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3 years ago

The moon has a mass of 7.34 . 102 kg and a radius of 1.74 . 106 meters. If you have a mass of 66 kg,

Physics

1 answer:

Ierofanga [76]3 years ago

3 0

Answer:

$F=1.06\times 10^{-18}\ N$

Explanation:

Given that,

Mass of the Moon, $M=7.34\times 10^2\ kg$

Mass of the person, m = 66 kg

The radius of Moon, $r=1.74\times 10^6\ m$

We need to find the force between the person and the Moon. The formula for the gravitational force is given by :

$F=G\dfrac{Mm}{r^2}\\\\F=6.67\times 10^{-11}\times \dfrac{7.34\times 10^2\times 66}{(1.74\times 10^6)^2}\\\\=1.06\times 10^{-18}\ N$

So, the required force is $1.06\times 10^{-18}\ N$ .

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A vehicle in for repair has a burst upper radiator tank. Technician A says thermostat that was stuck in the open position could

Morgarella [4.7K]

Answer:

Technician B

Explanation:

A thermostat being stuck in an open position allows coolant to constantly circulate through the radiator which will cause a drop in the temperature of the engine. Hence, thermostat being stuck in an open position has nothing to do with burst upper radiator.

On the other hand, a radiator cap controls the pressure within the radiator which needs to be kept constant in order prevent overheating or damage to the radiator itself.

Therefore, technician B is correct.

5 0

3 years ago

A 0.250 kgkg toy is undergoing SHM on the end of a horizontal spring with force constant 300 N/mN/m. When the toy is 0.0120 mm f

konstantin123 [22]

Answer:

(a) The total energy of the object at any point in its motion is 0.0416 J

(b) The amplitude of the motion is 0.0167 m

Explanation:

Given;

mass of the toy, m = 0.25 kg

force constant of the spring, k = 300 N/m

displacement of the toy, x = 0.012 m

speed of the toy, v = 0.4 m/s

(a) The total energy of the object at any point in its motion

E = ¹/₂mv² + ¹/₂kx²

E = ¹/₂ (0.25)(0.4)² + ¹/₂ (300)(0.012)²

E = 0.0416 J

(b) the amplitude of the motion

E = ¹/₂KA²

$A = \sqrt{\frac{2E}{K} } \\\\A = \sqrt{\frac{2*0.0416}{300} } \\\\A = 0.0167 \ m$

$E = \frac{1}{2} mv_{max}^2\\\\v_{max} = \sqrt{\frac{2E}{m} } \\\\v_{max} = \sqrt{\frac{2*0.0416}{0.25} } \\\\v_{max} = 0.577 \ m/s$

8 0

3 years ago

On Earth, a spring stretches by 5.0 cm when a mass of 3.0 kg is suspended from one end.

Neko [114]

Answer:

Mass = 18.0 kg

Explanation:

From Hooke's law,

F = ke

where: F is the force, k is the spring constant and e is the extension.

But, F = mg

So that,

mg = ke

On the Earth, let the gravitational force be 10 m/ $s^{2}$ .

3.0 x 10 = k x 5.0

30 = 5k

⇒ k = $\frac{30}{5}$ ................ 1

On the Moon, the gravitational force is $\frac{1}{6}$ of that on the Earth.

m x $\frac{10}{6}$ = k x 5.0

$\frac{10m}{6}$ = 5k

⇒ k = $\frac{10m}{30}$ ............. 2

Equating 1 and 2, we have;

$\frac{30}{5}$ = $\frac{10m}{30}$

m = $\frac{900}{50}$

= 18.0

m = 18.0 kg

The mass required to produce the same extension on the Moon is 18 kg.

8 0

3 years ago

Read 2 more answers

A 10 Newton effort force is required to lift a 10 kg box just off the ground. What is the effort required to lift the box off th

Karo-lina-s [1.5K]

2.0 N is the answer, I hope this helps :D

5 0

3 years ago

Read 2 more answers

A small mailbag is released from a helicopter that is descending steadily at 3 m/s.

mario62 [17]

<u>Answer:</u>

a) Speed of mailbag after 3 seconds = 32.4 m/s

b) Package is 44.1 meter below helicopter

c) If the helicopter was rising steadily at 3.00 m/s

Speed of mailbag after 3 seconds = 26.4 m/s

Package is 44.1 meter below helicopter

<u>Explanation:</u>

a) We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

Initial velocity = 3 m/s, acceleration = 9.8 $m/s^2$ and time = 3 seconds.

v = 3+9.8*3 = 32.4 m/s

Speed of mailbag after 3 seconds = 32.4 m/s

b) We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

Velocity of helicopter = 3 m/s, time taken = 3 seconds, acceleration = 0 $m/s^2$ .

$s= 3*3+\frac{1}{2} *0*3^2\\ \\ s=9m$

Distance traveled by helicopter = 9 meter.

Velocity of package = 3 m/s, time taken = 3 seconds, acceleration = 9.8 $m/s^2$ .

$s= 3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 53.1m$

Distance traveled by package = 53.1 meter.

So package is (53.1-9)meter below helicopter = 44.1 m

c) Initial velocity = -3 m/s, acceleration = 9.8 $m/s^2$ and time = 3 seconds.

v = -3+9.8*3 = 26.4 m/s

Speed of mailbag after 3 seconds = 26.4 m/s

Velocity of helicopter = -3 m/s, time taken = 3 seconds, acceleration = 0 $m/s^2$ .

$s= -3*3+\frac{1}{2} *0*3^2\\ \\ s=-9m$

Distance traveled by helicopter = 9 meter.

Velocity of package = -3 m/s, time taken = 3 seconds, acceleration = 9.8 $m/s^2$ .

$s= -3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 35.1m$

Distance traveled by package = 35.1 meter.

So package is (35.1+9)meter below helicopter = 44.1 m

4 0

3 years ago