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3 years ago

14

As waves get closer to a beach they _____.

1 answer:

krok68 [10]3 years ago

4 0

Answer:

Increase in energy

Explanation:

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17. Calculate the amount of gravitational potential energy at the top of one 4 points hill. The mass of the coaster is 500 kg. T

MissTica

Answer: D. 292,338 J

This is the correct answer :)

4 0

3 years ago

Dawn and Timothy are sailboating in Lake Obewon. Starting from rest near the shore, they accelerate with a uniform acceleration

castortr0y [4]

Answer:

The value is $s =63.18 \ m$

Explanation:

From the question we are told that

The uniform acceleration is $a = 0.39 \ m/s^2$

The time considered is $t = 18 \ seconds$

Generally the from kinematic equation we have that

$s = ut + \frac{1}{2} at^2$

Given that the boat started from rest , the initial velocity is u = 0 m/s

So

$s = 0 * 18 + \frac{1}{2} *0.39 * 18^2$

=> $s =63.18 \ m$

5 0

3 years ago

A ball thrown vertically upward is caught by the thrower after 2.00 s. Find (a) the initial velocity of the ball and (b) the max

ankoles [38]

Answer:

a) 9.8 m/s

b) 4.9 m

Explanation:

This problem is a good example of Vertical motion, where the main equations for this situation are:

$y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}$ (1)

$V^{2}={V_{o}}^{2}-2gy$ (2)

Where:

$y$ is the height of the ball at a given time

$y_{o}=0m$ is the initial height of the ball (assuming the hand of the thrower the origin of the system)

$V_{o}$ is the initial velocity of the ball

$V$ is the final velocity of the ball

$t=2s$ is the time it takes for the ball to make the complete movement (from the moment it is thrown until it falls back into the pitcher's hands)

$g=9.8 m/s^{2}$ is the acceleration due to gravity

Knowing this, let's begin with the answers:

<h3>a) Initial velocity </h3>

In order to find the initial velocity $V_{o}$ of the ball, we will use equation (1) and $t=2s$ , taking into account that $y=0 m$ and $y_{o}=0m$ at this given time:

$0=0+V_{o}t-\frac{1}{2}gt^{2}$ (3)

Isolating $V_{o}$ :

$V_{o}=\frac{1}{2}gt$ (4)

$V_{o}=\frac{1}{2}(9.8 m/s^{2})(2 s)$ (5)

Then:

$V_{o}=9.8 m/s$ (6)

<h3>b) Maximum height </h3>

In this part, we will use equation (2), knowing the value of the height is maximum when $V=0$ . So, we will name this height as $y_{max}$ :

$0={V_{o}}^{2}-2gy_{max}$ (7)

Isolating $y_{max}$ :

$y_{max}=\frac{{V_{o}}^{2}}{2g}$ (8)

$y_{max}=\frac{{(9.8 m/s)}^{2}}{2(9.8 m/s^{2})}$ (9)

Finally:

$y_{max}=4.9 m$

4 0

4 years ago

Two long, parallel, current-carrying wires lie in an xy-plane. The first wire lies on the line y = 0.300 m and carries a current

Marrrta [24]

Answer:

Explanation:

Since the wires attract each other , the direction of current will be same in both the wires .

Let I be current in wire which is along x - axis

force of attraction per unit length between the two current carrying wire is given by

$\frac{\mu_0}{4\pi}$ x $\frac{2 I_1\times I_2}{d}$

where I₁ and I₂ are currents in the wires and d is distance between the two

Putting the given values

285 x 10⁻⁶ = 10⁻⁷ x $\frac{2\times25.5\times I_2}{.3}$

I₂ = 16.76 A

Current in the wire along x axis is 16.76 A

To find point where magnetic field is zero due the these wires

The point will lie between the two wires as current is in the same direction.

Let at y = y , the neutral point lies

k 2 x $\frac{16.76}{y}$ = k 2 x $\frac{25.5}{.3-y}$

25.5y = 16.76 x .3 - 16.76y

42.26 y = 5.028

y = .119

= .12 m

3 0

3 years ago

Velocity is different from speed because it has size and what?

koban [17]

Answer:

direction

Explanation:

6 0

3 years ago

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