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vredina [299]
2 years ago
15

A spring has K=24 N/m, we put in its end an object with 4 kg, if we pull the object 0.5 m then leave it, what would be the speed

in 0.3 m from the balance point.
Physics
1 answer:
adoni [48]2 years ago
8 0

Answer:

Velocity- v=10m/s south Speed- v=10m/s  

Explanation:

the salar measurement of how fast an object is moving.

Velocity- v=10m/s south :

The vector rate of change of position.

Speed- v=10 m/s   :  the salar measurement of how fast an object is moving.

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The universal law of gravitation states that the force of attraction between two objects depends on which quantities?
Ronch [10]

Answer:

D. the masses of the objects and the distance between them

Explanation:

Gravitation is a force, a force doesn't care about the shape or density of objects, only about their masses... and distances.

And you can get it using the following equation:

f = \frac{Gm_{1}m_{2} }{d^{2} }

Where :

G is the universal gravitational constant : G = 6.6726 x 10-11N-m2/kg2

m represent the mass of each of the two objects

d is the distance between the centers of the objects.

4 0
3 years ago
Water flows through a 2.5cm diameter pipe at a rate of
My name is Ann [436]

Answer:

From the Bernoulli energy principle,

ΔP + 1/2ρΔv² = 0 -------------------------- eqn 1

where

ΔP = pressure drop = P2 - P1 = (1 - 0.25)x10⁵ N/m =7.5 x 10⁴N/m

Δv²= velocity change = v₂² - v₁²

ρ = water density = 1kg/m3

Recall volumetric flow rate, Q=A v = constant

A = cross sectional area = πr²=πd²/4

d=pipe diameter at point 2 = 2.5cm = 0.025m and Q =0.20m³/min = 0.00333m³/s

So A= 0.000491m²

we can get v2 = Q/A = 6.79m/s

From eqn 1, v₁² = 2(P2 - P1)/ρ + v₂²

v₁² =  (2 x 7.5 x 10⁴)/1000 + 6.79²

v₁² = 196

v₁ = 14m/s

we can now get the area of the constriction point 2, A₁ = Q/v₁

A₁ = 0.000238m² and the diameter now will be d₁

d₁² = 4 x A₁ / π

     = 4 x 0.000238/3.14 = 0.000303m²

d₁ = √0.000303 = 0.0174m

Therefore, the diameter of  a constriction in the pipe  at the new pressure = 0.0174m = 1.74cm

6 0
3 years ago
If a simple machine reduces the strength of a force, what must be increased.
svetoff [14.1K]
"The distance that the force moves" is the one among the following choices given in the question that must be increased, if a simple machine reduces the strength of a force. The correct option among all the options that are given in the question is the first option or option "A". I hope the answer helped you.

6 0
3 years ago
A point charge of 4.0 µC is placed at a distance of 0.10 m from a hard rubber rod with an electric field of 1.0 × 103 . What is
schepotkina [342]
4 for the first one and 5.2 for the second one.
3 0
3 years ago
Read 2 more answers
You travel at a speed of 24 m/s. How much time will it take you to travel 127m (round to two decimal places if needed)
shtirl [24]

Answer: 5.29 seconds

Explanation:

if the speed is 24 meters per second than it would take you 5.29 seconds or 127meters/24m/s to get 5.29s

4 0
3 years ago
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