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3 years ago

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A pulse with a height of +0.5 meter encounters a second pulse with a height of +2.3 meters. A. The two pulses interfere: B. The

resulting height of the medium when the pulses interfere will be: ____m

1 answer:

maksim [4K]3 years ago

3 0

CAN SOMEONE IN THE COMMETS ON EDG HELP ME WITH THIS IM NOT DOING THIS 4 POINTS I ACTULLY NEED HELP AND STRUGGLING U GUYS CAN CAN PUT A 1.0 STAR I REALLY DONT CARE BUT CAN SOMEONE HELP PLZ

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PLEASE HELP! 7TH GRADE SCIENCE!

professor190 [17]

They travel at different speeds

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3 years ago

A 100-kg object is moving at 20 m/s when a force brings the object to rest 3 points

zhenek [66]

Answer:

0.5

Explanation:

By reducing the force, the time taken to stop the object increases, as they are inversely proportional.

Therefore, when a doubling of the time is observed, a halving of the force should be as well.

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4 years ago

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Describe how an electromagnet can make a motor run.

irina [24]

An electric motor converts electrical energy into physical movement. Electric motors generate magnetic fields with electric current through a coil. The magnetic field then causes a force with a magnet that causes movement or spinning that runs the motor.

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3 years ago

Is there a point between a 10 nc charge and a 20 nc charge at which the electric field is zero?

KonstantinChe [14]

Let 10 nC and 20 nC charge is placed at "d" distance

now let say we will have a point at distance "x" from 10 nC at which electric field will be zero

now we will have

$E_1 = E_2$

$\frac{kq_1}{x^2} = \frac{kq_2}{(d-x)^2}$

now plug in all data

$\frac{k(10 nC)}{x^2} = \frac{k(20 nC)}{(d-x)^2}$

$\frac{10}{x^2} = \frac{20}{(d-x)^2}$

square root both sides

$\frac{1}{x} = \frac{\sqrt2}{d - x}$

$d - x = \sqrt2 x$

$d = x(1 + \sqrt2)$

$x = \frac{d}{1 + \sqrt2}$

so yes there will exist a point between 10 nC and 20 nC charge at which electric field will be zero

3 0

4 years ago

Read 2 more answers

An object has the acceleration graph shown in (Figure 1). Its velocity at t=0s is vx=2.0m/s. Draw the object's velocity graph fo

timama [110]

Answer:

Explanation:

We may notice that change in velocity can be obtained by calculating areas between acceleration lines and horizontal axis ("Time"). Mathematically, we know that:

$v_{b}-v_{a} = \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt$

$v_{b} = v_{a}+ \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt$

Where:

$v_{a}$ , $v_{b}$ - Initial and final velocities, measured in meters per second.

$t_{a}$ , $t_{b}$ - Initial and final times, measured in seconds.

$a(t)$ - Acceleration, measured in meters per square second.

Acceleration is the slope of velocity, as we know that each line is an horizontal one, then, velocity curves are lines with slopes different of zero. There are three region where velocities should be found:

Region I (t = 0 s to t = 4 s)

$v_{4} = 2\,\frac{m}{s} +\int\limits^{4\,s}_{0\,s} {\left(-2\,\frac{m}{s^{2}} \right)} \, dt$

$v_{4} = 2\,\frac{m}{s}+\left(-2\,\frac{m}{s^{2}} \right) \cdot (4\,s-0\,s)$

$v_{4} = -6\,\frac{m}{s}$

Region II (t = 4 s to t = 6 s)

$v_{6} = -6\,\frac{m}{s} +\int\limits^{6\,s}_{4\,s} {\left(1\,\frac{m}{s^{2}} \right)} \, dt$

$v_{6} = -6\,\frac{m}{s}+\left(1\,\frac{m}{s^{2}} \right) \cdot (6\,s-4\,s)$

$v_{6} = -4\,\frac{m}{s}$

Region III (t = 6 s to t = 10 s)

$v_{10} = -4\,\frac{m}{s} +\int\limits^{10\,s}_{6\,s} {\left(2\,\frac{m}{s^{2}} \right)} \, dt$

$v_{10} = -4\,\frac{m}{s}+\left(2\,\frac{m}{s^{2}} \right) \cdot (10\,s-6\,s)$

$v_{10} = 4\,\frac{m}{s}$

Finally, we draw the object's velocity graph as follows. Graphic is attached below.

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4 years ago