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Nutka1998 [239]
3 years ago
11

A spring stretches by 15cm when a mass of 300g hangs down from it,if the spring is then stretched an additional 10cm and release

d, calculate;the spring constant,the angular velocity, amplitude of oscillation, maximum velocity, maximum acceleration of the mass,period, frequency​
Physics
1 answer:
bearhunter [10]3 years ago
4 0

Answer:

0.1 m

Explanation:

It is given that,

Mass of the object, m = 350 g = 0.35 kg

Spring constant of the spring, k = 5.2 N/m

Amplitude of the oscillation, A = 10 cm = 0.1 m

Frequency of a spring mass system is given by :

Time period:

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a positively charged body makes contact with a body. after a while, the charged body becomes neutralised. state 3 main condition
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Answer:

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5 0
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An automobile starter motor has an equivalent resistance of 0.055 Ï and is supplied by a 12.0 v battery which has a 0.0305 Ï int
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12 V is the f.e.m. \epsilon of the battery. The potential difference that is applied to the motor is actually the fem minus the voltage drop on the internal resistance r:
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this is equal to the voltage drop on the resistance of the motor R:
RI
so we can write:
\epsilon - Ir = RI
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8 0
2 years ago
Plane a travels at 900km/h and plane b travels at 250/5.which plane travels faster
vesna_86 [32]

Explanation:

We have,

Speed of plane a is 900 km/h

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5 0
2 years ago
Why is understanding the concept of forces, friction and gravity important?
nydimaria [60]
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8 0
3 years ago
A 10.0-kg box starts at rest on a level floor. An external, horizontal force of 2.00 × 102 N is applied to the box for a distanc
Harman [31]

Answer:

vf = 11.2 m/s

Explanation:

m = 10 Kg

F = 2*10² N

x = 4.00 m

μ = 0.44

vi = 0 m/s

vf = ?

We can apply Newton's 2nd Law

∑ Fx = m*a   (→)

F - Ffriction = m*a  ⇒  F - (μ*N) = F - (μ*m*g) = m*a   ⇒  a = (F - μ*m*g)/m

⇒    a = (2*10² N - 0.44*10 Kg*9.81 m/s²)/10 Kg = 15.6836 m/s²

then , we use the equation

vf² = vi² + 2*a*x    ⇒    vf = √(vi² + 2*a*x)

⇒   vf = √((0)² + 2*(15.6836 m/s²)*(4.00m)) = 11.2 m/s

7 0
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