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azamat
3 years ago
6

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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Is this a balanced equation? 6O2+ C6H12O6---&gt; 6 CO2 + 6H2O <br><br> ]
grigory [225]

Yes, this equation is balanced.

You can check it using the law of conservation of mass, i.e., the total mass of the products formed must be equal to the total mass of the reactants.

7 0
3 years ago
To test a slide at an amusement park, a block of wood with mass 3.00 kgkg is released at the top of the slide and slides down to
dem82 [27]

Answer:

511.1 J

Explanation:

We are given that

Mass of wood block=m=3 kg

Vertical distance,h=23 m

Horizontal distance =x=30 m

Distance traveled in downward direction y=40 m

Initial velocity,u=0

y=ut+\frac{1}{2} gt^2

Where g=9.8 m/s^2

40=0+\frac{1}{2}(9.8)t^2=4.9t^2

t^2=\frac{40}{4.9}

t=\sqrt{\frac{40}{4.9}}=2.86 s

x=v_x\times t

30=v_x(2.86)

v=v_x=\frac{30}{2.86}=10.49 m/s

By work energy theorem

Change in kinetic energy=Work done= mgh-W

\frac{1}{2}mv^2-\frac{1}{2}mu^2=3\times 9.8\times 23-W

\frac{1}{2}(3)(10.49)^2-0=676.2-W

W=676.2-\frac{1}{2}(3)(10.49)^2=511.1 J

Hence, the work done due to friction on the block as it slides down the ramp=511.1 J

6 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

(c) The maximum speed attained by the object during its motion is 0.577 m/s

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

(c) the maximum speed attained by the object during its motion

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
Listed following are three possible models for the long-term expansion (and possible contraction) of the universe in the absence
vazorg [7]

Answer:

do your best

Explanation:

<u>babyboiiii∛√</u>

8 0
3 years ago
a car goes from a velocity zero to a velocity of 15 meters per second East in 2.1 seconds. What is the car's acceleration?
leva [86]
31.5 would be the acceleration.
3 0
3 years ago
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