Ask question

Ask question

All categories

3 years ago

11

A moving roller coaster speeds up with constant acceleration for 2.3\,\text{s}2.3s2, point, 3, start text, s, end text until it

reaches a velocity of 35\,\dfrac{\text{m}}{\text{s}}35 s m 35, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction. During this time, the roller coaster travels 41\,\text{m}41m41, start text, m, end text. We want to find the initial velocity of the roller coaster before it starts to accelerate. Which kinematic formula would be most useful to solve for the target unknown?

1 answer:

Ad libitum [116K]3 years ago

7 0

Answer:

Δx=(v+v0/2)t

Explanation:

We can figure out which kinematic formula to use by choosing the formula that includes the known variables, plus the target unknown.

In this problem, the target unknown is the initial velocity v_0v

0

v, start subscript, 0, end subscript of the roller coaster.

You might be interested in

A circular loop of wire has radius of 9.50 cmcm. A sinusoidal electromagnetic plane wave traveling in air passes through the loo

m_a_m_a [10]

Answer:

The induced emf is $\epsilon = 0.1041 \ V$

Explanation:

From the question we are told that

The radius of the circular loop is $r = 9.50 \ cm = 0.095 \ m$

The intensity of the wave is $I = 0.0215 \ W/m^2$

The wavelength is $\lambda = 6.90\ m$

Generally the intensity is mathematically represented as

$I = \frac{ c * B^2 }{ 2 * \mu_o }$

Here $\mu_o$ is the permeability of free space with value

$\mu_o = 4 \pi *10^{-7} N/A^2$

B is the magnetic field which can be mathematically represented from the equation as

$B = \sqrt{ \frac{ 2 * \mu_o * I }{ c} }$

substituting values

$B = \sqrt{ \frac{ 2 * 4\pi *10^{-7} * 0.0215 }{ 3.0*10^{8}} }$

$B = 1.342 *10^{-8} \ T$

The area is mathematically represented as

$A = \pi r^2$

substituting values

$A = 3.142 * (0.095)^2$

$A = 0.0284$

The angular velocity is mathematically represented as

$w = 2 * \pi * \frac{c}{\lambda }$

substituting values

$w = 2 * 3.142 * \frac{3.0*10^{8}}{ 6.90 }$

$w = 2.732 *10^{8} rad \ s^{-1}$

Generally the induced emf is mathematically represented as

$\epsilon = N * B * A * w * sin (wt )$

At maximum induced emf $sin (wt) = 1$

So

$\epsilon = N * B * A * w$

substituting values

$\epsilon = 1 * 1.342 *10^{-8} * 0.0284 *2.732 *10^{8}$

$\epsilon = 0.1041 \ V$

7 0

3 years ago

Which elements are found in all four components of Earth?

kifflom [539]

Answer:

oxygen silicon aluminun iron

3 0

3 years ago

How do you convert seconds to minutes, seconds to hours , minutes to seconds and hours to seconds

Bess [88]

Answer:

You could memorize conversions, or use conversion charts, or do a quick internet search for help.

Explanation:

I'll do a few of the conversions for you

<u>Seconds to minutes: </u>

there are 60 seconds in one minute. So if you are wondering how many seconds are in 3 and 1/2 minutes, you would do this conversion:

60 sec/min <em>times </em>x sec/3.5 min

which can be written as 60 x 3.5 = 210. so "x" seconds would be 210 seconds in 3 1/2 minutes.

<u>you could do the same thing in the opposite direction for minutes to seconds:</u>

1 min has 60 seconds. So in 7.25 minutes, how many seconds are there?

1 min/60 sec <em>times </em>7.25 min/x sec

which can be written as 7.25 x 60 = 435. so "x" seconds would be 435 seconds in 7.25 minutes.

<u>hours to seconds:</u>

this one is slightly more complicated

In one hour there are 60 minutes, and in one minute there are 60 seconds.

so to convert from hours to seconds you would do this conversion:

1 hr/60 min times 1 min/60 sec. then the "min" would cancel out, and you would be left with the label "hr/sec". to do the math, it would be 1 hr / 60 x 60.

60x60 = 3600. so you would have 1 hr/3600 sec. So in one hour there are 3600 seconds.

so if you want to know how many seconds are in 6.75 hours:

6.75 hr/x sec <em>times </em>3600 sec/1 hr

6.75 x 3600 = 24,300 so there are 24,300 seconds in 6.75 hours.

I hope this helps :)

4 0

4 years ago

Two point charges are separated by 10 cm, with an attractive force between them of 15 N. Find the force between them when they a

suter [353]

Answer:

(a) the force is 8.876 N

(b) the magnitude of each charge is 4.085 μC

Explanation:

Part (a)

Given;

coulomb's constant, K = 8.99 x 10⁹ N.m²/C²

distance between two charges, r = 10 cm = 0.1 m

force between the two charges, F = 15 N

when the distance between the charges changes to 13 cm (0.13 m)

force between the two charges, F = ?

Apply Coulomb's law;

$F = \frac{Kq_1q_2}{r^2} \\\\let \ Kq_1q_2 = C\\\\F =\frac{C}{r^2} \\\\C = Fr^2\\\\F_1r_1^2 = F_2r_2^2\\\\F_2 =\frac{F_1r_1^2}{r_2^2} \\\\F_2 = \frac{15*0.1^2}{0.13^2} \\\\F_2 = 8.876 \ N$

Part (b)

the magnitude of each charge, if they have equal magnitude

$F = \frac{KQ^2}{r^2}$

where;

F is the force between the charges

K is Coulomb's constant

Q is the charge

r is the distance between the charges

$F = \frac{KQ^2}{r^2} \\\\Q = \sqrt{\frac{Fr^2}{K} } \\\\Q = \sqrt{\frac{15*(0.1)^2}{8.99*10^9} } = 4.085 *10^{-6} \ C\\\\Q = 4.085 \ \mu C$

4 0

3 years ago

Is water wet my friends have been arguing about this for about 3 months now but now it is time to hear other people

nirvana33 [79]

Water is not wet, wet is a state of being. Water, however, can make things wet.

8 0

3 years ago

Read 2 more answers