The work of brain mind interface is to

Graphs are representations of equations.<br> A. True<br> B. False

zhuklara [117]

Answer:

Hi there!

The answer is: A. True

6 0

2 years ago

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player kicks a soccer ball from ground level and sends it flying at an angle of 30 degrees at a speed of 26 m/s. What is the hor

seropon [69]

Answer

22.5 m/s

Explanation

We shall use the trigonometric ratio cosine to find the horizontal component.

cos = adjacent/hypotenuse

Adjacent is the horizontal and hypotenuse is the fly speed.

cos 30° = horizontal / 26

horizontal velocity = 26 × cos 30°

= 26 × 0.866

= 22.5166

= 22.5 m/s

7 0

2 years ago

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Which of the following distinguishes electromagnetic waves from mechanical waves

Ivan

Answer:

Explanation:

1. Mechanical waves require material medium for their propagation while electromagnetic waves do not require material medium for their propagation.

2. Mechanical waves can either be transverse or longitudinal while electromagnetic waves are transverse.(Transverse waves are waves in which the vibration of the particules of the medium is perpendicular to the direction of the motion of wave. E.g water waves, waves of a plucked string and all electromagnetic waves RIVUXG . Longitudinal waves are waves whose vibration are parallel to the direction of the motion of the medium e.g waves in strings, sound waves.e.t.c)

6 0

2 years ago

A photographer in a helicopter ascending vertically at a constant rate of accidentally drops a camera out the window when the he

maksim [4K]

Answer:

The time it takes for the camera to reach the ground is 5 s.

Explanation:

To solve this problem, we will use the free fall cinematic equation.

Since the helicopter ascends with constant speed, the camera falls to the ground only by the effect of gravity on it.

The speed at which the helicopter ascends is not specified in the statement, but according to a similar problem, we will use 12.5 $\frac{m}{s}$ .

First, we must calculate the time and the maximum height at which the camera arrives after leaving the helicopter.

To calculate the maximum height to which it arrives, we will use the formula of vertical shot (since the camera leaves the helicopter with a speed upwards of 12,5 $\frac{m}{s}$ ).

$V_{f} ^{2}$ = $V_{0} ^{2}$ - 2 * g * h

Where:

$V_{f}$ : final speed at maximum height.

$V_{0}$ : initial speed when it falls from the helicopter.

g: gravity taken at 9.8 $\frac{m}{s^{2} }$

h: height reached from 60 m when leaving the helicopter

as $V_{f}$ =0

0= $(12,5\frac{m}{s}) ^{2}$ - 2 * $9,8\frac{m}{s^{2} }$ * h

clear h:

h= $(12,5 \frac{m}{s^{2} } )^{2}$ / (2 * $9,8 \frac{m}{s^{2} }$ )

h=7,97 m

Then we must calculate the time it takes to reach its maximum height:

$V_{f}$ = $V_{0}$ - g * t

t: time it takes to arrive from the moment it leaves the helicopter at its maximum height.

as $V_{f}$ =0

0=12.5 $\frac{m}{s}$ - 9.8 $\frac{m}{s^{2} }$ * t

clearing t

t=12.5 $\frac{m}{s}$ / 9.8 $\frac{m}{s^{2} }$

t=1.27 s.

Now we can calculate the time it takes to fall from the maximum height of 67.97 m.

The equation we will use is Y= $v_{0}*t+(\frac{g*t^{2} }{2} )$

where:

t: time it takes for the camera to fall.

Y: height from where the camera falls concerning the ground.

$v_{0}$ : initial speed of the camera at the time of starting the fall.

g: acceleration of gravity, estimated at 9.8 $\frac{m}{s^{2} }$

Step 1: As the helicopter ascends with constant speed, the initial speed of the camera at the moment of falling is 0.

$v_{0}$ =0

So the first term of our equation is nullified.

Step 2: To calculate the time it takes to fall, we clear "t" of the equation:

Y= $\frac{(g*t^{2})}{2}$

Y*2=(g* $t^{2}$ )

$\frac{Y*2}{g}$ = $t^{2}$

$\sqrt{\frac{Y*2}{g} }$ =t

Step 3: I replace the values with the incognites and get "t".

t= $\sqrt{\frac{67,97m*2}{9,8\frac{m}{s^{2} } } }$

t=3,73 s

The total time it takes for the camera to fall from the moment it leaves the helicopter is the sum of the time it takes to reach the maximum point of height and the time it takes to fall to the ground from that height.

t= 1,27 s + 3,73 s = 5 s

Have a nice day!

3 0

3 years ago

Calculate the most charge that a 100pF capacitor (with a 1.0mm plate separation) can store if the capacitor breaks down with an

Misha Larkins [42]

Answer:

0.6 μC

Explanation:

C = capacitance of the capacitor = 100 x 10⁻¹² F

d = separation between the plates of capacitor = 1 mm = 1 x 10⁻³ m

E = Electric field = 6 x 10⁶ N/C

Q = Amount of charge

V = Potential difference

Potential difference is given as

V = E d

Amount of charge stored is given as

Q = CV

hence

Q = C E d

inserting the values

Q = (100 x 10⁻¹²) (6 x 10⁶) (1 x 10⁻³)

Q = 6 x 10⁻⁷ C

Q = 0.6 μC

5 0

2 years ago