All categories

2 years ago

Find the magnitude of this vector:

Physics

1 answer:

xeze [42]2 years ago

3 0

Answer: = 25. 5 m

Explanation:

2-D vectors can be solved using the Pythagorean theorem:
$a^{2} +b^{2} =c^{2}$
The hypotenuse, or longest side is not given and that is what must be solved for
Direction can be omitted as the question only asks for magnitude
Magnitude: is the distance or quantity in which an object moves during motion

Solve for C using Pythagorean theorem:

1. $\sqrt{a^{2}+b^{2} }=c$

2. $\sqrt{22.2 m ^{2} +12.6m^{2} } =c$

3. $25.52646 m= c$

round to lowest number of significant digits when dividing and multiplying
rounded to 3 significant digits

∴ Magnitude of the vector is 25.5 m

You might be interested in

A ball is launched from a 300m cliff and lands 380m away from the cliff in 9 seconds. Calculate the initual speed and the angle

Marysya12 [62]

Answer:

Explanation:

Given

Maximum height H = 300m

Range (horizontal distance) = 380m

Required

Initial speed U and the angle of the ball when it was launched.

Range = U√2H/g

380 = U√2(300)/9.8

380 = U√600/9.8

380 = 7.8246U

U = 380/7.8246

U = 48.57m/s

The initial speed is 48.57m/s

b) Using the formula for calculating time of flight;

T = 2Usin theta/g

9 = 2(48.57)sin theta/9.8

9*9.8 = 97.14sin theta

88.2 = 97.14sin theta

88.2/97.14 = sin theta

sin theta = 0.9079

theta = sin^-1(0.9079)

theta = 65.23°

hence the angle when the ball was launched is 65.23°

6 0

3 years ago

Suppose a plane accelerates from rest for 32.3, achieving a takeoff speed of 47.1 m/s after traveling a distance of 607 m down t

V125BC [204]

Answer:

217.28 m/s

Explanation:

u = 0, t 32.3 s, v = 47.1 m/s, s = 607 m

Let a be the acceleration.

Use third equation of motion.

v^2 = u^2 + 2 a s

47.1 x 47.1 = 0 + 2 a x 607

a = 1.83 m/s^2

For small plane

a = 1.83 m/s^2 , v = 28.2 m/s, u = 0, Let teh distance be s.

Use third equation of motion

28.2^2 = 0 + 2 x 1.83 x s

s = 217.28 m/s

6 0

3 years ago

A water storage tank has the shape of a cylinder with diameter 14 ft. It is mounted so that the circular cross-sections are vert

Allisa [31]

Answer:

$\%A_F=77.335\%$

Explanation:

Given:

diameter of tank, $d=14\ ft$

level of the tank filled in its horizontal position, $h=13\ ft$

Now refer the schematic that show water with blue colour.

The triangle ORQ is symmetric about OS as it comes from center O on the cord QR at S.

$RO=QO=7\ ft$ (∵ radius of the cylinder)

$RS=\sqrt{RO^2-OS^2}$

$RS=3.6055\ ft$

Now the area of triangle ORQ:

$A_t=\frac{1}{2}\times QR\times OS$

$A_t=RS\times OS$

$A_t=3.6055\times 6$

$A_t=21.6333\ ft^2$

Now the angle ROS:

$\cos\theta=\frac{OS}{OR}$

$\theta=\cos^{-1}(\frac{6}{7} )$

$\theta=31.0027^{\circ}$

<u>Therefore the reflex angle ROQ:</u>

$\rangle ROQ=360^{\circ}-\theta^{\circ}$

$\rangle ROQ=360^{\circ}-31.0027^{\circ}$

$\rangle ROQ=328.9973^{\circ}$

Now the area of sector ROQPR:

We have the area of full circle, $A=\pi.r^2$

where:

r = radius of the circle

hence for sector:

$A_S=\pi\times 7^2\times \frac{328.9973}{360}$

$A_S=140.6811\ ft^2$

Now the cross sectional area filled with water:

$A_F=A_S+A_t$

$A_F=140.6811-21.6333$

$A_F=119.0478\ ft^2$

Total cross sectional area of tank:

$A=\pi.r^2$

$A=\pi\times 7^2$

$A=153.9380\ ft^2$

Now the percentage of total capacity used:

$\%A_F=\frac{A_F}{A}\times 100\%$

$\%A_F=\frac{119.0478}{153.9380} \times 100$

$\%A_F=77.335\%$

5 0

3 years ago

When a generator rotates a coil of wire in a magnetic field, which of the following is produced?

igomit [66]

Electric current is the answer well it should be i just did this qeustion on my class

4 0

3 years ago

Brandon hits a golf ball with an initial velocity of 30 m/s at an angle of 30 above the horizontal. How long is it in the air?

siniylev [52]

Given :

Brandon hits a golf ball with an initial velocity of 30 m/s at an angle of 30 above the horizontal.

To Find :

How long is it in the air.

Solution :

We know, the formula of time of flight is :

$T = \dfrac{2usin\ \theta}{g}\\\\T = \dfrac{2\times 30\times sin\ 30^o}{9.8}\\\\T = 3.06\ seconds$

Therefore, the ball is in air for 3.06 seconds.

7 0

2 years ago