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vivado [14]
3 years ago
15

H=ac+mn solve for m

Physics
1 answer:
Vinvika [58]3 years ago
7 0

the answer is m= H/n - ac/n

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How do i find stretch? The problem in questioning has already given me the elastic energy and k-value, but I have no idea how to
finlep [7]

Answer:

Stretch can be obtained using the Elastic potential energy formula.

The expression to find the stretch (x) is x=\sqrt{\frac{2\times EPE}{k}}

Explanation:

Given:

Elastic potential energy (EPE) of the spring mass system and the spring constant (k) are given.

To find: Elongation in the spring (x).

We can find the elongation or stretch of the spring using the formula for Elastic Potential Energy (EPE).

The formula to find EPE is given as:

EPE=\frac{1}{2}kx^2

Rewriting the above expression in terms of 'x', we get:

x=\sqrt{\frac{2\times EPE}{k}}

Example:

If EPE = 100 J and spring constant, k = 2 N/m.

Elongation or stretch is given as:

x=\sqrt{\frac{2\times EPE}{k}}\\\\x=\sqrt{\frac{2\times 100}{2}}\\\\x=\sqrt{100}=10\ m

Therefore, the stretch in the spring is 10 m.

So, stretch in the spring can be calculated using the formula for Elastic Potential Energy.

6 0
3 years ago
A damped harmonic oscillator consists of a mass on a spring, with a small damping force that is proportional to the speed of the
exis [7]

Answer:

2.19 N/m

Explanation:

A damped harmonic oscillator is formed by a mass in the spring, and it does a harmonic simple movement. The period of it is the time that it does one cycle, and it can be calculated by:

T = 2π√(m/K)

Where T is the period, m is the mass (in kg), and K is the damping constant. So:

2.4 = 2π√(0.320/K)

√(0.320/K) = 2.4/2π

√(0.320/K) = 0.38197

(√(0.320/K))² = (0.38197)²

0.320/K = 0.1459

K = 2.19 N/m

4 0
3 years ago
WRONG ANSWERS WILL BE REPORTED
denpristay [2]

Answer:

1 = C

2 = B

Explanation:

6 0
3 years ago
Which of the following is most likely not a case of uniform circular motion
Ivan

Answer:

Motion of a racing car on a circular track

Explanation:

Uniform circular motion means the motion of the object is in a circle with a CONSTANT SPEED.

The racing car will accelerate during its motion. Hence, it is not a uniform circular motion.

7 0
3 years ago
A Pitot-static tube is mounted on a 2.5 cm pipe where oil (???? = 860 kg/m3, ???? = 0.0103 kg/m·s) is flowing. The Pitot tube is
emmasim [6.3K]

Answer:

\dot{V}=0.0733 \,m^3.s^{-1}

Explanation:

Given:

density, \rho=860\,kg.m^{-3}

diameter of the pipe, d=2.5\times 10^{-2}m

pressure difference, \Delta P=95.8\times 10^{5}\,Pa

In case of  pitot tube, the velocity is given by:

v=\sqrt{\frac{2.\Delta P}{\rho} }

v=\sqrt{\frac{2\times 95.8\times 10^{5}}{860} }

v=149.26\,m.s^{-1}

Now we know that volumetric flow rate is given as:

\dot{V}=a.v

where :

a= cross sectional area of the pipe

v= velocity of flow

\dot{V}=(\pi\times \frac{(2.5\times 10^{-2})^2}{4} ) \times 149.26

\dot{V}=0.0733 \,m^3.s^{-1}

4 0
3 years ago
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