A wave produced by a doorbell

A particle starts from rest and has an acceleration function 5 − 10t m/s2 . (a) What is the velocity function? (b) What is the p

frez [133]

Explanation:

It is given that,

A particle starts from rest and has an acceleration function as :

$a(t)=(5-10t)\ m/s^2$

(a) Since, $a=\dfrac{dv}{dt}$

v = velocity

$dv=a.dt$

$v=\int(a.dt)$

$v=\int(5-10t)(dt)$

$v=5t-\dfrac{10t^2}{2}=5t-5t^2$

(b) $v=\dfrac{dx}{dt}$

x = position

$x=\int v.dt$

$x=\int (5t-5t^2)dt$

$x=\dfrac{5}{2}t^2-\dfrac{5}{3}t^3$

(c) Velocity function is given by :

$v=5t-5t^2$

$5t-5t^2=0$

t = 1 seconds

So, at t = 1 second the velocity of the particle is zero.

7 0

2 years ago

Which of the following is least likely to be a scientific experiment?

nikitadnepr [17]

D should be the answer

3 0

3 years ago

An electron experiences a magnetic force with a magnitude of 4.90×10−15 n when moving at an angle of 60.0 ∘ with respect to a ma

Leya [2.2K]

Missing questions: "find the speed of the electron".

Solution:
the magnetic force experienced by a charged particle in a magnetic field is given by
$F=qvB \sin \theta$
where
q is the particle charge
v its velocity
B the magnitude of the magnetic field
$\theta$ the angle between the directions of v and B.

Re-arranging the formula, we find:
$v= \frac{F}{qB \sin \theta}$
and by substituting the data of the problem (the charge of the electron is $q=1.6 \cdot 10^{-19} C$ ), we find the velocity of the electron:
$v= \frac{F}{qB \sin \theta}= \frac{4.90 \cdot 10^{-15}N}{(1.6 \cdot 10^{-19}C)(3.70 \cdot 10^{-3} T)(\sin 60^{\circ})}=9.56 \cdot 10^6 m/s$

4 0

3 years ago

1) A thin ring made of uniformly charged insulating material has total charge Q and radius R. The ring is positioned along the x

allochka39001 [22]

Answer:

(A) considering the charge "q" evenly distributed, applying the technique of charge integration for finite charges, you obtain the expression for the potential along any point in the Z-axis:

$V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }$

With $(\epsilon_{0})$ been the vacuum permittivity

(B) The expression for the magnitude of the E(z) electric field along the Z-axis is:

$E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }$

Explanation:

(A) Considering a uniform linear density $λ_{0}$ on the ring, then:

$dQ=\lambda dl$ (1)⇒ $Q=\lambda_{0} 2\pi R$ (2)⇒ $\lambda_{0}=\frac{Q}{2\pi R}$ (3)

Applying the technique of charge integration for finite charges:

$V(z)= 4\pi (ε_{0})\int\limits^a_b {\frac{1}{ r' }} \, dQ$ (4)

Been r' the distance between the charge and the observation point and a, b limits of integration of the charge. In this case a=2π and b=0.

Using cylindrical coordinates, the distance between a point of the Z-axis and a point of a ring with R radius is:

$r'=\sqrt{R^{2} +Z^{2}}$ (5)

Using the expressions (1),(4) and (5) you obtain:

$V(z)= 4\pi (\epsilon_{0})\int\limits^a_b {\frac{\lambda_{0}R}{ \sqrt{R^{2} +Z^{2}} }} \, d\phi$

Integrating results:

$V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }$ (S_a)

(B) For the expression of the magnitude of the field E(z), is important to remember:

$|E| =-\nabla V$ (6)

But in this case you only work in the z variable, soo the expression (6) can be rewritten as:

$|E| =-\frac{dV(z)}{dz}$ (7)

Using expression (7) and (S_a), you get the expression of the magnitude of the field E(z):

$E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }$ (S_b)

4 0

3 years ago

I need help ! if anyone answers this its worth 40 points please help !!

Sliva [168]

Double displacement...I think

3 0

3 years ago