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3 years ago

9

In a longitudinal wave, particles of the medium vibrate in a direction that is perpendicular that the wave travels? True or Fals

e plzz answer right will give brainliest

2 answers:

Korolek [52]3 years ago

7 0

It’s false they travel parallel to the direction of the wave

vodomira [7]3 years ago

3 0

A transverse wave is a wave in which particles of the medium vibrate perpendicular to the direction that the wave travels.

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A world class sprinter is travelling with speed 12.0 m/s at the end of a 100 meter race. Suppose he decelerates at the rate of 2

Solnce55 [7]

Answer:

after 6 second it will stop

he travel 36 m to stop

Explanation:

given data

speed = 12 m/s

distance = 100 m

decelerates rate = 2.00 m/s²

so acceleration a = - 2.00 m/s²

to find out

how long does it take to stop and how far does he travel

solution

we will apply here first equation of motion that is

v = u + at ......1

here u is speed 12 and v is 0 because we stop finally

put here all value in equation 1

0 = 12 + (-2) t

t = 6 s

so after 6 second it will stop

and

for distance we apply equation of motion

v²-u² = 2×a×s ..........2

here v is 0 u is 12 and a is -2 and find distance s

put all value in equation 2

0-12² = 2×(-2)×s

s = 36 m

so he travel 36 m to stop

3 0

2 years ago

An echo is not heard in a small room.why ? give reason<br>

JulsSmile [24]

Answer:

They can't hear an echo in small room because in it the sound can't be reflected back. For an echo of a sound to be heard,the minimum distance between the source of sound and the walls of the room should be 17.2 m.

hopw it helps

7 0

2 years ago

Read 2 more answers

When a wheel is rotated through an angle of 21◦ , a point on the circumference travels through an arc length of 2.1 m. When the

LiRa [457]

Answer:

R = 5.73 m

Explanation:

For an angle of rotation through 21 degree we know that

arc length is given as

$angle \times Radius = Arc$

now we know that

Arc = 2.1 m

Angle = 21 degree

$Angle = 21\times \frac{\pi}{180}$

so now we have

$21\times \frac{\pi}{180} = \frac{2.1}{R}$

$R = \frac{2.1 \times 180}{21 \times \pi}$

$R = 5.73 m$

6 0

2 years ago

An airplane of mass 1.60 ✕ 104 kg is moving at 66.0 m/s. The pilot then increases the engine's thrust to 7.70 ✕ 104 N. The resis

Ivan

(a) No, because the mechanical energy is not conserved

Explanation:

The work-energy theorem states that the work done by the engine on the airplane is equal to the gain in kinetic energy of the plane:

$W=\Delta K$ (1)

However, this theorem is only valid if there are no non-conservative forces acting on the plane. However, in this case there is air resistance acting on the plane: this means that the work-energy theorem is no longer valid, because the mechanical energy is not conserved.

Therefore, eq. (1) can be rewritten as

$W=\Delta K + E_{lost}$

which means that the work done by the engine (W) is used partially to increase the kinetic energy of the airplane ( $\Delta K$ ) and part is lost because of the air resistance ( $E_{lost}$ ).

(b) 77.8 m/s

First of all, we need to calculate the net force acting on the plane, which is equal to the difference between the thrust force and the air resistance:

$F=7.70\cdot 10^4 N - 5.00 \cdot 10^4 N=2.70\cdot 10^4 N$

Now we can calculate the acceleration of the plane, by using Newton's second law:

$a=\frac{F}{m}=\frac{2.70\cdot 10^4 N}{1.60\cdot 10^4 kg}=1.69 m/s^2$

where m is the mass of the plane.

Finally, we can calculate the final speed of the plane by using the equation:

$v^2- u^2 = 2aS$

where

$v=?$ is the final velocity

$u=66.0 m/s$ is the initial velocity

$a=1.69 m/s^2$ is the acceleration

$S=5.00 \cdot 10^2 m$ is the distance travelled

Solving for v, we find

$v=\sqrt{u^2+2aS}=\sqrt{(66.0 m/s)^2+2(1.69 m/s^2)(5.00\cdot 10^2 m)}=77.8 m/s$

8 0

2 years ago

14. Convert 22 degrees celsius to fahrenheit.<br> Please show your work

forsale [732]

(22°C × 9/5) + 32 = 71.6°F

6 0

3 years ago

Read 2 more answers