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Naddik [55]
3 years ago
10

You’re standing at the highest point on the Moon, 10,786 mm above the level of the Moon’s mean radius. You’ve got a golf club an

d a golf ball.
Part A

(How fast would you need to hit the ball horizontally so it goes into a circular orbit?)

Express your answer with the appropriate units.

Part B

(If you hit the ball vertically with the same speed, to what height above you would it rise?)

Express your answer with the appropriate units.
Physics
1 answer:
Anna71 [15]3 years ago
6 0

Answer:

A)   v = 1,675 10³ m / s  , B)    r₂ = 11,673 10⁶ m

Explanation:

A) This exercise we must use Newton's second law, where the forces of gravity are the Moon

        F = m a

acceleration is centripetal

        a = v² / r

force is the force of universal attraction

         F = G m M / r²

we substitute

        G m M / r² = m v² / r

        v² = G M / r

distance

        r = R_moon + h

        r = 1.74 10⁶ +1.0786 10⁴

        r = 1,750786 10⁶ m

we calculate

        v = √ (6.67 10⁻¹¹ 7.36 10²² / 1.75 10⁶)

        v = √ (2,8052 10⁶)

        v = 1,675 10³ m / s

B) let's use energy conservation

    Starting point. In the mountain

          Em₀ = K + U = ½ m v² + G m M / r

    Final point. Where the speed is zero

          Em_{f} = U = G mM / r₂

           Em₀ = Em_{f}

           ½ m v² + G m M / r = G mM / r₂

           1 / r₂ = (½ v₂ + G M / r) / GM

let's calculate

 1 / r₂ = (½ (1,675 10³)² + 6.67 10⁻¹¹ 7.36 10²² / 1.75 10⁶) /(6.67 10⁻¹¹ 7.36 10²²)

           1 / r₂ = (1,4028 10⁶ + 2,805 10⁶) / 49.12 10¹¹

           1 / r₂ = 8.5664 10⁻⁷

            r₂ = 11,673 10⁶ m

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A cart loaded with bricks has a total mass of 9.13 kg and is pulled at constant speed by a rope. The rope is inclined at 24.7 ◦
blagie [28]

Answer:

W = 0.63 KJ

Explanation:

Work (W) is defined as the point product of force (F) by the distance (d)the body travels due to this force.  

W= F*d *cosα Formula (1)  

F : force (N)

d : displacement (m)

α : angle between force and displacement

Newton's second law:

∑F = m*a Formula (2)  

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration  (m/s²)

We define the x-axis in the direction parallel to the movement of the cart on the ramp and the y-axis in the direction perpendicular to it.

Forces acting on the cart

W: Weight of the cart : In vertical direction

FN : Normal force : perpendicular to the floor

f : Friction force: parallel to the floor

T : tension Force,  inclined at  θ=24.7° above the horizontal

Calculated of the W

W= m*g

W= 9.13 kg* 9.8 m/s² = 89.47 N

x-y components o the  tension force (T)

Tx = Tcosθ = T*cos 24.7° (N)

Ty = Tsin θ = T*sin 24.7°  (N)

Calculated of the FN  

We apply the formula (2)  

∑Fy = m*ay ay = 0  

FN +Ty- W = 0  

FN = W-Ty  

FN =  89.47-T*sin 24.7°

Calculated of the friction force (f)

f = μk*FN

f =(0.597)*(  89.47-T*sin 24.7° )

f= 53.41-0.249T

Calculated of the tension force of the rope (f)

We apply the formula (2) :

∑Fx = m*ax  ,  ax= 0 ,because the speed of the cart  is constant

Tx - f = 0

T*cos 24.7°-( 53.41 - 0.249T )= 0

T*cos 24.7° + 0.249T = 53.41

(1.1575)T = 53.41

T= (53.41) / (1.1575)

T= 46.14 N

Work done on the cart by the rope

We apply the formula (1)

W=T*d *cosα

W= (46.14 N)*(15.1 m) *(cos24.7)

W = 632.97 (N*m) = 632.97 (J)

W = 0.63 KJ

6 0
4 years ago
car was moving in a straight road of length 320 km it covered 240 km with an average velocity 75 km/hr then it ran out of fuel a
Stella [2.4K]

The average velocity of the car for the whole journey is 69.57 km/h.

The given parameters:

  • <em>Length of the road, L = 320 km</em>
  • <em>Distance covered = 240 km at 75 km/h</em>
  • <em>time spent refueling, t₂ = 0.6 hr</em>
  • <em>Final velocity, = 100 km/hr</em>

The time spent by the before refueling is calculated as follows;

t = \frac{d}{v} \\\\t_1 = \frac{240}{75} \\\\t_1 = 3.2 \ hours

The time spent by the car for the remaining journey;

t_3 = \frac{320 - 240}{100} \\\\t_3 = 0.8 \ hr

The total time of the journey is calculated as follows;

t = t_1 + t_2 + t_3\\\\t = 3.2 \ hr \ + \ 0.6 \ hr \ + \ 0.8 \ hr\\\\t = 4.6 \ hours

The average velocity of the car for the whole journey is calculated as follows;

v = \frac{total \ distance }{total \ time} \\\\v = \frac{320}{4.6} \\\\v = 69.57 \ km/h

Learn more about average velocity here: brainly.com/question/6504879

6 0
3 years ago
El dormitorio de Pablo es rectangular, y sus lados miden 3 y 4 metros. Ha decidido dividirlo en dos partes triangulares con una
Paraphin [41]

Answer:

 c = 5 m

Explanation:

this exercise you want to divide the rectangular room into two triangular rooms

                 

the area of ​​triangles is

           A = ½ base height

           A = ½ 4 3

          A = 6 m²

the length of the curtain can be found using the Pythagorean theorem

           c² = b² + a²

           c = √ (4² + 3²)

           c = 5 m

this is the length of the curtain

5 0
3 years ago
The diffusion rate for a solute is 4.0 x 10^-11 kg/s in a solvent- filled channel that has a cross-sectional area of 0.50 cm^2 a
zlopas [31]

Answer:

s = 9.6\times 10^{-12}kg/s

Explanation:

Given:

Solute Diffusion rate  = 4.0 × 10⁻¹¹ kg/s

Area of cross-section = 0.50 cm²

Length of channel  =0.25 cm

Now for the new channel

Area of cross-section = 0.30 cm²

Length of channel  =0.10 cm

let the Solute Diffusion rate  of new channel = s

now equating the diffusion rate per unit volume for both the channels

\frac{4\times 10^{-11}}{0.50\times 0.25}=\frac{s}{0.30\times 0.10}

thus,

s = 9.6\times 10^{-12}kg/s

7 0
4 years ago
There is a bell at the top of a tower that is 45 m high. The bell weighs 190 N. The bell has ____________
salantis [7]

Answer:

The bell has a potential energy of 8550 [J]

Explanation:

Since the belt is 45 [m] above ground level, only potential energy is available. And this energy can be calculated by means of the following equation.

E_{p}= W*h\\E_{p} = 190*45\\E_{p}=8550[J]

8 0
3 years ago
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