Sharece knows that wave peaks and valleys can add and subtract. What would be the net effect if she was able to cross Wave 1 (a large-amplitude wave in a valley phase) with Wave 2 (a wave with slightly smaller amplitude than Wave 2, in a peak phase)?Sharece knows that wave peaks and valleys can add and subtract. What would be the net effect if she was able to cross Wave 1 (a large-amplitude wave in a valley phase) with Wave 2 (a wave with slightly smaller amplitude than Wave 2, in a peak phase)?

Answers

Answer 1

Answer:

Answer:

The two waves will add vectorially to produce a small amplitude wave in a valley phase.

Explanation:

The two waves will add vectorially to produce a small amplitude wave in a valley phase. This is because the amplitudes of the waves are slightly different and in opposite directions. When wave 1 cancels out all of wave 2, the resultant wave would be the slight difference between both waves, and it would be in the direction of wave 1 which is a valley phase.

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A water slide is constructed so that swimmers, starting from rest at the top of the slide, leave the end of the slide traveling horizontally. One person hits the water 5.00 m from the end of the slide in a time of 0.504 s after leaving the slide. Ignore friction and air resistance. Find the height H.

Answers

Answer:

4.93 m

Explanation:

According to the question, the computation of the height is shown below:

But before that first we need to find out the speed which is shown below:

As we know that

$Speed = (Distance)/(Time)$

$Speed = (5)/(0.504)$

= 9.92 m/s

Now

$v^2 - u^2 = 2* g* h$

$9.92^2 = 2* 9.98 * h$

98.4064 = 19.96 × height

So, the height is 4.93 m

We simply applied the above formulas so that the height i.e H could arrive

Final answer:

The height of the water slide is 5.04 meters.

Explanation:

The problem described in this question involves a water slide, where swimmers start from rest at the top and leave the slide traveling horizontally. To determine the height of the slide, we can use the equations of motion in the horizontal direction. The horizontal displacement (x) is given as 5.00 m and the time (t) is given as 0.504 s. Assuming no friction or air resistance, we can use the equation x = v*t, where v is the horizontal velocity. Rearranging the equation, we can solve for v, which is equal to x/t. Substituting the given values, we have v = 5.00 m / 0.504 s = 9.92 m/s. The horizontal velocity (v) is constant throughout the motion, so we can use the equation v = sqrt(2*g*H), where g is the acceleration due to gravity (9.8 m/s^2) and H is the height of the slide. Rearranging the equation, we can solve for H, which is equal to v^2 / (2*g). Substituting the known values, we have H = (9.92 m/s)^2 / (2*9.8 m/s^2) = 5.04 m.

1. If a net force of 412 N is required to accelerate an object at 5.82 m/s2, what must theobject's mass be?

Answers

Answer:

The mass of the object is approximately 70.79 kilograms

Explanation:

We use Newton's second law to solve this problem. This law states that the net force on an object equals the product of its mass times the acceleration:

$F_(net)=m\,a$

Therefore, for this case, since the net force on the object and its acceleration are given, we can use the equation above to solve for the unknown mass:

$F_(net)=m\,a\n412\,N=m\,(5.82\,(m)/(s^2) )\nm=(412\,N)/(5.82\,(m)/(s^2) ) \nm=70.79\.kg$

A softball player swings a bat, accelerating it from rest to 2.6 rev/srev/s in a time of 0.20 ss . Approximate the bat as a 0.90-kgkg uniform rod of length 0.95 mm, and compute the torque the player applies to one end of it.

Answers

Answer:

$\tau=22.13Nm$

Explanation:

information we have:

mass: $m=0.9kg$

lenght: $L=0.95m$

frequency: $f=2.6rev/s$

time: $t=0.2s$

and from the information we have we can calculate the angular velocity $\omega$ . which is defined as

$\omega=2\pi f$

$\omega=2\pi (2.6rev/s)\n\omega=16.336 rev/s$

----------------------------

Now, to calculate the torque

We use the formula

$\tau=I \alpha$

where $I$ is the moment of inertia and $\alpha$ is the angular acceleration

moment of inertia of a uniform rod about the end of it:

$I=(1)/(3)mL^2$

substituting known values:

$I=(1)/(3) (0.9kg)(0.95m)^2\nI=0.271kg/m^2$

for the torque we also need the acceleration $\alpha$ which is defined as:

$\alpha=(\omega)/(t)$

susbtituting known values:

$\alpha=(16.336rev/s)/(0.2s) \n\alpha=81.68rev/s^2$

and finally we substitute $I$ and $\alpha$ into the torque equation $\tau=I \alpha$ : $\tau=(0.271kg/m^2)(81.68rev(s^2)\n\tau=22.13Nm$

Final answer:

To calculate the torque, we need to use the formula: Torque = Moment of Inertia * Angular Acceleration. By approximating the bat as a uniform rod and using its length and mass, we can find the moment of inertia. Then, using the given angular velocity, we can calculate the angular acceleration. Finally, we can determine the torque by multiplying the moment of inertia by the angular acceleration.

Explanation:

To compute the torque the player applies to one end of the bat, we need to use the formula:

Torque = Moment of Inertia * Angular Acceleration

Given that the bat is approximated as a uniform rod and we know its length and mass, we can calculate the moment of inertia. Then, using the given angular velocity, we can compute the angular acceleration. Finally, we can find the torque by multiplying the moment of inertia by the angular acceleration.

Learn more about Torque here:

brainly.com/question/33222069

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Which correctly describes how the energy of a wave on the electromagnetic spectrum depends on wavelength and frequency?A.
Energy decreases with decreasing wavelength and decreasing frequency.
B.
Energy increases with decreasing wavelength and increasing frequency.
C.
Energy increases with decreasing wavelength and decreasing frequency.
D.
Energy decreases with increasing wavelength and increasing frequency.

Answers

Answer:

B. Energy increases with decreasing wavelength and increasing frequency.

Explanation:

A swimmer heads directly across a river, swimming at 1.00 m/s relative to still water. He arrives at a point 41.0 m downstream from the point directly across the river, which is 73.0 m wide. What is the speed of the river current?

Answers

Answer:

velocity of the river is equal to 0.56 m/s

Explanation:

given,

velocity of swimmer w.r.t still water = 1 m/s

width of river = 73 m

he arrives to the point = 41 m

$times = (distance)/(speed)$

$times = (73)/(1)$

t = 73 s

$velocity = (distance)/(time)$

= $(41)/(73)$

= 0.56 m/s

velocity of the river is equal to 0.56 m/s

A laser (electromagnetic wave) has the maximum electric field strength of 1.0x1011 V/m. What is the force the laser applies on a mirror (totally reflective) of 5.0 mm2 area? A. 2.76 x105N B. 1.21 x106N C. 1.94 x106N D.4.43 x105 N E. 7.82 x104N