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Serga [27]
3 years ago
9

Discuss the force that exists between the Earth and the moon by referring to the mass of each.

Physics
1 answer:
Fudgin [204]3 years ago
7 0
The word gravity is used to describe the gravitational pull (force) an object experiences on or near the surface of a planet or moon. The gravitational force is a force that attracts objects with mass towards each other. Any object with mass exerts a gravitational force on any other object with mass.

Hope it answers your question!

Brainliest would be nice but of course you don’t gotta :)
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You are holding a shopping basket at the grocery store with two 0.55-kg cartons of cereal at the left end of the basket. the bas
stepan [7]
Refer to the diagram shown below.

The basket is represented by a weightless rigid beam of length 0.78 m.
The x-coordinate is measured from the left end of the basket.

The mass at x=0 is 2*0.55 = 1.1 kg.
The weight acting at x = 0 is W₁ = 1.1*9.8 = 10.78 N

The mass near the right end is 1.8 kg.
Its weight is W₂ = 1.8*9.8 = 17.64 N

The fulcrum is in the middle of the basket, therefore its location is
x = 0.78/2 = 0.39 m.

For equilibrium, the sum of moments about the fulcrum is zero.
Therefore
(10.78 N)*(0.39 m) - (17.64 N)*(x-0.39 m) = 0
4.2042 - 17.64x + 6.8796 = 0
-17.64x = -11.0838
x = 0.6283 m

Answer:  0.63 m from the left end.

5 0
3 years ago
What is the lift (in newtons) due to Bernoulli's principle on a wing of area 76 m2 if the air passes over the top and bottom sur
AveGali [126]

Answer:

So lift will be 30.19632 N

Explanation:

We have given area of the wing a=76m^2

We know that density of air d=1.29kg/m^3

Speed at top surface v_2=290m/sec and speed at bottom surface v_1=150m/sec

According to Bernoulli's principle force is given by

F=A\times d\times \frac{v_2^2-v_1^2}{2}=76\times 1.29\times \frac{290^2-150^2}{2}=3019632N

4 0
3 years ago
What is the frequency of a transverse wave ?​
yuradex [85]

Answer:

The number of complete vibration or wave made in

one second is called frequency.

7 0
3 years ago
1. A perspex box has a 10 cm square base and contains water to a height of 10 cm. A piece of rock of mass 600g is lowered into t
AnnZ [28]

Answer:

(a) The volume of water is 100 cm³

(b) The volume of the rock is 20 cm³

(c) The density of the rock is 30 g/cm³

Explanation:

The given parameters of the perspex box are;

The area of the base of the box, A = 10 cm²

The initial level of water in the box, h₁ = 10 cm

The mass of the rock placed in the box, m = 600 g

The final level of water in the box, h₂ = 12 cm

(a) The volume of water in the box, 'V', is given as follows;

V = A × h₁

∴ The volume of water in the box, V = 10 cm² × 10 cm = 100 cm³

The volume of water in the box, V = 100 cm³

(b) When the rock is placed in the box the total volume, V_T, is given by the sum of the rock, V_r, and the  water, V, is given as follows;

V_T = V_r + V

V_T = A × h₂

∴ V_T = 10 cm² × 12 cm = 120 cm³

The total volume, V_T = 120 cm³

The volume of the rock, V_r = V_T - V

∴ V_r = 120 cm³ - 100 cm³ = 20 cm³

The volume of the rock, V_r = 20 cm³

(c) The density of the rock, ρ = (Mass of the rock, m)/(The volume of the rock)

∴ The density of the rock, ρ = 600 g/(20 cm³) = 30 g/cm³

8 0
3 years ago
A body of mass 120 g is taken vertically upwards to reach the height of 5m,claculate work done
Anuta_ua [19.1K]

FOLLOW ME FOR CLEARING YOUR NEXT DOUBT

3 0
3 years ago
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