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Nadusha1986 [10]

3 years ago

7

When sound waves travel through a medium, the particles of the medium move parallel to the wave motion. which type of wave displ

ays this behavior?
a. longitudinal wavesb) transverse wavesc) perpendicular wavesd) electromagnetic waves

1 answer:

kvv77 [185]3 years ago

8 0

The answer is A. Longitudinal waves.

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Which of the following is characteristic of proficient catching?

uysha [10]

Answer:

The correct answer is option D i.e. A and C

Explanation:

The correct answer is option D i.e. A and C

for proficient catching player must

- learn to absorbed the ball force

- moves the hang according to ball direction to hold the ball

- to catch ball at high height move the finger at higher position

- to catch ball at low height move the finger at lower position

5 0

3 years ago

Electromagnetic radiation that has a short wavelength will have a ________ frequency.

ArbitrLikvidat [17]

High frequency , it is because wavelength is inversely proportional to frequency

8 0

3 years ago

The equation Bold r (t )equals(8 t plus 9 )Bold i plus (2 t squared minus 8 )Bold j plus (6 t )Bold k is the position of a parti

KATRIN_1 [288]

Explanation:

It is given that, the position of a particle as as function of time t is given by :

$r(t)=(8t+9)i+(2t^2-8)j+6tk$

Let v is the velocity of the particle. Velocity of an object is given by :

$v=\dfrac{dr(t)}{dt}$

$v=\dfrac{d[(8t+9)i+(2t^2-8)j+6tk]}{dt}$

$v=(8i+4tj+6k)\ m/s$

So, the above equation is the velocity vector.

Let a is the acceleration of the particle. Acceleration of an object is given by :

$a=\dfrac{dv(t)}{dt}$

$a=\dfrac{d[8i+4tj+6k]}{dt}$

$a=(4j)\ m/s^2$

At t = 0, $v=(8i+0+6k)\ m/s$

$v(t)=\sqrt{8^2+6^2} =10\ m/s$

Hence, this is the required solution.

7 0

4 years ago

If a star is found directly above the sun on the h-r diagram, what can you conclude about its size?.

hammer [34]

Answer:

It is larger than the Sun.

Explanation:

Brainliest pls

:3

7 0

2 years ago

Model rocket engines are sized by thrust, thrust duration, and total impulse, among other characteristics. A size C5 model rocke

umka2103 [35]

Answer:

v_{f} = 115.95 m / s

Explanation:

This is an exercise of a variable mass system, let's form a system formed by the masses of the rocket, the mass of the engines and the masses of the injected gases, in this case the system has a constant mass and can be solved using the conservation the amount of movement. Which can be described by the expressions

Thrust = $v_{e} \frac{dM}{dt}$

$v_{f}$ -v₀ = v_{e} $ln ( \frac{M_{o} }{M_{f}} )$

where v_{e} is the velocity of the gases relative to the rocket

let's apply these expressions to our case

the initial mass is the mass of the engines plus the mass of the fuel plus the kill of the rocket, let's work the system in SI units

M₀ = 25.5 +12.7 + 54.5 = 92.7 g = 0.0927 kg

The final mass is the mass of the engines + the mass of the rocket

M_{f} = 25.5 +54.5 = 80 g = 0.080 kg

thrust and duration of ignition are given

thrust = 5.26 N

t = 1.90 s

Let's start by calculating the velocity of the gases relative to the rocket, where we assume that the rate of consumption is linear

thrust = v_{e} $\frac{M_{f} - M_{o} }{t_{f} - t_{o} }$

v_{e} = thrust $\frac{\Delta t}{\Delta M}$

v_{e} = 5.26 $\frac{1.90}{0.080 -0.0927}$

v_{e} = - 786.93 m / s

the negative sign indicates that the direction of the gases is opposite to the direction of the rocket

now we look for the final speed of the rocket, which as part of rest its initial speed is zero

v_{f}-0 = v_{e} $ln ( \frac{M_{o} }{M_{f} } )$

we calculate

v_{f} = 786.93 ln (0.0927 / 0.080)

v_{f} = 115.95 m / s

5 0

3 years ago