Ask question

Ask question

All categories

2 years ago

10

Assume that the car at point A and the one at point E are traveling along circular paths that have the same radius. If the car a

t point A now moves twice as fast as the car at point E, how is the magnitude of its acceleration related to that of car E.

1 answer:

vazorg [7]2 years ago

6 0

Answer:

The magnitude of car A acceleration is 4 times that of car E

Explanation:

Assuming they are traveling at a constant speed. Their (centripetal) acceleration is as the following

$a = \frac{v^2}{r}$

where v is the (linear) velocity and r is the radius. We can use this to calculate their ratio

$a_A / a_E = \frac{v_A^2/r_A}{v_E^2/r_E} = \left(\frac{v_A}{v_E}\right)^2\frac{r_E}{e_A}$

Since $v_A = 2V_E \rightarrow \frac{v_A}{v_E} = 2$ and $r_A = r_E$

$a_A / a_E = 2^2 = 4$

So the magnitude of car A acceleration is 4 times that of car E

You might be interested in

Two trains travel toward each other on the same track, beginning 100 miles apart. One train travels at 40 miles per hour; the ot

Paladinen [302]

D = distance between th two trains at the start of the motion = 100 miles

V = speed of the faster train towards slower train = 60 mph

v = speed of the slower train towards faster train = 40 mph

t = time taken by the two trains to collide = ?

time taken by the two trains to collide is given as

t = D/(V + v)

t = 100/(60 + 40) = 1 h

v' = speed of the bird = 90 mph

d = distance traveled by the bird

distance traveled by the bird is given as

d = v' t

d = 90 x 1

d = 90 miles

7 0

2 years ago

Read 2 more answers

A 2 nC point charge is at the origin, and a second 5 nC point charge is on the x-axis at x = 8 m. Find the electric field (magni

dimaraw [331]

Answer:

The magnitude of the electric field is 5.75 N/C towards positive x- axis.

Explanation:

Given that,

Point charge at origin = 2 nC

Second charge = 5 nC

Distance at x axis = 8 m

We need to calculate the electric field at the point x = 2 m

Using formula of electric field

$E=\dfrac{1}{4\pi\epsilon_{0}}(\dfrac{q_{1}}{r_{1}^2}+\dfrac{q_{2}}{r_{2}^2})$

Put the value into the formula

$E=9\times10^{9}\times(\dfrac{2\times10^{-9}}{2^2}+\dfrac{5\times10^{-9}}{(8-2)^2})$

$E=5.75\ N/C$

The direction is toward positive x- axis.

Hence, The magnitude of the electric field is 5.75 N/C towards positive x- axis.

7 0

2 years ago

A robot is exploring charon, the dwarf planet pluto's largest moon. Gravity on charon is 0.278 meters per second [m divided b

DochEvi [55]

In order to find the efficiency first we will find the Change in Potential energy of the small stone that robot picked up

First we will find the mass of the stone

As it is given that stone is spherical in shape so first we will find its volume

$V = \frac{4}{3}\pi r^3$

$V = \frac{4}{3}\pi *(\frac{0.06}{2})^3$

$V = 1.13 * 10^{-4} m^3$

Now it is given that it's specific gravity is 10.8

So density of rock is

$\rho = 10.8 * 10^3 kg/m^3$

mass of the stone will be

$m = \rho V$

$m = 10.8* 10^3 * 1.13 * 10^{-4}$

$m = 1.22 kg$

now change in potential energy is given as

$\Delta U = mgH$

here

g = gravity on planet = 0.278 m/s^2

H = height lifted upwards = 15 cm

$\Delta U = 1.22* 0.278 * 0.15$

$\Delta U = 0.051 J$

Now energy supplied by internal circuit of robot is given by

$E = Vit$

V = voltage supplied = 10 V

i = current = 1.83 mA

t = time = 12 s

$E = 10* 1.83 * 10^{-3} * 12$

$E = 0.22 J$

Now efficiency is defined as the ratio of output work with given amount of energy used

$\eta = \frac{\Delta U}{E}*100$

$\eta = \frac{0.051}{0.22} = 0.23$

so efficiency will be 23 %

5 0

2 years ago

Arunner has a speed of 23 m/s. They see the finish line and speed up to 27 m/s. This happens in 5 seconds. If the runner has a m

sp2606 [1]

The answer is 100

Have a great day

6 0

2 years ago

A wire loop of radius 0.37 m lies so that an external magnetic field of magnitude 0.35 T is perpendicular to the loop. The field

Galina-37 [17]

Answer:

168.57 mV

Explanation:

Initial magnetic flux = BA , B magnetic field and A is area of loop

= .35 x 3.14 x .37²

= .15 Weber

Final magnetic flux

= - .2 x 3.14 x .37²

= - .086 Weber

change in flux

.15 + .086

= .236 Weber

rate of change of flux

= .236 / 1.4

= .16857 V

= 168.57 mV

5 0

2 years ago