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timama [110]
1 year ago
5

A 60 kilogram astronaut weighs 96 newtons on the surface of the moon. calculate the acceleration due to gravity on the moon.

Physics
2 answers:
Charra [1.4K]1 year ago
8 0

Answer:

Acceleration due to gravity on moon, 1.6\ m/s^2

Explanation:

Mass of the astronaut, m = 60 kg

Weight of astronaut on the surafce of moon, W = 96 N

The weight of astronaut is given by :

W = m g

g=\dfrac{W}{m}

g=\dfrac{96\ N}{60\ kg}

g=1.6\ m/s^2

So, the acceleration due to gravity on the surface of moon is 1.6\ m/s^2

Also, the acceleration due to gravity on the surface of moon is (1/6)th of the acceleration due to gravity on the surface of earth.

Hence, this is the required solution.

34kurt1 year ago
6 0
Given: Mass m = 60 Kg

           Weight  W = 96 N

Required: Acceleration due to gravity, g = ?

Formula:  W = mg

                g = W/g

                g = 96 Kg.m/s²/60 Kg   (note: this is the derive unit for Newton "N")

                g = 1.6 m/s²

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The three point charges +4.0 μC, -5.0 μC, and -9.0 μC are placed on the x-axis at the points x = 0 cm, x = 40 cm, and x = 120 cm
ale4655 [162]

Answer:

 

Explanation:

4μC will attract -9μC towards the centre and -5μC will repel it away from the centre.  Both these forces are opposite to each other.

Force due to 4μC on -9μC towards the centre

= k x Q₁ Q₂/R² = 9 X 10⁹ X 4 X 10⁻⁶ X 9 X 10⁻⁶ / (1.2)² = 225 X 10⁻³ N/C

Force due to -5μC on -9μC  away from the centre

= 9 x 10⁹ x 5 x 10⁻⁶x 9 x 10⁻⁶/( 0.8)² = 632.8 x 10⁻³ .N/C

Ner field =407.8 N/C.

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2 years ago
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The loudness of a sound is inversely proportional to the square of your distance from the source of the sound. if your friend is
kirill [66]
Let loudness be L, distance be d, and k be the constant of variation such that the equation that would best represent the given above is,
                    L = k/(d^2)
For Case 1,
                       L1 = k/(d1^2)
For Case 2,
                       L2 = k/((d1/4)^2)
For k to be equal, L1 = 16L2. 
Therefore, the loudness at your friend's position is 16 times that of yours. 
5 0
2 years ago
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A 50-kg person stands 1.5 m away from one end of a uniform 6.0-m-long scaffold of mass 70.0 kg.
babymother [125]

Answer

given,

mass of the person, m = 50 Kg

length of scaffold = 6 m

mass of scaffold, M= 70 Kg

distance of person standing from one end = 1.5 m

Tension in the vertical rope = ?

now equating all the vertical forces acting in the system.

T₁ + T₂ = m g + M g

T₁ + T₂ = 50 x 9.8  + 70 x 9.8

T₁ + T₂ = 1176...........(1)

system is equilibrium so, the moment along the system will also be zero.

taking moment about rope with tension T₂.

now,

T₁ x 6 - mg x (6-1.5) - M g x 3 = 0

'3 m' is used because the weight of the scaffold pass through center of gravity.

6 T₁ = 50 x 9.8 x 4.5 + 70 x 9.8 x 3

6 T₁ = 4263

    T₁ = 710.5 N

from equation (1)

T₂ = 1176 - 710.5

 T₂ = 465.5 N

hence, T₁ = 710.5 N and T₂ = 465.5 N

4 0
2 years ago
Why is a solution of 4% acetic acid in 95% ethanol used to wash the crude aldol-dehydration product?
raketka [301]

A cold acetic acid solution is used to wash the residue of the reagent in preparation of an aldol condensation product after vacuum filtration.  The main reason in washing with the acetic acid rinse is to neutralize any sodium hydroxide.

5 0
2 years ago
In general it is best to conceptualize vectors as arrows in space, and then to make calculations with them using their component
Dimas [21]

(1) A - B

(2) B - C

(3) - A + B - C

(4) 3A - 2C

(5) - 2A + 3B - C

(6) 2A - 3 (B - C)

Answer:

(1)  (3,-5,-4)

(2) (-5, 4, 0)

(3) (-6, 4, 3)

(4) (-3, -2, -11)

(5) (-11, 14, 8)

(6) (17, -12, -6)

Explanation:

A⃗ =(1,0,−3)

B⃗ =(−2,5,1)

C⃗ =(3,1,1)

Vector additions and subtraction are done on a component by component basis, that is, only data from component î can be added to or subtracted from another Vector's component î. And so on for components j and k.

1) (A - B) = (1,0,−3) - (−2,5,1) = (1-(-2), 0-5, -3-1) = (3,-5,-4)

2)  (B - C) = (−2,5,1) - (3,1,1) = (-2-3, 5-1, 1-1) = (-5, 4, 0)

3) -A + B - C = -(1,0,−3) + (−2,5,1) - (3,1,1) = (-1-2-3, 0+5-1, 3+1-1) = (-6, 4, 3)

4) 3A - 2C = 3(1,0,−3) - 2(3,1,1) = (3,0,-9) - (6,2,2) = (3-6, 0-2, -9-2) = (-3, -2, -11)

5) -2A + 3B - C = -2(1,0,−3) + 3(−2,5,1) - (3,1,1) = (-2,0,6) + (-6,15,3) - (3,1,1) = (-2-6-3, 0+15-1, 6+3-1) = (-11, 14, 8)

6) 2A - 3 (B - C) = 2(1,0,−3) - 3[(−2,5,1) - (3,1,1)] = (2,0,-6) - 3(-5,4,0) = (2+15, 0-12, -6-0) = (17, -12, -6)

3 0
2 years ago
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