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amid [387]
3 years ago
12

The picture shows an orange falling from an orange tree

Physics
2 answers:
Alinara [238K]3 years ago
6 0

Answer:

A but not 100% certain.

Explanation:

you can eliminate C since it doesn't mention gravity, and B says gravity is pushing up on the orange. I think it is A because friction is going against the orange because it has to pass through the air, and that creates friction. I think that is the reason why things burn up when they enter the earths atmosphere.

il63 [147K]3 years ago
6 0

Answer:

Answer A is the correct one

Explanation:

The force of gravity is pulling down in the orange, while the force of resistance with the air is pushing up on the orange. This last force could be viewed as a type of "friction" force with the air.

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Sveta_85 [38]

Answer:

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Explanation:

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8 0
3 years ago
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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
Garret's swimming coach is giving him important and helpful tips to improve his swimming skills. Which gesture suggests that Gar
bogdanovich [222]

Answer: "important and helpful" Show garret is paying close attention to her coach. Hope this help!

7 0
3 years ago
Read 2 more answers
What is the maximum speed when the conditions are mass =450 kg, initial height= 30 m, and the roller coaster is initially at res
Zarrin [17]

Answer:

B. 24.2 m/s

Explanation:

Given;

mass of the roller coaster, m = 450 kg

height of the roller coaster, h = 30 m

The maximum potential energy of the roller coaster  due to its height is given by;

P.E_{max} = mgh\\\\PE_{max} = 450 *9.8*30\\\\PE_{max} = 132,300 \ J

P.E_{max} = K.E_{max} \ (law \ of \ conservation\ of \ energy)

K.E_{max} = \frac{1}{2}mv_{max}^2\\\\ v_{max}^2 = \frac{2K.E_{max}}{m}\\\\ v_{max}^2 = \frac{2*132300}{450}\\\\ v_{max}^2 =588\\\\v_{max} = \sqrt{588}\\\\  v_{max} = 24.2 \ m/s

Therefore, the maximum speed of the roller coaster is 24.2 m/s.

3 0
3 years ago
From the window of a house that is placed 15 m
kow [346]

Answer:

a) 52.915 m

b) The vertical velocity is approximately 21.092 m/s

The resultant velocity is approximately 26.5 m/s

Explanation:

a) The height of the window in the house from which the water was thrown = 15 m

The speed of the stream of water thrown = 20 m/s

The angle at which the water was thrown = 37° over the horizontal

The acceleration due to gravity, g = 10 m/s²

a) The distance from the base of the house at which the water will fall is given as follows;

y = y₀ + u·t·sin(θ) + 1/2·g·t²

Where;

y = The vertical height reached    

u = The initial velocity

t = Time of flight

From the point the steam of water is thrown, we get;

y₀ = 15 m

Therefore;

y = 15 + 20 × t × sin(37°) - 1/2 × 10 × t²

y = 15 + 20 × t × sin(37°) - 5 × t²

When y = 0, Ground level, we get

0 = 15 + 20 × t × sin(37°) - 5 × t²

5·t² - 20×sin(37°)×t -15 = 0

∴ t = (20 ×sin(37°) ± √((-20 × ·sin(37°))² - 4 × (5) × (-15)))/(2 × 5)

t ≈ 3.3128302, or t ≈ 0.906

Therefore, the time of flight of the water, t ≈ 3.3128302 seconds

The distance from the base of the house at which the water will fall = The horizontal distance travelled by the water, x

x = u·cos(θ)×t

∴ x = 20 × cos(37°) × 3.3128302 ≈ 52.915

The distance from the base of the house at which the water will fall = x ≈ 52.915 m

b) The velocity at which the water will reach the ground, 'v', is given as follows;

The vertical velocity, v_y = u·sin(θ)·t - g·t

At the ground, t ≈ 3.3128302 seconds

∴ v_y = 20 × sin(37) - 10 × 3.3128302 ≈ -21.092

The vertical velocity at which the water will reach the ground, v_y ≈ 21.092 m/s (downwards)

The resultant velocity, v = √(v_y² + vₓ²)

∴ v = √(21.092² + (0 × cos(37°))²) ≈ 26.5

The resultant velocity at which the water will reach the ground, v ≈ 26.5 m/s.

5 0
3 years ago
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