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Katen [24]
3 years ago
6

What is the displacement of a spring if it has a spring constant of 10 N/m, and a force of 2.5 N is applied?

Physics
2 answers:
pochemuha3 years ago
5 0
O.25 m is the displacement
disa [49]3 years ago
4 0

Answer:

Displacement, x = 0.25 m

Explanation:

It is given that,

Spring constant, k = 10 N/m

Force, F = 2.5 N

We have to find the displacement of a spring. It can be calculated using Hooke's law i.e.

F = - kx

x=\dfrac{F}{k}

x=\dfrac{2.5\ N}{10\ N/m}

x = 0.25 m

Hence, the displacement of the spring is 0.25 m.

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kogti [31]

Answer:

40 meters. look for the dot above the 20 on the x-axis and follow it over to the left.

Explanation:

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3 years ago
5p - 14 = 8 p + 4 find p
Digiron [165]

Answer:

Explanation:

5p - 14 = 8p + 4

5p = 8p + 18 <-- Moving constants to one side; add the same number of +14 to both sides.

-3p = 18. <-- The same thing with the variable itself.

p = -6 <-- Divide both sides by negative 3.

6 0
3 years ago
the maximum range of a projectile is 2÷√3 times its actual range what is the angle of the projection for the actual range​
Murrr4er [49]

Answer:

The actual angle is 30°

Explanation:

<h2>Equation of projectile:</h2><h2>y axis:</h2>

v_y(t)=vo*sin(A)-g*t

the velocity is Zero when the projectile reach in the maximum altitude:

0=vo-gt\\t=\frac{vo}{g}

When the time is vo/g the projectile are in the middle of the range.

<h2>x axis:</h2>

d_x(t)=vo*cos(A)*t\\

R=Range

R=d_x(t=2*\frac{vo}{g})

R=vo*cos(A)*2\frac{vo}{g} \\\\R=\frac{(vo)^{2}*2* sin(A)cos(A)}{g} \\\\R=\frac{(vo)^{2} sin(2A)}{g}

**sin(2A)=2sin(A)cos(A)

<h2>The maximum range occurs when A=45°(because sin(90°)=1)</h2><h2>The actual range R'=(2/√3)R:</h2>

Let B the actual angle of projectile

\frac{vo^{2} }{g} =(\frac{2}{\sqrt{3} }) \frac{vo^{2} *sin(2B)}{g}\\\\1= \frac{2 }{\sqrt{3}} *sin(2B)\\\\sin(2B)=\frac{\sqrt{3}}{2}\\\\

2B=60°

B=30°

7 0
3 years ago
Two loudspeakers emit sound waves along the x-axis. A listener in front of both speakers hears a maximum sound intensity when sp
grandymaker [24]

Answer:

frequency of the sound = f = 1,030.3 Hz

phase difference = Φ = 229.09°

Explanation:

Step 1: Given data:

Xini = 0.540m

Xfin = 0.870m

v = 340m/s

Step 2: frequency of the sound (f)

f = v / λ

λ = Xfin - Xini = 0.870 - 0.540 = 0.33

f = 340 / 0.33

f = 1,030.3 Hz

Step 3: phase difference

phase difference = Φ

Φ = (2π/λ)*(Xini - λ) = (2π/0.33)* (0.540-0.33) = 19.04*0.21 = 3.9984

Φ = 3.9984 rad * (360°/2π rad)

Φ = 229.09°

Hope this helps!

5 0
3 years ago
When a refrigerant enters the compressor, it is a ____ and when it leaves the compressor, it is a ____. A. low pressure low temp
saveliy_v [14]

Answer:

A

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The compressor receive hot refrigerant and raises the pressure and temperature even further as it is send to the condenser.

5 0
3 years ago
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