Ask question

Ask question

All categories

2 years ago

6

the two vectors and in Fig. 3-28 have equal magnitudes of 10.0 m and the angles are 30° and 105°. Find the (a) x and (b) y compo

nents of their vector sum r , (c) the magnitude of r , and (d) the angle r makes with the positive direction of the x axis.

1 answer:

leva [86]2 years ago

6 0

Answer:

Explanation:

For rectangular components:

Ax = Acostheta

= 10 x cos30

= 8.66

Ay = Asintheta

= 10 x sin30

=5.

Bx = Bcostheta

= 10 x cos105

= -2.58

By = Bsin105

= 10 x sin105

= 9.65

For magnitude of R:

A = Ax + Ay

= 8.66 + 5

= 13.66

B = Bx + by

= -2.58 + 9.65

= 7.07

R = sq A*2 + B*2

= 15.38

You might be interested in

What is the equivalent resistance of the

BigorU [14]

Answer:

Approximately $111\; {\rm \Omega}$ .

Explanation:

It is given that $R_{1} = 200\; {\Omega}$ and $R_{2} = 250\; {\Omega}$ are connected in a circuit in parallel.

Assume that this circuit is powered with a direct current power supply of voltage $V$ .

Since $R_{1}$ and $R_{2}$ are connected in parallel, the voltage across the two resistors would both be $V$ . Thus, the current going through the two resistors would be $(V / R_{1})$ and $(V / R_{2})$ , respectively.

Also because the two resistors are connected in parallel, the total current in this circuit would be the sum of the current in each resistor: $I = (V / R_{1}) + (V / R_{2})$ .

In other words, if the voltage across this circuit is $V$ , the total current in this circuit would be $I = (V / R_{1}) + (V / R_{2})$ . The (equivalent) resistance $R$ of this circuit would be:

$\begin{aligned} R &= \frac{V}{I} \\ &= \frac{V}{(V / R_{1}) + (V / R_{2})} \\ &= \frac{1}{(1/R_{1}) + (1 / R_{2})}\end{aligned}$ .

Given that $R_{1} = 200\; {\Omega}$ and $R_{2} = 250\; {\Omega}$ :

$\begin{aligned} R &= \frac{1}{(1/R_{1}) + (1 / R_{2})} \\ &= \frac{1}{(1/(200\: {\rm \Omega})) + (1/(250\; {\rm \Omega}))} \\ &\approx 111\; {\rm \Omega}\end{aligned}$ .

7 0

1 year ago

A charged particle is surrounded by an electric field and a magnetic field

zlopas [31]

Answer:

yeah

Explanation:

electric fields help the charged particles interact

and isn't magnetic field the same as electric field.

7 0

2 years ago

Read 2 more answers

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el

Tresset [83]

Complete Question

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

I = 1.2 A at time 5 secs.

Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.

Answer:

The charge is $Q =2.094 C$

Explanation:

From the question we are told that

The diameter of the wire is $d = 0.205cm = 0.00205 \ m$

The radius of the wire is $r = \frac{0.00205}{2} = 0.001025 \ m$

The resistivity of aluminum is $2.75*10^{-8} \ ohm-meters.$

The electric field change is mathematically defied as

$E (t) = 0.0004t^2 - 0.0001 +0.0004$

Generally the charge is mathematically represented as

$Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt$

Where A is the area which is mathematically represented as

$A = \pi r^2 = (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2$

So

$\frac{A}{\rho} = \frac{3.3 *10^{-6}}{2.75 *10^{-8}} = 120.03 \ m / \Omega$

Therefore

$Q = 120 \int\limits^{t}_{0} { E(t) } \, dt$

substituting values

$Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt$

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.$

From the question we are told that t = 5 sec

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.$

$Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }$

$Q =2.094 C$

5 0

3 years ago

If earth had no atmosphere, would a falling object ever reach terminal velocity?

stepan [7]

No, because terminal velocity is when the acceleration of the Earth’s gravity is balanced by the air resistance of the atmosphere.

5 0

2 years ago

What is the height of a building that an object is dropped from if it has a mass of 3 kg and hits the ground with a velocity of

Marizza181 [45]

Answer:

125 m

Explanation:

m = 3kg

v = 50m/s

u = 0m/s

a = +g = 10m/s²

s = H = ?

using the formula,

v² = u² + 2as

50² = 0² + 2(10)(H)

2500 = 20H

H = 2500/20

H = 125m

8 0

1 year ago