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Solnce55 [7]
4 years ago
6

A roller coaster, traveling with an initial speed of 15 meters per second, decelerates uniformly at –7.0 m/s2 to a full stop. Ap

proximately how far does the roller coaster travel during its deceleration?
Physics
1 answer:
liubo4ka [24]4 years ago
4 0

Answer:

 Roller coaster will travel 16.07 meter before coming to stop.

Explanation:

We have equation of motion, v^2=u^2+2as, where u is the initial velocity, u is the final velocity, s is the displacement and a is the acceleration.

In this case we have final velocity = 0 m/s, initial velocity = 15 m/s, and acceleration = -7 m/s^2.

0^2=15^2-2*7*s\\ \\ 14s=225\\ \\ s=16.07m

So, roller coaster will travel 16.07 meter before coming to stop.

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Two sinusoidal waves are moving through a medium in the positive x-direction, both having amplitudes of 7.00 cm, a wave number o
lys-0071 [83]

Answer:

0.99 m

Explanation:

Parameters given:

Amplitude, A = 7.00cm

Wave number, k = 3.00m^-1

Angular Frequency, ω = 2.50Hz

Period = 6.00 s

Phase, ϕ = π/12 rad

Note: All parameters are the same for both waves except the phase.

Wave 1 has a wave function:

y1(x, t) = Asin(kx - ωt)

y1(x, t) = 7sin(3x - 2.5t)

Wave 2 has a wave function:

y2(x, t) = Asin(kx - ωt + ϕ)

y2(x, t) = 7sin(3x - 2.5t + π/12)

π is in radians.

When Superposition occurs, the new wave is represented by:

y(x, t) = 7sin(3x - 2.5t) + 7sin(3x - 2.5t + π/12)

y(x, t) = 7[sin(3x - 2.5t) + sin(3x - 2.5t + π/12)]

Using trigonometric function:

sin(a) + sin(b) = 2cos[(a - b)/2]sin[(a + b)/2]

Where a = 3x - 2.5t, b = 3x - 2.5t + π/12

We have that:

y(x, t) = (2*7)[cos(π/24)sin(3x - 2.5t + π/24)]

Therefore, when x = 0.53cm and t = 2s,

y(x, t) = (2*7)[cos(π/24)sin{(3*0.53) - (2.5*2)+ π/24}]

y(x, t) = 14 * 0.9914 * 0.0713

y(x, t) = 0.99 m

The height of the resultant wave is 0.99cm

5 0
3 years ago
a certain projetor uses a concave mirror for projecting an object's image on a screen .it produces on image that is 4 times bigg
IrinaK [193]

Answer:

f = 1 m

Explanation:

The magnification of the lens is given by the formula:

M = \frac{q}{p}

where,

M = Magnification = 4

q = image distance = 5 m

p = object distance = ?

Therefore,

4 = \frac{5\ m}{p}\\\\p = \frac{5\ m}{4}\\\\p = 1.25\ m

Now using thin lens formula:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}\\\\\frac{1}{f} = \frac{1}{1.25\ m}+\frac{1}{5\ m}\\\\\frac{1}{f} = 1\ m^{-1}\\\\

<u>f = 1 m</u>

6 0
3 years ago
Is there air resistance in space?........HELP!!!!!
Leokris [45]
There's no air in space, so there's no air resistance there.
5 0
4 years ago
Read 2 more answers
When a wave strikes a new medium, three things can happen: reflection, transmission, and absorption. Consider an echo. Do you th
kolbaska11 [484]

Answer:

YES

Explanation:

An echo may be defined as a sound which is repeated because of the sound waves that are produced are reflected back after striking a surface. Sound waves can smoothly bounce off the hard objects in the same manner as a rubber ball bounces back the ground.

When a sound wave strikes a hard surface, the sound waves gets reflected back and bounces back to the observer and produces an echo. If the sound waves strikes a soft surface it absorbs the sound.

Although the direction of a sound changes but the echo sounds in the same way as the original sound.

7 0
3 years ago
A car of mass 2200 kg collides with a truck of mass 4500 kg, and just after the collision the car and truck slide along, stuck t
katovenus [111]

Answer:

37.7m/s: principle of conservation of momentum

Explanation:

The principle to make use of is the principle of conservation of momentum which States that the sum of momentum of bodies before collision is equal to the sum of momentum of bodies after collision. This bodies will move with the same velocity after collision.

Momentum = Mass × velocity

For car of mass 2200kg moving with velocity 33m/s:

Momentum of car before collision = 2200×33

= 72,600kgm/s

For the truck of mass 4500kg;

Momentum = 4500 ×(22-(-18)

= 4500×40

= 180000kgm/s

After collision, their momentum is:

Momentum after collision = (2200+4500)v

= 6700v

Using the principle above to get the common velocity v we have

72600+180000 = 6700v

252600 = 6700v

v = 252600/6700

v = 37.7m/s

7 0
4 years ago
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