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3 years ago

A woman of mass 57 kg jumps off the bow of

Physics

1 answer:

dmitriy555 [2]3 years ago

5 0

Here we will use the momentum conservation

Since there is no external force on this system so momentum of woman + canoe system will always remains conserved

So here initial momentum of the system is zero as they are at rest so final momentum will always be zero

$m_1 v_1 + m_2v_2 = 0$

$57(1.5) + 42(v) = 0$

$85.5 + 42v = 0$

$v = -\frac{85.5}{42}$

$v = - 2.03 m/s$

so the canoe will go back with speed 2.03 m/s

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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.

$- E_f = \frac{1}{2}mv_{max}^2$ 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:

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

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3 years ago

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When an astronomer sees certain stars and galaxies that look much redder than expected, what conclusion might the astronomer dra

Ratling [72]

It all comes to the doppler effect, the red shift means that the galaxy is moving away from us. The redshift is a result from the doppler effect, so as the galaxy moves away the wavelength expands, increasing the wavelength which responds to the red light.

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4 years ago

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A uniform meterstick of mass 0.20 kg is pivoted at the 40 cm mark. where should one hang a mass of 0.50 kg to balance the stick?

Tcecarenko [31]

The weight of the meterstick is:
$W=mg=0.20 kg \cdot 9.81 m/s^2 = 1.97 N$
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
$d_1 = 0.50 m - 0.40 m=0.10 m$
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
$M_w = W d_1 = (1.97 N)(0.10 m)=0.20 Nm$

To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
$(mg) d_2 = 0.20 Nm$
from which we find the value of d2:
$d_2 = \frac{0.20 Nm}{mg}= \frac{0.20 Nm}{(0.5 kg)(9.81 m/s^2)}=0.04 m$

So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.

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4 years ago

It may seem dangerous to be in a car during a thunderstorm, but it's actually relatively safe. Since the car is essentially a me

balu736 [363]

Answer:

The charges from the thunderstorm flow through the conductive metal

of which the vehicle is made and distribute themselves on the outside surface of the vehicle

Explanation:

It is actually safer to stay inside a car during a thunderstorm rather than standing outside the car. The reason is this, thunder passes electrical charges through a conductor. The body of the vehicle is made of a metal which is a good conductor of electricity. The charges will redistribute themselves on the body of the vehicle (a metallic conductor of electricity) hence the occupants of the car are relatively safe.

The reasons described above makes those inside the vehicle relatively safe compared to a person standing outside.

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If a bean of mass 2.0g jumps 1.0cm from your hand into the air, how much potential energy has it gained in reaching its highest

kirill [66]

$PE = mg\Delta h = 0.002 \, kg \cdot 9.8 \, m/s^2 \cdot 0.01 \, m = ~2 \cdot 10^{-4} \, J$

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3 years ago