Ask question

Ask question

All categories

2 years ago

14

A car is brought to rest uniformly in 6 seconds. The initial velocity of the car was 24 m/s. How far does the car travel while d

ecelerating?

1 answer:

pav-90 [236]2 years ago

4 0

Answer:

v=u+at

24=0+at

24=a×6

a=4m/s

hence

s=ut+at^2÷2

s=36m

Explanation:

since the car is brought to rest the u=0

You might be interested in

PLEASE HELP ME (stop putting links ) Two objects m1 and m2, each with a mass of 5 kg and 6 kg separated by a distance. A third o

BabaBlast [244]

Answer:

Explanation:

Newton's Gravitation Law

$\displaystyle \frac{GmM}{d^2}$

where G is a constant, M and M the masses e d the distance betwen masses.

$\displaystyle G\frac{5\cdot2}{x^2}=G\frac{6\cdot 2}{(2x+1)^2} \quad \sqrt{6}x=(2x+1)\sqrt {5} \quad x=\frac{\sqrt{5}}{\sqrt{6}-2\sqrt{5}}$

7 0

2 years ago

3. Maverick and Goose are flying a training mission in their F-14. They are

Elanso [62]

Answer:

A. The bomb will take <em>17.5 seconds </em>to hit the ground

B. The bomb will land <em>12040 meters </em>on the ground ahead from where they released it

Explanation:

Maverick and Goose are flying at an initial height of $y_0=1500m$ , and their speed is v=688 m/s

When they release the bomb, it will initially have the same height and speed as the plane. Then it will describe a free fall horizontal movement

The equation for the height y with respect to ground in a horizontal movement (no friction) is

$y=y_0 - \frac{gt^2}{2}$ [1]

With g equal to the acceleration of gravity of our planet and t the time measured with respect to the moment the bomb was released

The height will be zero when the bomb lands on ground, so if we set y=0 we can find the flight time

The range (horizontal displacement) of the bomb x is

$x = v.t$ [2]

Since the bomb won't have any friction, its horizontal component of the speed won't change. We need to find t from the equation [1] and replace it in equation [2]:

Setting y=0 and isolating t we get

$t=\sqrt{\frac{2y_0}{g}}$

Since we have $y_0=1500m$

$t=\sqrt{\frac{2(1500)}{9.8}}$

$t=17.5 sec$

Replacing in [2]

$x = 688\ m/sec \ (17.5sec)$

$x = 12040\ m$

A. The bomb will take 17.5 seconds to hit the ground

B. The bomb will land 12040 meters on the ground ahead from where they released it

6 0

2 years ago

What is pitch related to in terms of sound? Does it affect the speed? If so, how?

Juli2301 [7.4K]

Pitch is the impression the listener gets of the <em>frequency</em> of the sound.

The speed of the sound is <em>not</em> related to its pitch/frequency.

If the speed and frequency were related, that would be a real problem. Bands, orchestras, and choirs could not exist ! All the instruments in the orchestra could play a note together, at the same time. But then the higher instruments ... the flute, trumpet, violins, high guitar strings and high piano keys ... would travel to you fast, and the lower instruments ... the trombone, tuba, double bass, bass drum, low guitar strings and the low piano keys ... would travel to you slow. They all played the note at the same time, but by the time you heard it, it would be all smeared out ... every instrument arriving at your ear at a different time !

5 0

3 years ago

A cyclist maintains a constant velocity of 6.1 m/s headed away from point A. At some initial time, the cyclist is 242 m from poi

KengaRu [80]

Answer:

553.1m

Explanation:

When an object moves at constant velocity we can express this movement like V=x/t, where V is the velocity, x is the displacement and t is the time spent on it.

In that way, the expression x=V.t give us the displacement from t=0s until t=51s, but we have to sum the initial distance from the point A.

x=242m +V.t = 242m + (6.1m/s x 51s) = 553.1m

7 0

3 years ago

Three point charges have equal magnitudes, two being positive and one negative. These charges are fixed to the corners of an equ

Irina-Kira [14]

Answer:

Magnitude of the net force on q₁-

Fn₁=1403 N

Magnitude of the net force on q₂+

Fn₂= 810 N

Magnitude of the net force on q₃+

Fn₃= 810 N

Explanation:

Look at the attached graphic:

The charges of the same sign exert forces of repulsion and the charges of opposite sign exert forces of attraction.

Each of the charges experiences 2 forces and these forces are equal and we calculate them with Coulomb's law:

F= (k*q*q)/(d)²

F= (9*10⁹*3*10⁻⁶*3*10⁻⁶)(0.01)² =810N

Magnitude of the net force on q₁-

Fn₁x= 0

Fn₁y= 2*F*sin60 = 2*810*sin60° = 1403 N

Fn₁=1403 N

Magnitude of the net force on q₃+

Fn₃x= 810- 810 cos 60° = 405 N

Fn₃y= 810*sin 60° = 701.5 N

$Fn_{3} = \sqrt{405^{2}+701.5^{2} }$

Fn₃ = 810 N

Magnitude of the net force on q₂+

Fn₂ = Fn₃ = 810 N

6 0

3 years ago