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3 years ago

What is the speed of an object at rest

1 answer:

4 0

The speed of an object is measured by calculating the distance it travels per unit time.
The formula in Physics for calculating speed is = distance/ time.

Since an object at rest is not covering any distance, the speed is always assumed to be 0 m/s.

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Plz HELP

zubka84 [21]

The 2nd has the most porosity

Infiltration

That's all I can answer for you. Hope it helps

3 0

4 years ago

The builders of the pyramids used a long ramp to lift 20000-kg (20.0-ton) blocks. If a block rose 0.840 m in height while travel

olga_2 [115]

Given Information:

Mass = m = 20000 kg

Height of ramp = h = 0.840 m

Length of ramp = L = 20 m

Required Information:

Uphill Force = F = ?

Answer:

F = 8232 N

Explanation:

The force can be found using the equation

F = mgsinθ

Where m is the mass of block, g is the acceleration due to gravity

Since the ramp can be modeled as an inclined plane so it can formed into a triangle, recall that in right angle triangle sinθ is equal to opposite over hypotenuse. The opposite is the height and hypotenuse is the ramp surface.

sinθ = h/L = 0.840/20 = 0.042

F = 20000*9.8*0.042

F = 8232 N

Therefore, an uphill force of 8232 N would be needed.

8 0

4 years ago

Gravity prevents planets from flying off at a

Bad White [126]

Answer:elliptical orbit

Explanation:

7 0

3 years ago

Se dispara un proyectil con una velocidad inicial de 50 m/s y un ángulo de 30°, por encima de la horizontal. Calcular: a) Posici

dmitriy555 [2]

Answer:

a) Posición y velocidad después de los 6s

i) Posición = -26.58m

ii) velocidad = -33.86m/s

b) Tiempo para alcanzar la altura máxima

= 2.55s

c) Alcance horizontal

= 220.7m

Explanation:

a) Posición y velocidad después de los 6s

i) Posición

y = (usinθ)t – 1/2 gt²

u = 50m/s

θ = 30°

g = 9.81m/s²

t = 6s

y = (50 × sin 30)6 - 1/2 × 9.81 × 6²

y = 150 - 176.58

y = -26.58m

ii) velocidad

v = u sinθ–gt

u = 50m/s

θ = 30°

g = 9.81m/s²

t = 6s

v = 50 × sin 30 - 9.81 × 6

v = 25 - 58.86

v = -33.86m/s

b) Tiempo para alcanzar la altura máxima

usinθ /g

u = 50m/s

θ = 30°

g = 9.81m/s²

= 50 × sin 30/ 9.81

= 25/9.81

= 2.5484199796s

≈ 2.55s

c) Alcance horizontal

R = u²sin2θ/g

u = 50m/s

θ = 30°

g = 9.81m/s²

R = 50² ×( sin 2 ×30°) /9.81

R = 220.69964419m

≈ 220.7m

7 0

3 years ago

A large goose lands in a lake and bobs up and down for a short time. A fisherman notices that the first wave crest created by th

sleet_krkn [62]

Answer:

11.85m

Explanation:

For the development of this problem we must resort to the concepts of the basic kinematic equations that define the distance of a body, such as speed for a certain time, that is,

$D=v*t$