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tino4ka555 [31]
3 years ago
12

Which types of changes must follow the law of conservation of mass?

Physics
2 answers:
Naddika [18.5K]3 years ago
5 0

Answer:

Both physical and chemical changes

PSYCHO15rus [73]3 years ago
3 0

Answer:

Both physical and chemical changes follow the law since when the system is closed to all transfers of matter and energy, the mass of the system must remain constant over time, irrespective of the state.

Explanation:

pls mark brainliest

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The amplitude of the wave would be 5 meters.
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How to find the total displacement of an object ?
OLga [1]

Answer:

it can be calculated by measuring the final distance away from a point, and then subtracting the initial distance

8 0
3 years ago
A certain spring stores 10.0J of potential energy when it isstretched by 2.00cm from its equilibrium position.How much potential
qaws [65]

Answer:

Answered

Explanation:

x= 0.02 m

E_p= 10.0 J

E_p= 0.5kx^2

10= 0.5k(0.02)^2

solving we get

K= 50.0 N/m

Now

E'_p= 0.5kx'^2

E'_p= 0.5×50×(0.04)^2

E'_p=40 J

b) potential energy is a scalar quantity and it only depends magnitude and not direction so it will remain same in compression and expansion both

c) 20 J = 0.5×50,000×x^2

solving

x= 0.028 m

d) k is 50.0 N/m  from above calculation

3 0
3 years ago
The frequency of a physical pendulum comprising a nonuniform rod of mass 1.15 kg pivoted at one end is observed to be 0.658 Hz.
S_A_V [24]

Answer:

The rotational inertia of the pendulum around its pivot point is 0.280\,kg\cdot m^{2}.

Explanation:

The angular frequency of a physical pendulum is measured by the following expression:

\omega = \sqrt{\frac{m\cdot g \cdot d}{I_{o}} }

Where:

\omega - Angular frequency, measured in radians per second.

m - Mass of the physical pendulum, measured in kilograms.

g - Gravitational constant, measured in meters per square second.

d - Straight line distance between the center of mass and the pivot point of the pendulum, measured in meters.

I_{O} - Moment of inertia with respect to pivot point, measured in kg\cdot m^{2}.

In addition, frequency and angular frequency are both related by the following formula:

\omega =2\pi\cdot f

Where:

f - Frequency, measured in hertz.

If f = 0.658\,hz, then angular frequency of the physical pendulum is:

\omega = 2\pi \cdot (0.658\,hz)

\omega = 4.134\,\frac{rad}{s}

From the formula for the physical pendulum's angular frequency, the moment of inertia is therefore cleared:

\omega^{2} = \frac{m\cdot g \cdot d}{I_{o}}

I_{o} = \frac{m\cdot g \cdot d}{\omega^{2}}

Given that m = 1.15\,kg, g = 9.807\,\frac{m}{s^{2}}, d = 0.425\,m and \omega = 4.134\,\frac{rad}{s}, the moment of inertia associated with the physical pendulum is:

I_{o} = \frac{(1.15\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.425\,m)}{\left(4.134\,\frac{rad}{s} \right)^{2}}

I_{o} = 0.280\,kg\cdot m^{2}

The rotational inertia of the pendulum around its pivot point is 0.280\,kg\cdot m^{2}.

8 0
3 years ago
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