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3 years ago

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A person walks 9.0 km directly east and then turns left and heads directly north for 12.0 km. What is his displacement from the

starting position? give the magnitude and direction relative to due east. (treat this as a positive x-axis)

1 answer:

Readme [11.4K]3 years ago

5 0

Explanation:

Given that,

A person walks 9.0 km directly east and then turns left and heads directly north for 12.0 km.

We need to find his displacement from the starting position.

We know that,

Displacement = shortest path covered

$D=\sqrt{9^2+12^2} \\\\D=15\ km$

For direction,

$\theta=\tan^{-1}(\dfrac{d_y}{d_x})\\\\\theta=\tan^{-1}(\dfrac{12}{9})\\\\= $$53.13^{\circ}$

Hence, this is the required solution.

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Why is water considered a good solvent?

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Answer:

D: Water is polar, so it can form attractions to and pull apart molecules of many types of substances.

Explanation:

Water is an excellent solvent simply because its polar water molecules usually form hydrogen bonds with ions and polar molecules, thereby allowing the ionic and polar covalent compounds to easily disperse easily in water.

Thus, the correct option is; D

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A rod of length 35.50 cm has linear density (mass per length) given by λ = 50.0 + 23.0x where x is the distance from one end, an

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Answer:

(a)20.65g

(b)0.19m

Explanation:

(a) The total mass would be it's mass per length multiplied by the total lenght

0.355(50 + 23*0.355) = 20.65 g

(b) The center of mass would be at point c where the mass on the left and on the right of c is the same

Hence the mass on the left side would be half of its total mass which is 20.65/2 = 10.32 g

$c(50 + 23c) = 10.32$

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Choose all of the true statements regarding the relationship between voltage, resistance, and current.

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To initiate a nuclear reaction, an experimental nuclear physicist wants to shoot a proton into a 5.50-fm-diameter 12C nucleus. T

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Answer:

$V_1= 3.4*10^7m/s$

Explanation:

From the question we are told that

Nucleus diameter $d=5.50-fm$

a 12C nucleus

Required kinetic energy $K=2.30 MeV$

Generally initial speed of proton must be determined,applying the law of conservation of energy we have

$K_2 +U_2=K_1+U_1$

where

$K_1$ =initial kinetic energy

$K_2$ =final kinetic energy

$U_1$ =initial electric potential

$U_2$ =final electric potential

mathematically

$U_2 = \frac{Kq_pq_c}{r_2}$

where

$r_f$ =distance b/w charges

$q_c$ =nucleus charge $=6(1.6*10^-^1^9C)$

$K$ =constant

$q_p$ =proton charge

Generally kinetic energy is know as

$K=\frac{1}{2} mv^2$

Therefore

$U_2 = \frac{Kq_pq_c}{r_2} + K_2=\frac{1}{2} mv_1^2 +U_1$

Generally equation for radius is $d/2$

Mathematically solving for radius of nucleus

$R=(\frac{5.50}{2}) (\frac{1*10^-^1^5m}{1fm})$

$R=2.75*10^-^1^5m$

Generally we can easily solving mathematically substitute into v_1

$q_p$ $=6(1.6*10^-^1^9C)$

$K_1=9.0*10^9 N-m^2/C^2$

$U_1= 0$

$R=2.75*10^-^1^5m$

$K=2.30 MeV$

$m= 1.67*10^-^2^7kg$

$V_1= (\frac{2}{1.67*10^-^2^7kg})^1^/^2 (\frac{(9.0*10^9 N-m^2/C^2)*(6(1.6*10^-^1^9C)(1.6*10^-^1^9C)}{2.75*10^-^1^5m+2.30 MeV(\frac{1.6*10^-^1^3 J}{1 MeV}) }$

$V_1= 3.4*10^7m/s$

Therefore the proton must be fired out with a speed of $V_1= 3.4*10^7m/s$

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Answer:

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