter²/4 = π * 1.1²/4 = 0.000095 m²

The tube has an inner diameter of 0.85 cm (0.0085 m), hence the inner area is:

A₂ = π * diameter²/4 = π * 0.0085²/4 = 0.0000567 m²

The conductor area = outer area - inner area = 0.000095 m² - 0.0000567 m² = 0.0000383 m²

Hence, the resistance is:

R=\frac{\rho L}{A} =\frac{1.7*10^{-8}*0.2}{0.0000383}=0.00009\ ohm

7 0
3 years ago
A 47.5-turn circular coil of radius 5.25 cm can be oriented in any direction in a uniform magnetic field having a magnitude of 0
Free_Kalibri [48]

Answer:

Torque, 5.91\times 10^{-3}\ N-m

Explanation:

Given that,

The number of turns in the coil, N = 47.5

Radius of the coil, r = 5.25 cm

Uniform magnetic field, B = 0.535 T

Current in the coil, I = 26.9 mA

The magnitude of the maximum possible torque exerted on the coil isg given by :

\tau=NIAB\\\\\tau=47.5\times 26.9\times 10^{-3}\times \pi (5.25\times 10^{-2})^2\times 0.535 \\\\\tau=0.00591\ N-m\\\\\tau=5.91\times 10^{-3}\ N-m

So, the magnitude of the maximum possible torque exerted on the coil is 5.91\times 10^{-3}\ N-m.

3 0
3 years ago
2. Two projectiles thrown from the same point at angles 0,=60° and 02=30° with the horizontal,
Liono4ka [1.6K]

Answer:

The answer is below

Explanation:

The maximum height (h) of a projectile with an initial velocity of u, acceleration due to gravity g and at an angle θ with the horizontal is given as:

h=\frac{u^2sin^2\theta}{2g}

Given that the two projectile has the same height.

For\ the\ first\ projectile\ with\ an\ angle\ \60^0 \ with\ the\ horizontal\ and initial\ velocity\ u_1:\\\\h=\frac{u_1^2sin^260}{2g} =\frac{0.75u_1^2}{2g} \\\\For\ the\ second\ projectile\ with\ an\ angle\ \30 \ with\ the\ horizontal\ and initial\ velocity\ u_2:\\\\h=\frac{u_2^2sin^230}{2g} =\frac{0.25u_2^2}{2g}\\\\\frac{0.25u_2^2}{2g}=\frac{0.75u_1^2}{2g}\\\\0.25u_2^2=0.75u_1^2\\\\\frac{u_1^2}{u_2^2} =\frac{0.25}{0.75} \\\\\frac{u_1^2}{u_2^2}=\frac{1}{3} \\\\\frac{u_1}{u_2}=\sqrt\frac{1}{3}

8 0
3 years ago
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