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3 years ago

PLEASE HELP ASAP!!! NO LINKS! Which type of wave does the illustration depict?

Physics

1 answer:

Xelga [282]3 years ago

3 0

Answer:

Transverse wave

Explanation:

The wave is moving forwards from the hand to the point of attachment

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A model rocket is launched from point A with an initial velocity v of 86 m/s. If the rockets descent parachute does not deploy a

krek1111 [17]

Height of the rocket will be h(t)=−12gt2+v0tsinθ+h0 where g=9.8 m/s2 v0=86 m/s h0=0 m θ= angle formed with the vertical

That's a parabola. You'll solve that for h(tf)=0 to find the time of flight. The horizontal component of the rocket's velocity will be vx=v0cosθ. You know that x=vxtf=104 m where tf is the time of flight. You can use that relationship to write an expression for tf in terms of v0 and θ. Substitute that into the first equation and solve for θ. Once you've got the parabola figured out, you can easily find the maximum height by finding the vertex, and you've already found the duration of the flight.

8 0

3 years ago

Read 2 more answers

A bungee jumped of mass m jumps off a brodge. Assume that the bungee cord behaves like an ideal spring of srping constant k when

anzhelika [568]

Answer:

The magnitude of the acceleration is at its maximum value.

Acceleration = $\frac{-Ak}{m} (m/s)^{2}$

Explanation:

During simple harmonic motions such as the oscillation of a spring, at the two extremes of the motion, the acceleration of the mass becomes maximum. This is because this is the point where the force on the body is maximum.

The angular velocity $\omega=\sqrt{\frac{k}{m} }$

At points of maximum extension x, the maximum acceleration is given by the formula.

$a=-A\omega^{2}$

substituting in the value for $\omega$ , this implies that

$a= -A\times (\sqrt{\frac{k}{m} } )^{2}$

⇒ $a=\frac{-Ak}{m} (m/s)^{2}$

4 0

3 years ago

What is Archimedes principle?

trasher [3.6K]

Archimedes' principle states that a body immersed in a fluid is subjected to an upwards force equal to the weight of the displaced fluid. This is a first condition of equilibrium. We consider that the above force, called force of buoyancy, is located in the centre of the submerged hull that we call centre of buoyancy.

6 0

3 years ago

A hiker throws a ball at an angle of 21.0° above the horizontal from a hill 21.0 m high. The hiker’s height is 1.750 m. The magn

nydimaria [60]

Answer:

38.8 m

Explanation:

h = -(21 + 1.75) = - 22.75 m

g = - 9.8 m /s^2

Ux = 14.004 m/s

Uy = + 5.376 m/s

Let the ball hits the ground in time t and at a distance d from the base of hill.

Use second equation of motion

h = Uyt + 0.5 at^2

- 22.75 = 5.376 t - 0.5 x 9.8 t^2

4.9 t^2 - 5.376 t - 22.75 = 0

$t=\frac{5.376\pm \sqrt{5.376^{2}+4\times 4.9\times 22.75}}{9.8}$

By solving

t = 2.77 second

So, horizontal distance

d = Ux t

d = 14.004 x 2.77 = 38.8 m

6 0

3 years ago

A crate weighing 1000 Newton's needs to be lifted at a constant speed from the ground to a loading dock 1.5 meters high in 5.0 s

miskamm [114]

Answer:

the Power required is 300 W

Explanation:

We know that the definition of work is given by the following expresion:

$W=F*d$

the distance is 1.5 meters and the Force is equal to 1000 N.

Replacing this values in the work expresion

$w= 1000 *1.5 = 1500 J$

And the power is equal to the amount of work done in a given time, therefore we have:

$Power = \frac{work}{time} \\$

Time in seconds

$Power = 1500/5\\\\Power = 300 watts$

3 0

3 years ago