4. The result is that a(n)<br> electric current is induced<br> PLA

The amplitude of the wave in the picture below is... Select one: a. 1 cm b. 1.5 cm c. 2 cm d. 3 cm

antiseptic1488 [7]

Answer:

1cm

Explanation:

The amplitude of a wave is the distance from the center line (in this case the 1 cm marker on the vertical ruler) to the highest or lowest point. For this image the highest point is 2 cm, 2cm-1cm=1cm. The lowest point is 0cm, 1cm-0cm=1cm.

8 0

2 years ago

Which describes a p-type semiconductor?

Digiron [165]

It has holes (an electron deficiency).

6 0

3 years ago

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A ball of mass m is thrown into the air in a 45° direction of the horizon, after 3 seconds the ball is seen in a direction 30° f

Rzqust [24]

Answer:

Velocity (magnitude) is 98.37 m/s

Explanation:

We use the vertical component of the initial velocity, which is:

$v_{0y}=v_0*sin(45)=\frac{\sqrt{2} }{2}v_0$

Using kinematics expression of vertical velocity (in y direction) for an accelerated motion (constant acceleration, which is gravity):

$v_{y}=v_{0y}+a*t=\frac{\sqrt{2} }{2}v_0-9.8t$

Now we need to find $v_y$ as a function of $v_0$ . We use the horizontal velocity, which is always the same as follow:

$v_x=v_0cos(45\º)=\frac{\sqrt{2} }{2}v_0=v_{t=3}*cos(30\º) \\$

We know the angle at 3 seconds:

$v_y(t=3)=v_{t=3}*sin(30\º)\\v_{t=3}=\frac{v_y}{sin(30\º)}$

Substitute $v_{t=3}$ in $v_x$ and then solve for $v_y$

$\frac{\sqrt{2} }{2}v_0=\frac{v_y*cos(30\º) }{sin(30\º)} \\v_y=\frac{\sqrt{6} }{6}v_0$

With this expression we go back to the kinematic equation and solve it for initial speed

$\frac{\sqrt{6} }{6} v_0 =\frac{\sqrt{2} }{2}v_0-29.4\\v_0(\frac{\sqrt{6}-3\sqrt{2}}{6} )=-29.4\\v_0=98.37 m/s$

3 0

4 years ago

The most recently discovered system close to Earth is a pair of brown dwarfs known as Luhman 16. It has a distance of 6.5 light-

just olya [345]

Answer:

1.99 parsecs.

Explanation:

We have been given that the most recently discovered system close to Earth is a pair of brown dwarfs known as Luhman 16. It has a distance of 6.5 light-years.

We know that one light year equals to 0.306601 parsecs. To convert 6.5 light-years to parsecs, we will multiply 0.306601 by 6.5.

$6.5\text{ Light-years}=6.5\times 0.306601\text{ Parsecs}$

$6.5\text{ Light-years}=1.9929065\text{ Parsecs}$

$6.5\text{ Light-years}\approx 1.99\text{ Parsecs}$

Therefore, Luhman 16 is approximately 1.99 parsecs away from the Earth.

5 0

4 years ago

While buying a hot plate you notice the resistance of the hot plate is 25.0 . when connected to a 120. v source, how much curren

klemol [59]

The basic relationship between voltage, resistance and current of an electrical device is given by Ohm's law:
$V=IR$
where
V is the voltage
I is the current
R is the resistance

The hot plate in our problem is connected to a source of V=120 V and it has a resistance of $R=25.0 \Omega$ , therefore we can rearrange the previous equation to calculate the current through the device:
$I= \frac{V}{R}= \frac{120 V}{25.0 \Omega}=4.8 A$

4 0

4 years ago

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4. The result is that a(n) electric current is induced PLA