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4 years ago

5

Two protons move at speed v, one along the x axis and one along the y axis. What is the relative speed of the two protons? Evalu

ate your answer for the special case v = 3/5 c.

1 answer:

forsale [732]4 years ago

8 0

Answer:

0.85 c

Explanation:

When the two travel at right angles to each other , their relative velocity is given by the hypotenuse formed by the x and y direction velocities.

Relative velocity of the two protons is given as

$\sqrt{v^{2}+v^{2}} = (\sqrt{2} ) v$

If v = 3/5 c , then the relative speed = $\sqrt{2}$ (3/5c) =0.85 c

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Which is an IUPAC name for a covalent compound

Elena L [17]

In naming covalent compound (binary) based in IUPAC naming, we have 4 rules to be followed:

1. The first element of the formula will use the normal name of the given element. for example: CO2 ( Carbon Dioxide), Carbon is the element name of the first element of the formula.

2. The second element is named as if they are treated like an anion but put in mind that these are no ions in a covalent compound but we put -ide on the second element as if it is an anion.

3. Prefixes are used to indicate the number of atom of the elements in the compound. for example: mono- 1 atom, di- 2atoms, tri- 3 atoms and etc

4. Prefix "mono"is never used in naming the first element. For example: Carbon dioxide, there should be no monocarbon dioxide.

3 0

3 years ago

Read 2 more answers

An artillery shell is launched on a flat, horizontal field at an angle of α = 40.8° with respect to the horizontal and with an i

Lina20 [59]

Answer:

1317.4 m

Explanation:

We are given that

Angle= $\alpha=40.8^{\circ}$

Initial speed = $v_0=346m/s$

We have to find the horizontal distance covered by the shell after 5.03 s.

Horizontal component of initial speed= $v_{ox}=v_0cos\theta=346cos40.8=261.9m/s$

Vertical component of initial speed= $v_{oy}=346sin40.8=226.1m/s$

Time=t=5.03 s

Horizontal distance = $Horizontal\;velocity\times time$

Using the formula

Horizontal distance= $261.9\times 5.03$

Horizontal distance=1317.4 m

Hence, the horizontal distance covered by the shell=1317.4 m

8 0

3 years ago

How much force is needed to accelerate a 1000Kg car at a rate of 3m/s2?

ikadub [295]

According to Newton (2nd law), Force = (mass) x (acceleration)

Substitute what we know : Force = (1,000 kg) x (3 m/s²)

Do the arithmetic: Force = 3,000 kg-m/s² = 3,000 newtons

6 0

3 years ago

Read 2 more answers

A point charge with charge q1 = 3.40 μC is held stationary at the origin. A second point charge with charge q2 = -4.90 μC moves

expeople1 [14]

Answer:

-0.79 J

Explanation:

We are given that

$q_1=3.4\mu C=3.4\times 10^{-6} C$

$1\mu C=10^{-6} C$

$q_2=-4.9\mu C=-4.9\times 10^{-6} C$

$x_1=0.125,y_1=0$

$x_2=0.280,y_2=0.235$

We have to find the work done by the electric force on the moving point charge.

$r_1=\sqrt{x^2_1+y^2_1}=\sqrt{(0.125)^2+0}=0.125$

$r_2=\sqrt{(0.280)^2+(0.235)^2}=0.366$

Work done, $W=kq_1q_2(\frac{1}{r_1}-\frac{1}{r_2})$

Where $k=9\times 10^9$

Using the formula

$W=9\times 10^9\times 3.4\times 10^{-6}\times(-4.9\times 10^{-6})(\frac{1}{0.125}-\frac{1}{0.366})$

$W=-0.79 J$

5 0

4 years ago

The FM radio band in most places goes from frequencies of about 89 MHz to 106 MHz. How long is the wavelength of the radiation a

ohaa [14]

Answer:

2.83 m

Explanation:

The relationship between frequency and wavelength for an electromagnetic wave is given by

$\lambda=\frac{c}{f}$

where

$\lambda$ is the wavelength

$c=3.0\cdot 10^8 m/s$ is the speed of light

$f$ is the frequency

For the FM radio waves in this problem, we have:

$f_1=89 MHz=89\cdot 10^6 Hz$ is the minimum frequency, so the maximum wavelength is

$\lambda_2=\frac{c}{f_1}=\frac{3\cdot 10^8}{89\cdot 10^6}=3.37 m$

The maximum frequency is instead

$f_2=106 MHz=106\cdot 10^6 Hz$

Therefore, the minimum wavelength is

$\lambda_1=\frac{c}{f_2}=\frac{3\cdot 10^8}{106\cdot 10^6}=2.83 m$

So, the wavelength at the beginning of the range is 2.83 m.

8 0

3 years ago