Ask question

Ask question

All categories

3 years ago

7

Find the uncertainty in position (in nm) associated with an electron that is moving with a velocity of 572 km/s. The uncertainty

in the velocity is 5.00%. The electron rest mass is 9.11 x 10-31 kg.

1 answer:

Salsk061 [2.6K]3 years ago

7 0

Answer:

c

Explanation:

You might be interested in

Suppose that a 117.5 kg football player running at 6.5 m/s catches a 0.43 kg ball moving at a speed of 26.5 m/s with his feet of

Harrizon [31]

Answer:

<em>a) 6.57 m/s</em>

<em>b) 53.75 J </em>

<em>c) 6.37 m/s</em>

<em>d) -98.297 J</em>

Explanation:

mass of player = $m_{p}$ = 117.5 kg

speed of player = $v_{p}$ = 6.5 m/s

mass of ball = $m_{b}$ = 0.43 kg

velocity of ball = $v_{b}$ = 26.5 m/s

Recall that momentum of a body = mass x velocity = mv

initial momentum of the player = mv = 117.5 x 6.5 = 763.75 kg-m/s

initial momentum of the ball = mv = 0.43 x 26.5 = 11.395 kg-m/s

initial kinetic energy of the player = $\frac{1}{2} mv^{2}$ = $\frac{1}{2}$ x 117.5 x $6.5^{2}$ = 2482.187 J

a) according to conservation of momentum, the initial momentum of the system before collision must equate the final momentum of the system.

for this first case that they travel in the same direction, their momenta carry the same sign

$m_{p}$ $v_{p}$ + $m_{b}$ $v_{b}$ = ( $m_{p}$ + $m_{b}$ )v

where v is the final velocity of the player.

inserting calculated momenta of ball and player from above, we have

763.75 + 11.395 = (117.5 + 0.43)v

775.145 = 117.93v

v = 775.145/117.93 = <em>6.57 m/s</em>

b) the player's new kinetic energy = $\frac{1}{2} mv^{2}$ = $\frac{1}{2}$ x 117.5 x $6.57^{2}$ = 2535.94 J

change in kinetic energy = 2535.94 - 2482.187 = <em>53.75 J gained</em>

c) if they travel in opposite direction, equation becomes

$m_{p}$ $v_{p}$ - $m_{b}$ $v_{b}$ = ( $m_{p}$ + $m_{b}$ )v

763.75 - 11.395 = (117.5 + 0.43)v

752.355 = 117.93v

v = 752.355/117.93 =<em> 6.37 m/s</em>

d) the player's new kinetic energy = $\frac{1}{2} mv^{2}$ = $\frac{1}{2}$ x 117.5 x $6.37^{2}$ = 2383.89 J

change in kinetic energy = 2383.89 - 2482.187 = -98.297 J

that is <em>98.297 J lost</em>

3 0

4 years ago

A projectile is shot vertically upward with a given initial velocity. It reaches a maximum height of 72.0 m. On a second shot, t

exis [7]

Answer:

Explanation:

Given

maximum height reached $h=72\ m$

suppose Projectile is launched vertically with initial velocity u

applying equation of motion

$v^2-u^2=2 as$

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

Here final velocity is zero so

$s=frac{u^2}{2g}$

for twice initial velocity

$s'=\frac{(2u)^2}{2g}$

$s'=4\times \frac{u^2}{2g}$

$s'=4\times 72$

$s'=288\ m$

5 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

How much current will flow through your body if the resistance between your hands is 30,000 ohms and you are connected to 120 vo

Law Incorporation [45]

Resistance of our body is given as

$R = 30,000 ohm$

voltage applied across the body is

$V = 120 V$

now by ohm's law current pass through our body is given by

$i = \frac{V}{R}$

$i = \frac{120}{30,000}[\tex][tex]i = 4 * 10^{-3} A$

So current from our body will be 4 * 10^-3 A

4 0

4 years ago

As an electron approaches a proton, the electron's force of attraction...

Tems11 [23]

Answer: B. Increases

Explanation:

An electron is a negatively charged particle, while a proton is positively charged.

Opposites attract so as it approaches the proton the force of attraction will increase.

7 0

3 years ago