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Talja [164]
3 years ago
15

Suppose that scientists observe violent gas eruptions on a planet with an acceleration due to gravity of 3.9 m/s2. The jets thro

w sand and dust about 75 m above the surface. What is the speed i of the material just as it leaves the surface?
Physics
1 answer:
Lina20 [59]3 years ago
4 0

Answer:

The velocity with which the sand throw is 24.2 m/s.

Explanation:

acceleration due to gravity, a =  3.9 m/s2

height, h = 75 m

final velocity, v = 0

Let the initial  velocity at the time of throw is u.

Use third equation of motion

v^{2} = u^{2} - 2 g h \\\\0 = u^{2} -2\times 3.9\times 75\\\\u=24.2m/s

The velocity with which the sand throw is 24.2 m/s.

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Un automovil transita por una curva en forma de U y recorre una distancia de 400m en 30s sin embargo su posición final está a so
pochemuha

Answer:

Definimos:

Rapidez media es igual al cociente entre la distancia recorrida y el tiempo que se tarda en recorrer esa distancia.

En este caso la distancia recorrida es 400m, y el tiempo que se tarda es 30s, entonces la rapidez media va a ser:

RM = 400m/30s = 13.33 m/s

La velocidad media por otro lado, es igual al cociente entre el desplazamiento y el tiempo necesario para desplazarse.

El desplazamiento es igual a la distancia entre la posición final y la posición inicial, que en este caso eso 40m, y el tiempo necesario sigue siendo 30s, entonces la velocidad media va a ser:

VM = 40m/30s = 1.33 m/s

8 0
3 years ago
A golf club rotates 215 degrees and has a length (radius) equal to 29 inches. The time it took to swing the club was 0.8 seconds
vichka [17]

Answer:

The average linear velocity (inches/second) of the golf club is 136.01 inches/second

Explanation:

Given;

length of the club, L = 29 inches

rotation angle, θ = 215⁰

time of motion, t = 0.8 s

The angular speed of the club is calculated as follows;

\omega = (\frac{\theta}{360} \times 2\pi, \ rad) \times \frac{1}{t} \\\\\omega =  (\frac{215}{360} \times 2\pi, \ rad) \times \frac{1}{0.8 \ s} \\\\\omega = 4.69 \ rad/s

The average linear velocity (inches/second) of the golf club is calculated as;

v = ωr

v = 4.69 rad/s  x  29 inches

v = 136.01 inches/second

Therefore, the average linear velocity (inches/second) of the golf club is 136.01 inches/second

8 0
2 years ago
What is 1.0 x 10^9 in standard form?
jasenka [17]
1.0 x 10^9= 1000000000
3 0
3 years ago
Describe electric potential energy
nirvana33 [79]
Is potential energy that results from conservative Coulomb forces and is associated with the configuration of a particular set of point charges within a defined system
3 0
3 years ago
Glycerin is poured into an open U-shaped tube until the height in both sides is 20cm. Ethyl alcohol is then poured into one arm
Liula [17]

Answer:

7.5 cm

Explanation:

In the figure we can see a sketch of the problem. We know that at the bottom of the U-shaped tube the pressure is equal in both branches. Defining \rho_A: Ethyl alcohol density and \rho_G: Glycerin density , we can write:

\rho_A\times g \times h_1 + \rho_G \times g \times h_2 = \rho_G \times g \times h_3

Simplifying:

\rho_A\times h_1 = \rho_G \times (h_3 - h_2) (1)

On the other hand:

h_1 + h_2 = \Delta h + h_3

Rearranging:

h_1 - \Delta h = h_3 - h_2 (2)

Replacing  (2) in (1):

\rho_A\times h_1 = \rho_G \times (h_1 - \Delta h)

Rearranging:

\frac{h_1 \times (\rho_A - \rho_G)}{- \rho_G} = \Delta h

Data:

h_1 = 20 cm; \rho_A = 0.789 \frac{g}{cm^3}; \rho_G = 1.26 \frac{g}{cm^3}

\frac{20 cm \times (0.789 - 1.26) \frac{g}{cm^3}}{- 1.26\frac{g}{cm^3}}  = \Delta h

7.5 cm =  \Delta h

7 0
3 years ago
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