How to write an advertisement about child labour

Viewers of Star Trek hear of an antimatter drive on the Starship Enterprise. One possibility for such a futuristic energy

Dominik [7]

a) 0.261 T

b) this field strength is obtainable with today's technology

Explanation:

a)

The force experienced by a charged particle moving perpendicular to a magnetic field is given by

$F=qvB$

where

q is the charge

v is the velocity

B is the strength of the field

This force is perpendicular to the motion of the particle, which therefore moves in a circular path; and so, this force acts as centripetal force, so we can write:

$qvB=m\frac{v^2}{r}$

where

m is the mass of the particle

r is the radius of the circle

In this problem, we have:

$q=1.6\cdot 10^{-19}C$ (magnitude of the charge of antiprotons)

$v=5.00 \cdot 10^7 m/s$ (velocity)

$m=1.67\cdot 10^{-27}kg$ (mass of antiprotons)

r = 2.00 m (radius)

Therefore, we can re-arrange the equation and solve to find B, the magnetic field strength:

$B=\frac{mv}{qr}=\frac{(1.67\cdot 10^{-27})(5.00\cdot 10^7)}{(1.6\cdot 10^{-19})(2.00)}=0.261 T$

B)

The strength of the magnetic field calculated in part A) is

$B=0.261 T$

This is indeed a very strong magnetic field. In fact, by comparison, the Earth's magnetic field has a strength of about

$B_{earth}=5\cdot 10^{-5} T$

However, there are current technologies available that are able to produce such strong fields. For instance, the superconducting magnets in the LHC (Large Hadron Collider) are able to produce magnetic fields of strength up to 8 Tesla (8 T).

Therefore, we can say that this field strength is obtainable with today's technology.

5 0

3 years ago

For the population, scores on the test are normally distributed with μ = 70 and σ = 15. The sample of n = 25 students had a mean

AysviL [449]

Answer:

$z=\frac{75-70}{\frac{15}{\sqrt{25}}}=1.67$

$p_v =2*P(t_{24}>1.67)=0.108$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis, and the the actual mean is not significant different from 70.

Explanation:

1) Data given and notation

$\bar X=75$ represent the sample mean

$\sigma=15$ represent the standard deviation for the population

$n=25$ sample size

$\mu_o =70$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to determine if the mean is different from 70, the system of hypothesis would be:

Null hypothesis: $\mu = 70$

Alternative hypothesis: $\mu \neq 70$

We know the population deviation, so for this case is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{75-70}{\frac{15}{\sqrt{25}}}=1.67$

Calculate the P-value

First we need to calculate the degrees of freedom given by:

$df=n-1=25-1=24$

Since is a two tailed test the p value would be:

$p_v =2*P(t_{24}>1.67)=0.108$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis, and the the actual mean is not significant different from 70.

3 0

4 years ago

The magnetic field lines outside a bar magnet

n200080 [17]

The lines of magnetic field from a bar magnet form closed lines. By convention, the field direction is taken to be outward from the North pole and in to the South pole of the magnet.

8 0

3 years ago

An aluminum wire having a cross-sectional area equal to 2.20 10-6 m2 carries a current of 4.50 A. The density of aluminum is 2.7

Kazeer [188]

Answer:

The drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

Explanation:

We can find the drift speed by using the following equation:

$v = \frac{I}{nqA}$

Where:

I: is the current = 4.50 A

n: is the number of electrons

q: is the modulus of the electron's charge = 1.6x10⁻¹⁹ C

A: is the cross-sectional area = 2.20x10⁻⁶ m²

We need to find the number of electrons:

$n = \frac{6.022\cdot 10^{23} atoms}{1 mol}*\frac{1 mol}{26.982 g}*\frac{2.70 g}{1 cm^{3}}*\frac{(100 cm)^{3}}{1 m^{3}} = 6.03 \cdot 10^{28} atom/m^{3}$

Now, we can find the drift speed:

$v = \frac{I}{nqA} = \frac{4.50 A}{6.03 \cdot 10^{28} atom/m^{3}*1.6 \cdot 10^{-19} C*2.20 \cdot 10^{-6} m^{2}} = 2.12 \cdot 10^{-4} m/s$

Therefore, the drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

I hope it helps you!

4 0

3 years ago

A rock is thrown upward from level ground in such a way that the maximumheight of its flight is equal to its horizontal rangeR.

Ne4ueva [31]

The range of a projectile motion is given by:

$\frac{u_o^2 sin2\theta}{g}$

where, $u$ is the initial speed of the projectile, $\theta$ is the angle of the projectile and $g$ is the acceleration due to gravity.

The maximum height reached is given by:

$\frac{u_o^2 sin^2\theta}{2g}$

Part a

It is given that the maximum height reached is equal to the horizontal range. we need to find the angle of the projectile.

Equating the two:

$\frac{u_o^2 sin2\theta}{g}=\frac{u_o^2 sin^2\theta}{2g}\\ \Rightarrow 2 sin2\theta =sin^2\theta\\ \Rightarrow 2\times 2 sin\theta cos\theta=sin^2 \theta\\ \Rightarrow tan\theta =4\\ \Rightarrow \theta=tan^{-1}4=75.96^o$

Hence, the projectile was thrown at an initial angle of $\theta=75.96^o$ .

Part b

we need to find the angle for which range would be maximum and then write this maximum range in terms of original range.

So, we know that range is given by:

$R= \frac{u_o^2sin2\theta}{g}$

It would be maximum when $sin2\theta=1\Rightarrow 2\theta=90^o\Rightarrow \theta=45^o$

Hence, $R_{max}=\frac{u_o^2}{g}$

Original range,

$R=\frac{u_o^2sin2\times75.96^o}{g}\\ \Rightarrow R=\frac{u_o^2}{g}\times 0.47\\ \Rightarrow R=R_{max}0.47\\ \Rightarrow R_{max}=2.125R$

Part c:

In the part a, we know that the angle of the projectile is independent of the $g$ i.e. the acceleration due to gravity and this is the only factor that varies with the different planets. Hence, the answer would remain same.

7 0

3 years ago