Ask question

Ask question

All categories

3 years ago

11

Over a short interval the coordinate of a car in the meters isgiven by x(t) = 27t - 4.0 t3 where time t is in seconds,at the end

of 1.0s the acceleration of the car is
1. 27 m/s2
2. 4.0 m/s2
3. - 4.0 m/s2
4. -12 m/s2
5. -24 m/s2

1 answer:

wariber [46]3 years ago

4 0

Answer:

5. -24 m/s²

Explanation:

Acceleration: This can be defined as the rate of change of velocity.

The S.I unit of acceleration is m/s².

mathematically,

a = dv/dt ............................ Equation 1

Where a = acceleration, dv/dt = is the differentiation of velocity with respect to time.

But

v = dx(t)/dt

Where,

x(t) = 27t-4.0t³...................... Equation 2

Therefore, differentiating equation 2 with respect to time.

v = dx(t)/dt = 27-12t²............. Equation 3.

Also differentiating equation 3 with respect to time,

a = dv/dt = -24t

a = -24t .................... Equation 4

from the question,

At the end of 1.0 s,

a = -24(1)

a = -24 m/s².

Thus the acceleration = -24 m/s²

The right option is 5. -24 m/s²

You might be interested in

When two point charges are a distance d part, the electric force that each one feels from the other has magnitude F. In order to

Contact [7]

Answer:

Because the force is inversely proportional to the square of the distance

Explanation:

The magnitude of the electrostatic force between two charged particles is given by

$F=k\frac{q_1 q_2}{d^2}$

where

k is the Coulomb's constant

q1, q2 are the magnitudes of the two charges

d is the distance between the two charges

We observe that the magnitude of the force is inversely proportional to the square of the distance.

Therefore, when the distance changes to

$d'=\frac{d}{\sqrt{2}}$

The force will double:

$F'=k\frac{q_1 q_2}{(d/\sqrt{2})^2}=2(k\frac{q_1 q_2}{d^2})=2F$

5 0

4 years ago

The intensity of the sound from a certain source is measured at two points along a line from the source. The points are separate

Artemon [7]

Answer:

The source is at a distance of 4.56 m from the first point.

Solution:

As per the question:

Separation distance between the points, d = 11.0 m

Sound level at the first point, L = 66.40 dB

Sound level at the second point, L'= 55.74 dB

Now,

$L = 10log_{10}\frac{I}{I_{o}}$

$I = I_{o}10^{\frac{L}{10}} = I_{o}10^{0.1L} = 10^{- 12}\times 10^{0.1\times 66.40} = 10^{- 5.36}$

$L' = 10log_{10}\frac{I'}{I_{o}}$

$I' = I_{o}10^{\frac{L'}{10}} = 10^{- 12}\times 10^{0.1\times 55.74} = 10^{- 6.426}$

where

$I_{o} = 10^{- 12} W/m^{2}$

I = Intensity of sound

Now,

$I = \frac{P}{4\pi R^{2}}$

Similarly,

$I' = \frac{P}{4\pi (R + 11.0)^{2}}$

Now,

$\frac{I}{I'} = \frac{(R + 11.0)^{2}}{R^{2}}$

$\frac{10^{- 5.36}}{10^{- 6.426}} = \frac{(R + 11.0)^{2}}{R^{2}}$

$R ^{2} + 22R + 121 = 11.64R^{2}}$

$10.64R ^{2} - 22R - 121 = 0$

Solving the above quadratic eqn, we get:

R = 4.56 m

8 0

3 years ago

A car moving in a straight line starts at x = 0 at t = 0 . It passes the point x = 30.0 m with a speed of 10.0 m/s at t = 3.00 s

Bogdan [553]

Answer:

a) $v=20.3m/s$

b) $a=2.35m/s^2$

Explanation:

From the exercise we know:

$x_{1}=30m$ ; $v_{1}=10m/s$ ; $t_{1}=3s$

$x_{2}=375m$ ; $v_{2}=50m/s$ ; $t_{2}=20s$

The formula for average velocity is:

$v=\frac{x_{2}-x_{1}}{t_{2}-t_{1}}$

a) $v=\frac{375m-30m}{20s-3s}=20.3m/s$

The formula for average acceleration is:

$a=\frac{v_{2}-v_{1}}{t_{2}-t_{1}}$

b) $a=\frac{(50-10)m/s}{(20-3)s}=2.35m/s^2$

4 0

3 years ago

Alexis pulls with 3.0 Newton’s on a toy doll toward east. Brianna pulls on the same toy toward the south with 2.0 Newton’s. Calc

jeka57 [31]

Brianna pulls on the same toy toward the south with 2.0 ... Calculate the magnitude of the resultant force on the toy

6 0

3 years ago

Read 2 more answers

HELP PLEASE NOWWWW !!!! ( 1 ) Jill and Scott both road their bikes for 30 minutes. Jill traveled 5 kilometers and Scott trave

katovenus [111]

#1

Jill and Scott both moves for 30 minutes

now if Jill cover 5 km distance and Scott cover 10 km distance

now we know that the formula of speed is given as

$speed = \frac{distance}{time}$

now we will have

speed of Jill

$v = \frac{5}{0.5} = 10 km/h$

speed of Scott

$v = \frac{10}{0.5} = 20 km/h$

so correct answer here is

<em>Scott had the faster speed since he rode at 20 k/h while Jill only traveled 10 km/h.</em>

<em>#2</em>

distance travelled by each car is given as

$d = 400 miles$

now here it is given that

time taken by green car

$t_1 = 10 hours$

time taken by yellow car

$t_2 = 8 hours$

now we can find the speed of two cars

$speed = \frac{distance}{time}$

speed of green car

$v = \frac{400}{10} = 40 mph$

speed of yellow car

$v = \frac{400}{8} = 50 mph$

so correct answer will be

<em>The yellow car was faster. Yellow traveled at a speed of 50 mph while green was traveling at an average of 40 mph.</em>

6 0

3 years ago

Read 2 more answers