All categories

3 years ago

A hiker walks 15km due east then heads due north for 8km . What is the direction of the resultant vector?

Physics

1 answer:

4vir4ik [10]3 years ago

8 0

his displacements are given as

$d_1 = 15 km$ East

$d_2 = 8 km$ North

now we can find the direction of displacement by using the concept

$\theta = tan^{-1}\frac{\deta y}{\delta x}$

$\theta = tan^{-1}\frac{8}{15}$

$\theta = 28.1 degree$

magnitude of displacement is given as

$d = \sqrt{x^2 + y^2}$

$d = \sqrt{15^2 + 8^2}$

$d= 17 km$

<em>so the displacement is 17 km at angle 28.1 degree North of East</em>

You might be interested in

Help me pls, the lesson is simple machines

Y_Kistochka [10]

Answer:

1.pulley,screw,inclined plane

2.using simple machine properly with right way and precautions

Explanation:

3.cuz if they are not used properly they can cause injuries to organs

8 0

2 years ago

Stereotypes are widely held beliefs that people have certain traits because

Ray Of Light [21]

Answer:

Stereotypes are widely held beliefs that people have certain traits because they belong to a particular group stereotypes are often in accurate and frequently portray the members of less powerful less controlling groups or negatively and members of more powerful more controlling groups.

4 0

3 years ago

An electron that has a velocity with x component 2.4 x 106 m/s and y component 3.6 x 106 m/s moves through a uniform magnetic fi

likoan [24]

Answer:

(a) 7.315 x 10^(-14) N

(b) - 7.315 x 10^(-14) N

Explanation:

As you referred at the final remark, the electron and proton undergo a magnetic force of same magnitude but opposite direction. Using the definition of magnetic force, a cross product must be done. One technique is either calculate the magnitude of the velocity and magnetic field and multiplying by sin (90°), but it is necessary to assure both vectors are perpendicular between each other ( which is not the case) or do directly the cross product dealing with a determinant (which is the most convenient approach), thus,

(a) The electron has a velocity defined as: $\overrightarrow{v}=(2.4x10^{6} i + 3.6x10^{6} j) \frac{[m]}{[s]}\\\\$

In respect to the magnetic field; $\overrightarrow{B}=(0.027 i - 0.15 j) [T]$

The magnetic force can be written as;

$\overrightarrow{F} = q(\overrightarrow{v} x \overrightarrow{B})\\ \\\\\overrightarrow{F}= q \left[\begin{array}{ccc}i&j&k\\2.4x10^{6}&3.6x10^{6}&0\\0.027&-0.15&0\end{array}\right]$

Bear in mind $q =-1.6021x10^{-19} [C]$

thus,

$\overrightarrow{F}= q \left[\begin{array}{ccc}i&j&k\\2.4x10^{6}&3.6x10^{6}&0\\0.027&-0.15&0\end{array}\right]\\\\\\\overrightarrow{F}= q(2.4x10^{6}* (-0.15)- (0.027*3.6x10^{6}))\\\\\\\overrightarrow{F}= -1.6021x10^{-19} [C](-457200) [T]\frac{m}{s}\\\\\overrightarrow{F}=(7.3152x10^{-14}) k [\frac{N*m/s}{C*m/s}]\\\\|F|= \sqrt{ (7.3152x10^{-14})^{2}[N]^2 *k^{2}}\\\\F=7.3152x10^{-14} [N]$

Note: The cross product is operated as a determinant. Likewise, the product of the unit vector k is squared and that is operated as dot product whose value is equal to one, i.e, $k^{2}=k\cdot k = 1$

(b) Considering the proton charge has the same magnitude as electron does, but the sign is positive, thus

$\overrightarrow{F}= q \left[\begin{array}{ccc}i&j&k\\2.4x10^{6}&3.6x10^{6}&0\\0.027&-0.15&0\end{array}\right]\\\\\\\overrightarrow{F}= q(2.4x10^{6}* (-0.15)- (0.027*3.6x10^{6}))\\\\\\\overrightarrow{F}= 1.6021x10^{-19} [C](-457200) [T]\frac{m}{s}\\\\\overrightarrow{F}=(-7.3152x10^{-14}) k [\frac{N*m/s}{C*m/s}]\\\\|F|= \sqrt{ (-7.3152x10^{-14})^{2}[N]^2 *k^{2}}\\\\F=-7.3152x10^{-14} [N]$

Note: The cross product is operated as a determinant. Likewise, the product of the unit vector k is squared and that is operated as dot product whose value is equal to one, i.e, $k^{2}=k\cdot k = 1$

Final remarks: The cross product was performed in R3 due to the geometrical conditions of the problem.

6 0

3 years ago

HELP! how did michael jordan influence future basketball players to make prodcuts?

nataly862011 [7]

Answer:

By encouraging them.

Explanation:

#CARRYONLEARNING

5 0

2 years ago

Read 2 more answers

The first and second coils have the same length, and the third and fourth coils have the same length. They differ only in the cr

stealth61 [152]

Answer:

$\frac{R_2}{R_1}=\frac{A_1}{A_2}\\\frac{R_4}{R_3}=\frac{A_3}{A_4}$

Explanation:

The resistance of a conductor is directly proportional to its length and is inversely proportional to its cross-sectional area, this dependence is given by:

$R=\frac{\rho L}{A}$

$\rho$ is the material's resistance, L is the legth and A is the cross-sectional area.

For the first and second coils, we have:

$R_1=\frac{\rho L}{A_1}\\R_2=\frac{\rho L}{A_2}\\\rho L=R_1A_1\\\rho L=R_2A_2\\R_1A_1=R_2A_2\\\frac{R_2}{R_1}=\frac{A_1}{A_2}$

For the third and fourth coils, we have:

$R_3=\frac{\rho L'}{A_3}\\R_4=\frac{\rho L'}{A_4}\\\rho L'=R_3A_3\\\rho L'=R_4A_4\\R_3A_3=R_4A_4\\\frac{R_4}{R_3}=\frac{A_3}{A_4}$

6 0

3 years ago