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tresset_1 [31]
2 years ago
10

What type of waves moves energy forward, but the source moves up and down?

Physics
1 answer:
stich3 [128]2 years ago
4 0
I'm pretty sure its transverse waves 
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You drive a car 640 m to the east, then 340 m to the north what is the magnitude of your displacement
mixer [17]

Magnitude of displacement = \sqrt{640^2 + 340^2}

Adding the squares gives displacement = \sqrt{525,200}

Displacement =  \sqrt{525,200} ≈ 724.7m

6 0
3 years ago
a force 2.4E2 N exists between a positive charge of 8E-5 C and a positive charge of 3E-5 C. What distance separates the charges?
Verdich [7]

The distance between the two charges is 0.3 m

Explanation:

The electrostatic force between two charged objects is given by Coulomb's law:

F=k\frac{q_1 q_2}{r^2}

where:

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q_1, q_2 are the charges of the two objects

r is the separation between the two charges

In this problem, we are given the following:

q_1 = 8\cdot 10^{-5} C

q_2 = 3\cdot 10^{-5} C

F=2.4\cdot 10^2 N

Therefore, we can rearrange the equation to solve for r, the distance between the two charges:

r=\sqrt{\frac{kq_1 q_2}{F}}=\sqrt{\frac{(8.99\cdot 10^9)(8\cdot 10^{-5})(3\cdot 10^{-5})}{2.4\cdot 10^2}}=0.3 m

Learn more about electrostatic force:

brainly.com/question/8960054

brainly.com/question/4273177

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6 0
3 years ago
A disk rotates around an axis through its center that is perpendicular to the plane of the disk. The disk has a line drawn on it
natka813 [3]

Answer:

t = \frac{\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}

Explanation:

The rotated angle is given by:

\theta=\omega0*t+1/2*\alpha*t^2

Since this is a quadratic equation it can be solved using:

x=\frac{-b \± \sqrt{b^2-4*a*c}  }{2*a}

Rewriting our equation:

1/2*\alpha*t^2+\omega0*t-\theta=0

t = \frac{\±\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}

Since \sqrt{\omega0^2+2*\theta*\alpha} >\omega0 we discard the negative solution.

t = \frac{\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}

8 0
3 years ago
Plasticity refers to the
olasank [31]
the brain's ability to change throughout the lifespan as a result of experience
5 0
3 years ago
A projectile is fired with an initial speed of 65.2 m/s at an angle of 34.5º above the horizontal on a long flat firing range. (
Margarita [4]

Answer:

A) h = 69.58 m

D) v = 58.12 \frac{m}{s}  (Speed magnitude)

α = 22.49° (Speed direction above the horizontal)

Explanation:

Conceptual analysis:

To solve this problem we consider the following concepts:

1) The projectile in its movement describes a curved line called a parabola, therefore two coordinates are required to fix the position at each instant of time, since the movement is performed in the X-Y plane.

The initial velocity (Vo) is tangent to the trajectory at the initial point and can be broken down into two components, one vertical (Voy) and one horizontal(Vox):

V_{ox} =  V_{o}Cos\alpha _{o} Formula (1)

V_{oy} =  V_{o}Sin\alpha _{o} Formula (2)

Where:

Vo: Initial velocity in m/s

\alpha_{o}: Initial angle above the horizontal in grades

2) The formula to calculate its velocity at any vertical position(y) is as follows:

v_{y}^{2} = v_{oy}^{2} -2gy Formula (3)

Where:

v_{f}^{2}: Final speed component in vertical direction in m/s

v_{oy}^{2}: Initial speed component in vertical direction in m/s

g: acceleration due to gravity in m/s2

3) The formulas to calculate the projectile velocity components at any time (t) are:

v_{x} = v_{ox} Because the movement is uniform in the x direction (constant speed)

v_{y} = v_{oy}-g*t Formula (4)  Because the movement is uniformly accelerated in the y direction

Known information:

We know the following data:

v_{o} = 65.2 \frac{m}{s}

\alpha _{o} = 34.5º above the horizontal

g=9.8 \frac{m}{s^{2}}

Development of the problem:

Initial speed components(Vox, Voy), (Formula (1), Formula (2)

v_{ox} =  65.2*Cos34.5 = 53.7\frac{m}{s}

v_{oy} =  65.2*Sin34.5 = 36.93\frac{m}{s}

A) Maximum height (h):

When the projectile reaches its maximum height (h) ,the speed component Vy = 0, then, we replace this value in the Formula (3):

0=(36.93)^2-2*9.8*h

h=\frac{ (36.93)^2}{2*9.8} = 69.58 m

D) Speed of the projectile 1.50s after firing

We replace t=1.5 s in the formula(4)

v_{y} = 36.93-9.8*1.5 = 22.23 \frac{m}{s}

v_{x} = v_{ox} = 53.7 \frac{m}{s}

v = \sqrt{53.7^{2}+22.23^{2}} = 58.12 \frac{m}{s}  (Speed magnitude)

\alpha = tan^{-1} (\frac{v_{y}}{v_{x}}) =  tan^{-1} (\frac{22.23}{53.7})

α = 22.49° (Speed direction above the horizontal)

6 0
3 years ago
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