The earth's magnetic field deflects the flow of current from?

Mercury is in the 80th position in the periodic table. How many protons does it have?

Verdich [7]

Mercury has 80 protons. Ironic?

7 0

3 years ago

A 1 kg mass is attached to a spring with spring constant 7 Nt/m. What is the frequency of the simple harmonic motion? What is th

Scorpion4ik [409]

1. 0.42 Hz

The frequency of a simple harmonic motion for a spring is given by:

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$

where

k = 7 N/m is the spring constant

m = 1 kg is the mass attached to the spring

Substituting these numbers into the formula, we find

$f=\frac{1}{2\pi}\sqrt{\frac{7 N/m}{1 kg}}=0.42 Hz$

2. 2.38 s

The period of the harmonic motion is equal to the reciprocal of the frequency:

$T=\frac{1}{f}$

where f = 0.42 Hz is the frequency. Substituting into the formula, we find

$T=\frac{1}{0.42 Hz}=2.38 s$

3. 0.4 m

The amplitude in a simple harmonic motion corresponds to the maximum displacement of the mass-spring system. In this case, the mass is initially displaced by 0.4 m: this means that during its oscillation later, the displacement cannot be larger than this value (otherwise energy conservation would be violated). Therefore, this represents the maximum displacement of the mass-spring system, so it corresponds to the amplitude.

4. 0.19 m

We can solve this part of the problem by using the law of conservation of energy. In fact:

- When the mass is released from equilibrium position, the compression/stretching of the spring is zero: $x=0$ , so the elastic potential energy is zero, and all the mechanical energy of the system is just equal to the kinetic energy of the mass:

$E=K=\frac{1}{2}mv^2$

where m = 1 kg and v = 0.5 m/s is the initial velocity of the mass

- When the spring reaches the maximum compression/stretching (x=A=amplitude), the velocity of the system is zero, so the kinetic energy is zero, and all the mechanical energy is just elastic potential energy:

$E=U=\frac{1}{2}kA^2$

Since the total energy must be conserved, we have:

$\frac{1}{2}mv^2 = \frac{1}{2}kA^2\\A=\sqrt{\frac{m}{k}}v=\sqrt{\frac{1 kg}{7 N/m}}(0.5 m/s)=0.19 m$

5. Amplitude of the motion: 0.44 m

We can use again the law of conservation of energy.

- $E_i = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2$ is the initial mechanical energy of the system, with $x_0=0.4 m$ being the initial displacement of the mass and $v_0=0.5 m/s$ being the initial velocity

$- E_f = \frac{1}{2}kA^2$ is the mechanical energy of the system when x=A (maximum displacement)

Equalizing the two expressions, we can solve to find A, the amplitude:

$\frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2=\frac{1}{2}kA^2\\A=\sqrt{x_0^2+\frac{m}{k}v_0^2}=\sqrt{(0.4 m)^2+\frac{1 kg}{7 N/m}(0.5 m/s)^2}=0.44 m$

6. Maximum velocity: 1.17 m/s

We can use again the law of conservation of energy.

- $E_i = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2$ is the initial mechanical energy of the system, with $x_0=0.4 m$ being the initial displacement of the mass and $v_0=0.5 m/s$ being the initial velocity

$- E_f = \frac{1}{2}mv_{max}^2$ is the mechanical energy of the system when x=0, which is when the system has maximum velocity, $v_{max}$

Equalizing the two expressions, we can solve to find $v_{max}$ , the maximum velocity:

$\frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2=\frac{1}{2}mv_{max}^2\\v_{max}=\sqrt{\frac{k}{m}x_0^2+v_0^2}=\sqrt{\frac{7 N/m}{1 kg}(0.4 m)^2+(0.5 m/s)^2}=1.17 m/s m$

4 0

3 years ago

Read 2 more answers

Which of the following mixture can be separate using decantation method?

vivado [14]

Answer:

Hi sorry for answering here but you didnt put the options there

Explanation:

I'll still try to answer though so maybe the mixture from one of the questions might be something like oil and water which don't mix and can be separated by decantation so something similar would work. Hope this helps

3 0

3 years ago

a 100g ice cube at 0 degrees celsius is placed in 650 grams of water at 25 degrees celsius. When the mixture reaches equillibriu

Artyom0805 [142]

Answer:

The latent heat of fusion of water is 334.88 Joules per gram of water.

Explanation:

Let the latent heat of ice be 'x' J/g

1) Thus heat absorbed by 100 gram of ice to get converted into water equals

$Q_1=100\times x$

2) heat energy required to raise the temperature of water from 0 to 25 degree Celsius equals

$Q_2=100\times 4.186\times 11=4604.6Joules$

Thus total energy needed equals $Q_1+Q_2=100x+4604.6$

3) Heat energy released by the decrease in the temperature of water from 25 to 11 degree Celsius is

$Q_3=650\times 4.186\times (25-11)\\\\Q_{3}=38092.6Joules$

Now by conservation of energy we have

$Q_1+Q_2=Q_3\\\\100x+4604.6=38092.6\\\\\therefore x=\frac{38092.6-4604.6}{100}=334.88J/g$

6 0

3 years ago

As a cold air mass advances on a warm air mass, what usually comes before it?

Marina CMI [18]

Answer: A cold front occurs when a cold air mass advances into a region occupied by a warm air mass. If the boundary between the cold and warm air masses doesn't move, it is called a stationary front.

Explanation: Two types of occluded front exist: the warm-type and the cold-type. They’re distinguished by the relative temperatures of the air mass ahead of the occlusion – in other words, the air mass ahead of the original warm front – and the air mass behind the cold front. If the air behind the cold front is colder than the air ahead of the occlusion, it shoves beneath that air (because it’s denser) to form a cold-type occluded front. If the air behind the cold front is warmer than the air ahead, it rides over it to form a warm-type occluded front – which appears to be the more common case. In either situation, the lighter warm air representing the air mass originally between the warm and cold fronts sits above the boundary between the two cooler air masses.

Hope this helps!!

8 0

3 years ago