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3 years ago

If the car went 30 km west in 25 min. and then 40 km east in 35 min., what would be its total displacement?

2 answers:

ozzi3 years ago

8 0

Displacement only contain the length of movement and the direction of that movement
So you can pretty much disregard the speed.

30 to the west than 40 to the east

The displacement will be 10 km to the east

hope this helps

Pani-rosa [81]3 years ago

6 0

Answer:

10 km east

Explanation:

Take the displacement towards east is positive and towards west is negative.

d1 = - 30 km, d2 = 40 km

So, net displacement, d = d + d2 = - 30 + 40 = 10 km

It is towards east.

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An aluminum wire with a diameter of 0.115 mm has a uniform electric field of 0.235 V/m imposed along its entire length. The temp

tekilochka [14]

Answer with Explanation:

We are given that

Diameter of coil=d=0.115mm

Radius, r= $\frac{d}{2}=\frac{0.115}{2}=0.0575mm=0.0575\times 10^{-3} m$

Using $1mm=10^{-3} m$

Electric field=E=0.235V/m

T=55 degree C

$T_0=20^{\circ} C$

$\rho_0=2.82\times 10^{-8}\Omega m$

$\alpha=3.9\times 10^{-3}/C$

(a).We know that

$\rho=\rho_0(1+\alpha(T-T_0))$

Substitute the values

$\rho=2.82\times 10^{-8}(1+3.9\times 10^{-3}(55-20))$

$\rho=3.2\times 10^{-8}\Omega m$

(b).Current density, $J=\frac{E}{\rho}$

Using the formula

$J=\frac{0.235}{3.2\times 10^{-8}}=7.3\times 10^6A/m^2$

c.Total current,I=JA

Where $A=\pi r^2$

$\pi=3.14$

Using the formula

$I=7.3\times 10^6\times 3.14\times (0.0575\times 10^{-3})^2$

I=0.076A

d.Length of wire=l=2m

$V=El$

Substitute the values

$V=0.235\times 2=0.47 V$

5 0

3 years ago

A deuteron particle consists of one proton and one neutron and has a mass of 3.34x10^-27 kg. A deuteron particle moving horizont

nikklg [1K]

_dThe radius of curvature of a subatomic particle under a magnetic field is given by the following formula:

$r=\frac{mv}{qB}$

Where:

$\begin{gathered} r=\text{ radius} \\ v=\text{ velocity} \\ q=\text{ charge} \\ B=\text{ magnetic field} \end{gathered}$

We can determine the quotient between the velocity and the charge of the deuteron particle from the formula. First, we divide both sides by the mass:

$\frac{r_d}{m_d}=\frac{v}{q_B_}$

Now, we multiply both sides by the magnetic field "B":

$\frac{Br_d}{m_d}=\frac{v}{q}$

Since the charge of the deuterion is the same as the charge of the proton and the velocity we are considering are the same this means that the quotient between velocity and charge is the same for both particles. Therefore, we can apply the formula for the radius again, this time for the proton: