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Alex_Xolod [135]

3 years ago

10

Alfredo leaves camp and, using a compass, walks 4 km E, then 6 km S, 3 km E, 5 km N, 10 km W, 8 km N, and, finally, 3 km S. At t

he end of two days, he is planning his trip back. To get back to camp, Alfredo should travel a distance of ----------?
km in a direction
---------------? ° south of east.

1 answer:

kari74 [83]3 years ago

8 0

Answer: 3km E, 1km N

Explanation:

4E + 6S + 3E + 5N + 10W + 8N + 3S= 3W, 1S

3W, 1S= 3E, 1N

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RideAnS [48]

the answer is (b) 0.6

3 0

3 years ago

Alpha particles, each having a charge of +2e and a mass of 6.64 ×10-27 kg, are accelerated in a uniform 0.50 T magnetic field to

sergij07 [2.7K]

Answer:

$KE=1.2036\times 10^{-12}\ J$

Explanation:

Given:

charge on the alpha particle, $q=2e=3.2\times 10^{-19}\ C$

mass of the alpha particle, $m=6.64\times 10^{-27}\ kg$

strength of a uniform magnetic field, $B=0.5\ T$

radius of the final orbit, $r=0.5\ m$

<u>During the motion of a charge the magnetic force and the centripetal forces are balanced:</u>

$q.v.B=m.\frac{v^2}{r}$

$m.v=q.B.r$

where:

v = velocity of the alpha particle

$v=\frac{q.B.r}{m}$

$v=\frac{3.2\times 10^{-19}\times 0.5\times 0.5}{6.64\times 10^{-27}}$

$v=1.2048\times 10^{7}\ m.s^{-1}$

Here we observe that the velocity of the aprticle is close to the velocity of light. So the kinetic energy will be relativistic.

<u>We firstly find the relativistic mass as:</u>

$m'=\frac{1}{\sqrt{1-\frac{v^2}{c^2} } } \times m$

$m'=\frac{6.64\times 10^{-27}}{\sqrt{1-\frac{(1.2048\times 10^7)^2}{(3\times 10^8)^2} } }$

$m'=6.6533\times10^{-27}\ kg$

now kinetic energy:

$KE=m'.c-m.c$

$KE=6.6533\times 10^{-27}\times (3\times 10^8)^2-6.64\times 10^{-27}\times (3\times 10^8)^2$

$KE=1.2036\times 10^{-12}\ J$

6 0

3 years ago

Which of the following choices correctly identifies the process created by moving a magnet through a conducting loop?

jolli1 [7]

Answer: D.) electromagnetic induction

Explanation: Electroctromagnetic induction may be explained as a process whereby electric current is induced or produced by difference in potential resulting from the movement of conductor across a magnetic field.

In simple terms, an electromotive force is induced when a magnet is moved through a conducting loop.

The electromotive force produced by moving a magnet through a conducting loop can be represented by the relation:

E = - N (dΦ / dt)

Where E = electromotive force in voltage

N = number of loop in conductor

dΦ = change in magnetic Flux

dt = change in time

3 0

4 years ago

A long jumper can jump a distance of 7.4 m when he takes off at an angle of 45° with respect to the horizontal. Assuming he can

GenaCL600 [577]

Answer:

0.02 m

Explanation:

R₁ = initial distance jumped by jumper = 7.4 m

R₂ = final distance jumped by jumper = ?

θ₁ = initial angle of jump = 45°

θ₂ = final angle of jump = 42.9°

$v$ = speed at which jumper jumps at all time

initial distance jumped is given as

$R_{1}=\frac{v^{2}Sin2\theta _{1} }{g}$

final distance jumped is given as

$R_{2}=\frac{v^{2}Sin2\theta _{2} }{g}$

Dividing final distance by initial distance

$\frac{R_{2}}{R_{1}}=\frac{Sin2\theta _{1}}{Sin2\theta _{2}}$

$\frac{R_{2}}{7.4}=\frac{Sin2(42.9)}{Sin2(45))}$

$R_{2} =7.38$

distance lost is given as

d = $R_{1} - R_{2}$

d = 7.4 - 7.38

d = 0.02 m

8 0

3 years ago

A stone is thrown downward with a speed of 8 m/s from a height of 14 m. (acceleration due to gravity: 10 m/s2) What is the speed

max2010maxim [7]

Answer:

The speed of the stone just before it hits the ground is 18.54 m/s

Explanation:

Given that,

Initial speed of the stone, u = 8 m/s

The stone is thrown downward from a height of 14 m

We need to find the speed of the stone just before it hits the ground. It can be calculated using third equation of motion as :

$v^2-u^2=2ah$

v is the speed of the stone just before it hits the ground

$v^2=2ah+u^2$

$v^2=2\times 10\times 14+(8)^2$

v = 18.54 m/s

So, the speed of the stone just before it hits the ground is 18.54 m/s. Hence, this is the required solution.

6 0

3 years ago