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Arte-miy333 [17]
2 years ago
8

How do you calculate gravity’s pull?

Physics
2 answers:
Ilya [14]2 years ago
5 0
Gravity pulls at a constant of 9.8 m/s
WINSTONCH [101]2 years ago
4 0

Answer:

The gravitational force between a mass and the Earth is the object's weight. Mass is considered a measure of an object's inertia, and its weight is the force exerted on the object in a gravitational field. On the surface of the Earth, the two forces are related by the acceleration due to gravity: Fg = mg.

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A hunter aims at a deer which is 40 yards away. Her cross- bow is at a height of 5ft, and she aims for a spot on the deer 4ft ab
shutvik [7]

Answer:

a)  θ₁ = 0.487º , b)   t = 0.400 s ,        x = 11.73 ft

Explanation:

For this exercise let's use the projectile launch relationships.

The initial height is I = 5 ft and the final height y = 4 ft

            y = y₀ + v_{oy} t - ½ g t²

The distance to the band is x = 40 yard (3 ft / 1 yard) = 120 ft

            x = v₀ₓ t

            t = x / v₀ₓ

We replace

             y –y₀ = v_{oy} x / v₀ₓ - ½ g x² / v₀ₓ²

             v_{oy} = v₀ sin θ

             v₀ₓ = vo cos θ

             

             y –y₀ = x tan θ - ½ g x² / v₀² cos² θ

                5-4 = 120 tan θ - ½ 32 120 / (300 2 cos2 θ)

                1 = 120 tan θ - 0.0213 sec² θ

Let's use the trigonometry relationship

               Sec² θ = 1 - tan² θ

                 1 = 120 tan θ - 0.0213 (1 –tan²θ)

                 0.0213 tan²θ + 120 tanθ -1.0213 = 0

                 

We change variables

          u = tan θ

          u² + 5633.8 u - 48.03 = 0

We solve the second degree equation

          u = [-5633.8 ±√(5633.8 2 + 4 48.03)] / 2

          u = [- 5633.8 ± 5633.82] / 2

           u₁ = 0.0085

           u₂= -5633.81

           u = tan θ

           θ = tan⁻¹ u

For u₁

           θ₁ = tan⁻¹ 0.0085

           θ₁ = 0.487º

For u₂

           θ₂ = -89.99º

The launch angle must be 0.487º

b) let's look for the time it takes for the arrow to arrive

         x = v₀ₓ t

         t = x / v₀ cos θ

         

         t = 120 / (300 cos 0.487)

         t = 0.400 s

The deer must be at a distance of

           v = 20 mph (5280 ft / 1 mi) (1 h / 3600s) = 29.33 ft / s

           x = v t

           x = 29.33 0.4

           x = 11.73 ft

3 0
3 years ago
Which best describes what occurs when an object takes in a wave as the wave hits it?.
statuscvo [17]

Answer:

Absorption

Explanation:

7 0
2 years ago
What is the angular momentum of a 0.25 kg mass rotating on the end of a piece of
finlep [7]

L = r x p = rmv = mr²ω

L = 0.25 x 0.75² x 12.5 = 1.758

4 0
2 years ago
PLEASE HELP. NOTICE HOW THEY PUT FUTURE
MrMuchimi

I think its <u>B</u> because it looks like it might indicate a future rain storm.

6 0
3 years ago
On earth, two parts of a space probe weigh 14500 N and 4800 N. These parts are separated by a center-to-center distance of 18 m
Nastasia [14]

Answer:

F = 1.489*10^{-7}  N

Explanation: Weight of space probes on earth is given by:W= m*g

W= weight of the object( in N)

m= mass of the object (in kg)

g=acceleration due to gravity(9.81 \frac{m}{s^{2} })

Therefore,

m_{1} = \frac{14500}{9.81}

m_{1} = 1478.08  kg

Similarly,

m_{2} = \frac{4800}{9.81}

m_{2} = 489.29  kg

Now, considering these two parts as uniform spherical objects

Also, according to Superposition principle, gravitational net force experienced by an object is sum of all individual forces on the object.

Force between these two objects is given by:

F =  \frac{Gm_{1} m_{2}}{R^{2} }

G= gravitational constant (6.67 * 10^{-11} m^{3} kg^{-1} s^{-2})

m_{1} , m_{2}= masses of the object

R= distance between their centres (in m)(18 m)

Substituiting all these values into the above formula

F = 1.489*10^{-7}  N

This is the magnitude of force experienced by each part in the direction towards the other part, i.e the gravitational force is attractive in nature.

7 0
3 years ago
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