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stiks02 [169]
3 years ago
8

Two golf balls are thrown with the same speed off the top of a large bridge at the same time. ball a is thrown straight downward

, and ball b is thrown horizontally off the bridge. which ball hits the ground first?
a. ball a
b. ball b
c. both balls hit at the same time.
d. we would need to know the speed.
Physics
1 answer:
iren2701 [21]3 years ago
8 0
I think that ball  a hit the ground because it says that it went straight down.
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A spelunker is surveying a cave. She follows a passage that takes her a distance 184 m straight west, then a distance 220 m in a
Sever21 [200]
Refer to the diagram shown below.

Define the unit vector i to point in the eastern direction, and the unit vector j to point in the northern direction.

The first distance is 184 m west. It is represented by
d₁ = -184 i

The second distance is 220 m at 30° south of east. It is
d₂ = 220(cos 30° i - sin 30° j) = 190.53 i - 110 j

The third distance is 104 m at 80 east of north. It is
d₃ = 104(sin 80° i + cos 80° j) =  102.42 i + 18.06 j

Let the fourth distance be 
d₄ = a i + b j

Because the traveler ends back at the original position, the vector sum of the distances is zero. It means that each component of the vector sum is zero.

The x-component yields
-184 + 190.53 + 102.42 + a = 0
a = -108.95

The y-component yields
0 - 110 + 18.06 + b = 0
b = 91.94

The magnitude of the fourth displacement is
√[(-108.95)² + 91.94² ] = 142.56 m

The direction is at an angle θ north of west, given by
θ = tan⁻¹ (91.94/108.95) = 40.2°

Answer:
The fourth displacement has a magnitude of 142.56 m. It is about 40° north of west.

7 0
3 years ago
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Use the drop-down menus to complete the statements. When the 5.0 kg cylinder fell 100 m, the final temperature of the water was
OLEGan [10]

Answer:

A. 26.17 B. 1.17 C. 30.86 D. 5.86

Explanation:

7 0
3 years ago
A convex spherical mirror having a radius of curvature 18 cm (focal length = 1/2 radius of curvature for a spherical mirror) pro
tekilochka [14]

Answer:

distance between object and image =  18.9 cm

Explanation:

given data

radius of curvature = 18 cm

focal length = 1/2 radius of curvature

magnification = 40%

to find out

distance between object and image

solution

we know lens formula that is

1/f = 1/v + 1/u     ....................1

here f = 18 /2 and v and u is object and image distance

and we know m = 40% = 0.40

so 0.40 = -v / u

so here v = - 0.40 u

so from equation 1

1/f = 1/v + 1/u

2/18 = - 1/0.40u + 1/u

u = -13.5 cm   ..................2

and

v = -0.40 (- 13.5)

v = 5.4 cm     ......................3

so from equation 2 and 3

distance between object and image =  5.4 + 13.5

distance between object and image =  18.9 cm

6 0
3 years ago
Un movil de masa 12Kg sobre el cual estan actuando varias fuerzas F_1=48N, F_2=60N y F_3=30N Calcular la aceleracion con la cual
Nikitich [7]

Answer:

Lamentablemente el problema está incompleto, pues no sabemos la dirección en la que se aplican las fuerzas. Por ello, voy a resolver el problema asumiendo dos casos. (abajo se puede ver una imagen donde se describe cada caso)

1) Todas las fuerzas están en la misma dirección.

Entonces la fuerza neta será la suma de las 3 fuerzas, entonces:

F = 48N + 60N + 30N = 138N

Y por la segunda ley de Newton sabemos que:

F = m*a

fuerza igual a masa por aceleración.

Entonces la aceleración está dada por:

a = F/m = 138N/12kg = 11.5 m/s^2

2) Segundo caso, suponemos que F1 es opuesta a F2 y F3

En este caso, la fuerza neta será:

F = F2 + F3 - F1 = 60N + 30N - 48N = 42N

En este caso, la aceleración será:

a = 42N/12kg = 3.5 m/s^2

7 0
2 years ago
Astronauts use a centrifuge to simulate the acceleration of a rocket launch. The centrifuge takes 40.0 s to speed up from rest t
Vinvika [58]

Answer

Time period T = 1.50 s

time t = 40 s

r = 6.2 m

a)

Angular speed ω = 2π/T

                              = \dfrac{2\pi }{1.5}  

                              = 4.189 rad/s

Angular acceleration α = \dfrac{\omega}{t}

                                      = \dfrac{4.189}{40}

                                      = 0.105 rad/s²

Tangential acceleration a = r α = 6.2 x 0.105 = 0.651 m/s²

b)The maximum speed.

       v = 2πr/T

          = \dfrac{2\pi \times 6.2}{1.5}

          = 25.97 m/s

So centripetal acceleration.

        a = \dfrac{v^2}{r}

          = \dfrac{25.97^2}{6.2}

          =  108.781 m/s^2

          = 11.1 g    

in combination with the gravitation acceleration.

a_{total} = \sqrt{(11.1g)^2+g^2}

a_{total}= 11.145 g

6 0
3 years ago
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