a 1250 kg car accelerates from rest to 6.13m/s over a distance of 8.58m calculate the average force of traction

A point charge q1=−4.00nc is at the point x=0.600 meters, y=0.800 meters, and a second point charge q2=+6.00nc is at the point x

Katen [24]

electric field between mid point of two charges is given by

$E = E_1 + E_2$

here we have

$E_1 = \frac{kq_1}{r^2}$

$E_1 = \frac{9*10^9 * 4 * 10^{-9}}{0.4^2}$

$E_1 = 225 N/c$

now similarly for other charge

$E_2 = \frac{kq_2}{r^2}$

$E_2 = \frac{9*10^9 * 6 * 10^{-9}}{0.4^2}$

$E_2 = 337.5 N/C$

so the net electric field will be given as

$E = 225 + 337.5$

$E = 565.5 N/C$

8 0

2 years ago

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The point on the graph that lies on the y-axis (vertical axis) is called the y-intercept. What does the y-intercept tell you abo

jekas [21]

Answer:

The starting position of the runner.

Explanation:

When you look at the graph, you can see that the first point on the graph is twenty on the y-axis.

The runner starts at twenty, and ends at thirty.

Therefore, the runner starts at twenty on the y-axis, so it's the starting position of the runner.

7 0

2 years ago

According to the diagram, in order for a solar eclipse to occur, the Earth, Moon, and Sun must A) form a right angle with the Mo

torisob [31]

Answer:

B) form a straight line with the Moon in the middle.

Explanation:

For the occurrence of a solar eclipse the earth and the moon and the sun must be in a straight line and moon should be in center of the earth so that it completely blocks the rays of the sun and the shadow falls on earth and the sun appears to form a ring and thus the eclipse takes place.

7 0

2 years ago

Two uniform, solid cylinders of radius R and total mass M are connected along their common axis by a short, light rod and rest o

sveta [45]

Explanation:

A) To prove the motion of the center of mass of the cylinders is simple harmonic:

System diagram for given situation is shown in attached Fig. 1

We can prove the motion of the center of mass of the cylinders is simple harmonic if

$a_{x} = -\omega^{2} x$

where aₓ is acceleration when attached cylinders move in horizontal direction:

<h3>PROOF:</h3>

rotational inertia for cylinders is given as:

$I=\frac{1}{2}MR^{2}$ -----(1)

Newton's second law for angular motion is:

∑τ = Iα ------(2)

For linear motion in horizontal direction it is:

∑Fₓ = Maₓ ------ (3)

By definition of torque:

τ = RF --------(4)

Put (4) and (1) in (2)

$RF=\frac{1}{2}MR^{2}\alpha$

from Fig 3 it can be seen that fs is force by which the cylinders roll without slipping as they oscillate

So above equation becomes

$f_{s}=\frac{1}{2}MR\alpha$ ------ (5)

As angular acceleration is related to linear by:

$a= R\alpha$

Eq (5) becomes

$f_{s}=\frac{1}{2}Ma_{x}$ ---- (6)

aₓ shows displacement in horizontal direction

From (3)

∑Fₓ = Maₓ

Fₓ is sum of fs and restoring force that spring exerts:

$\sum F_{x} = f_{s} - kx$ ----(7)

Put (7) in (3)

$f_{s} - kx = Ma_{x}$ [/tex] -----(8)

Using (6) in (8)

$\frac{1}{2}Ma_{x} - kx =Ma_{x}$

$a_{x} = \frac{2k}{3M} x$ --- (9)

For spring mass system

$a= -\omega^{2} x$ ----- (10)

Equating (9) and (10)

$\omega^{2} = \frac{2k}{3M}$

$\omega = \sqrt{ \frac{2k}{3M}}$

then (9) becomes

$a_{x} = - \omega^{2}x$

(The minus sign says that x and aₓ have opposite directions as shown in fig 3)

This proves that the motion of the center of mass of the cylinders is simple harmonic.

<h3 /><h3>B) Time Period</h3>

Time period is related to angular frequency as:

$T=\frac{2\pi }{\omega}$

$T = 2\pi \sqrt{\frac{3M}{2k}$

5 0

2 years ago

A police siren of frequency fsiren is attached to a vibrating platform. The platform and siren oscillate up and down in simple h

babunello [35]

Answer:

he maximum frequency occurs when the denominator is minimum

f’= f₀ $\frac{343}{343 + v_s}$

Explanation:

This is a doppler effect exercise, where the sound source is moving

f = fo $\frac{v}{v-v)s}$ when the source moves towards the observer

f ’=f_o $\frac{v}{v+v_{sy}}$ Alexandrian source of the observer

the maximum frequency occurs when the denominator is minimum, for both it is the point of maximum approach of the two objects

f’= f₀ $\frac{343}{343 + v_s}$

8 0

2 years ago