Find the dimension of volume

Vector C has a magnitude of 24.6 m and points in the − y ‑ direction. Vectors A and B both have positive y ‑ components, and mak

frez [133]

Answer:

A= 61.35

B= -44.40

Explanation:

1. Using the components method we have:

$A_{x}= A cos \alpha\\B_{x}= B cos \beta\\C_{x}= 0\\\\A_{y}= A sin \alpha\\B_{y}= B sin \beta\\C_{y}= 24.6\\$

Considering that the vector sum $A+B+C=0$ , then:

$|V|=\sqrt{V_{x}^{2} +V_{y}^{2} }=0$

Then:

$V_{x} ^{2} =0; V_{x} =0\\V_{y} ^{2} =0; V_{y} =0$

It means the value of x and y component is 0.

2. Determinate the equations that describe each component:

$V_{x}= A cos \alpha -B cos \beta=0 (1)\\V_{y}= A sin \alpha +B sin \beta - C=0 (2)$

Form Eq. (1):

$A=B \frac{cos \beta}{cos \alpha} (3)$

Replacing A in Eq. (2):

$(B\frac{cos \beta}{cos \alpha})(sin \alpha)+ B sin\beta-C=0\\(B\frac{cos \beta}{cos \alpha})(sin \alpha)+ B sin\beta-=C\\\\B(\frac{cos \beta. sin \alpha}{cos \alpha}+ sin\beta)=C\\B=C(\frac{cos \beta. sin \alpha}{cos \alpha}+ sin\beta)^{-1} (4)$

Replacing values of C, α and β in (4):

$B= 24.6 (\frac{(cos 27.7)(sin 44.9)}{cos 44.9}+sin 27.7)^{-1} \\B= -44.4$

Replacing value of B in (3)

$A=-44.40\frac{cos 27.7}{sin 49.9} \\A= 61.35$

5 0

3 years ago

A uniform magnetic field is perpendicular to the plane of a circular loop of diameter 13 cm formed from wire of diameter 2.6 mm

I am Lyosha [343]

Answer:

Rate of change of magnetic field is $3.466\times 10^3T/sec$

Explanation:

We have given diameter of the circular loop is 13 cm = 0.13 m

So radius of the circular loop $r=\frac{0.13}{2}=0.065m$

Length of the circular loop $L=2\pi r=2\times 3.14\times 0.065=0.4082m$

Wire is made up of diameter of 2.6 mm

So radius $r=\frac{2.6}{2}=1.3mm=0.0013m$

Cross sectional area of wire $A=\pi r^2=3.14\times0.0013^2=5.30\times 10^{-6}m^2$

Resistivity of wire $\rho =2.18\times 10^{-8}m$

Resistance of wire $R=\frac{\rho L}{A}=\frac{2.18\times 10^{-8}\times 0.4082}{5.30\times 10^{-6}}=1.67\times 10^{-3}ohm$

Current is given i = 11 A

So emf $e=11\times 1.67\times 10^{-3}=0.0183volt$

Emf induced in the coil is $e=-\frac{d\Phi }{dt}=-A\frac{dB}{dt}$

$0.0183=5.30\times 10^{-6}\times \frac{dB}{dt}$

$\frac{dB}{dt}=3.466\times 10^3=T/sec$

8 0

3 years ago

The tires on your truck have 0.35 m radius. In a straight line, you drive 2600 m. What is the angular displacement of the tire,

Travka [436]

1) In a circular motion, the angular displacement $\theta$ is given by
$\theta = \frac{S}{r}$
where S is the arc length and r is the radius. The problem says that the truck drove for 2600 m, so this corresponds to the total arc length covered by the tire: $S=2600 m$ . Using the information about the radius, $r=0.35 m$ , we find the total angular displacement:
$\theta = \frac{2600 m}{0.35 m} =7428 rad$

2) If we put larger tires, with radius $r=0.60 m$ , the angular displacement will be smaller. We can see this by using the same formula. In fact, this time we have:
$\theta = \frac{2600 m}{0.60 m}=4333 rad$

8 0

3 years ago

A graph of angular position v. time has the following equation:

Harman [31]

Answer:

you can simply answer by derivative = 3.5x^2+25x+250-y=0 you can derivate this eqn 7x +25-1=0 7x=24 yo u can divide you get it

5 0

3 years ago

Write the first equation of motion. Under what condition(s) is this equation valid?

Zepler [3.9K]

Explanation:

The first equation of motion in kinematics is given by :

$v=u+at$ .....(1)

u is initial speed

a is acceleration

v is final speed

t is time

Equation (1) is valid when the object is moving with constant acceleration. This equation gives relation between velocity and time.

3 0

3 years ago

Find the dimension of volume​

Find the dimension of volume