The temperature of water in a beaker is 45°C. What does this measurement represent?

15 points! An atomic nucleus initially moving at 420 m/s emits an alpha particle in the direction of its velocity, and the remai

alexandr1967 [171]

The alpha particle is emitted at 4235 m/s

Explanation:

We can use the law of conservation of momentum to solve the problem: the total momentum of the original nucleus must be equal to the total momentum after the alpha particle has been emitted. Therefore:

$p_i = p_f\\ Mu=m_1 v_1 + m_2 v_2 =$

where:

$M =222u$ is the mass of the original nucleus

$v=420 m/s$ is the initial velocity of the nucleus

$m_1 = 4 u$ is the mass of the alpha particle

$v_1$ is the final velocity of the alpha particle

$m_2 = 222u-4u = 218 u$ is the mass of the daughter nucleus

$v_2 = 350 m/s$ is the final velocity of the nucleus

Solving for $v_1$ , we find the final velocity of the alpha particle:

$v_1 = \frac{Mu-m_2 v_2}{m_1}=\frac{(222)(420)-(218)(350)}{4}=4235 m/s$

Learn more about momentum:

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4 0

3 years ago

Explain how climbing a mountain is similar to hiking from the equator to one of the poles

4vir4ik [10]

Answer:

Explanation:

SRRY God bless

7 0

3 years ago

Read 2 more answers

What is energy that comes from the movement of particles

e-lub [12.9K]

Thermal energy comes from the ,movement of particles which produces heat, the faster the movement is the more heat

5 0

3 years ago

A piano wire with mass 2.60g and length 84.0 cm is stretched with a tension of 25.0 N. A wave with frequency 120.0 Hz and amplit

likoan [24]

Answer:

Power will be 0.2023 watt

And when amplitude is halved then power will be 0.0505 watt

Explanation:

We have given mass of the Piano wire m = 2.60 gram = 0.0026 kg

Length of wire l = 84 cm = 0.84 m

So mass density $\mu =\frac{m}{l}=\frac{0.0026}{0.84}=0.0031kg/m$

Tension in the wire T = 25 N

Frequency f = 120 Hz

So angular frequency $\omega =2\pi f=2\times 3.14\times 120=753.6rad/sec$

And amplitude A = 1.6 mm = 0.0016 m

We have to find the generated power

Power is given by $P=\frac{1}{2}\sqrt{\mu T}\omega ^2A^2=\frac{1}{2}\times \sqrt{0.0031\times 25}\times 753.6^2\times 0.0016^2=0.2023watt$

From the relation we can see that power $P\ \propto\ A^2$

So if amplitude is halved then power will be $\frac{1}{4}$ times

So power will be equal to $\frac{0.2023}{2}=0.0505watt$

4 0

3 years ago

A wire that is 0.65 m long and carrying a current of 8.2 A is at right angles to a uniform magnetic field. The force on the wire

Luda [366]

Answer:

0.075 T

Explanation:

When a current-carrying wire is immersed in a region with magnetic field, the wire experiences a force, given by

$F=ILB sin \theta$

where

I is the current in the wire

L is the length of the wire

B is the strength of the magnetic field

$\theta$ is the angle between the direction of I and B

In this problem we have:

L = 0.65 m is the length of the wire

I = 8.2 A is the current in the wire

F = 0.40 N is the force experienced by the wire

$\theta=90^{\circ}$ since the current is at right angle with the magnetic field

Solving the formula for B, we find the strength of the magnetic field:

$B=\frac{F}{IL sin \theta}=\frac{0.40}{(8.2)(0.65)(sin 90^{\circ})}=0.075 T$

3 0

3 years ago