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Naddik [55]
3 years ago
10

A mass is oscillating up and down on a spring. In the above graph of

Physics
1 answer:
melomori [17]3 years ago
5 0

Answer:

<em>Amplitude= 8 m</em>

Explanation:

<u>The Amplitude of a Wave</u>

Sinusoidal Function  refers to a mathematical curve with a smooth and periodic oscillation. Its name comes from the sine function and is characterized by the amplitude or the maximum displacement or distance moved by a point on a vibrating body measured from its equilibrium position.

To calculate the amplitude from a graph, we measure the maximum point and the minimum point the wave reaches. Then we subtract both values and divide the result by 2.

The shown wave in the figure has a maximum value of 8 m and a minimum value of -8 m. The distance from the maximum to the minimum is 8-(-8)= 16 m, thus the amplitude is 16/2= 8m.

Amplitude= 8 m

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3 years ago
Numerical Problems
Evgesh-ka [11]

Answer:

a = 5 [m/s²]

Explanation:

To solve this problem we must use the following equation of kinematics.

v_{f}=v_{o}+a*t

where:

Vf = final velocity = 20 [m/s]

Vo = initial velocity = 10 [m/s]

t = time = 2 [s]

a = acceleration [m/s²]

Now replacing:

20 =10 +a*2\\10=2*a\\a=5[m/s^{2} ]

4 0
3 years ago
An open container holds ice of mass 0.555 kg at a temperature of -16.6 ∘C . The mass of the container can be ignored. Heat is su
s2008m [1.1K]

Answer: A. 23.59 minutes.

              B. 249.65 minutes

Explanation: This question involves the concept of Latent Heat and specific heat capacities of water in solid phase.

<em>Latent heat </em><em>of fusion </em>is the total amount of heat rejected from the unit mass of water at 0 degree Celsius to convert completely into ice of 0 degree Celsius (and the heat required for vice-versa process).

<em>Specific heat capacity</em> of a substance is the amount of heat required by the unit mass of a substance to raise its temperature by 1 kelvin.

Here, <u>given that</u>:

  • mass of ice, m= 0.555 kg
  • temperature of ice, T= -16.6°C
  • rate of heat transfer, q=820 J.min^{-1}
  • specific heat of ice, c_{i}= 2100 J.kg^{-1}.K^{-1}
  • latent heat of fusion of ice, L_{i}=334\times10^{3}J.kg^{-1}

<u>Asked:</u>

1. Time require for the ice to start melting.

2. Time required to raise the temperature above freezing point.

Sol.: 1.

<u>We have the formula:</u>

Q=mc\Delta T

Using above equation we find the total heat required to bring the ice from -16.6°C to 0°C.

Q= 0.555\times2100\times16.6

Q= 19347.3 J

Now, we require 19347.3 joules of heat to bring the ice to 0°C  and then on further addition of heat it starts melting.

∴The time required before the ice starts to melt is the time required to bring the ice to 0°C.

t=\frac{Q}{q}

=\frac{19347.3}{820}

= 23.59 minutes.

Sol.: 2.

Next we need to find the time it takes before the temperature rises above freezing from the time when heating begins.

<em>Now comes the concept of Latent  heat into the play, the temperature does not starts rising for the ice as soon as it reaches at 0°C it takes significant amount of time to raise the temperature because the heat energy is being used to convert the phase of the water molecules from solid to liquid.</em>

From the above solution we have concluded that 23.59 minutes is required for the given ice to come to 0°C, now we need some extra amount of energy to convert this ice to liquid water of 0°C.

<u>We have the equation:</u> latent heat, Q_{L}= mL_{i}

Q_{L}= 0.555\times334\times10^{3}= 185370 J

<u>Now  the time required for supply of 185370 J:</u>

t=\frac{Q_{L}}{q}

t=\frac{185370}{820}

t= 226.06 minutes

∴ The time it takes before the temperature rises above freezing from the time when heating begins= 226.06 + 23.59

= 249.65 minutes

8 0
3 years ago
Find the magnitude of the gravitational force a 63.5 kg person would experience while standing on the surface of Earth with a ma
Anastaziya [24]

Answer: 3976N

Explanation:

Using the formula for calculating gravitational force between two masses, we have

F = GMm/r^2

Where G is the gravitational constant

M and m are the masses

r is the distance between the masses

F= 6.673 × 10-¹¹ × 5.98 × 10²⁴ × 63.5/ (6.37 × 10^6)^2

F= 2.533×10^16/6.37×10^12

F= 0.3976×10⁴N

F= 3976N

3 0
3 years ago
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