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3 years ago

True or False: Modulated waves can be analog or digital.

Physics

1 answer:

Firdavs [7]3 years ago

6 0

Answer:

True

Explanation:

Modulation waves(schemes) can be analog or digital. An analog modulation scheme has an input wave that varies continuously like a sine wave.

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Four identical metallic objects carry the following charges 1.82 6.65 4.80 and 9.30 C The objects are brought simultaneously int

GuDViN [60]

Answer:

a) 5.64 C

b) 3.5*10¹⁹ protons

Explanation:

Since the four metallic objects are identical, and total charge must be conserved, this means that after brought simultaneously into contact so that each touches the others, once separated, total charge must be the same than before being brought in contact.
But due they are identical, after charges were able to transfer freely between them, the four objects must have the same final charge, i.e. the fourth part of the total charge, as follows:

$Q_{n} = \frac{Q_{tot}}{4} = \frac{22.57C}{4} = 5.64 C (1)$

This charge will be divided between n protons, since the charge is positive.
Since each proton carries a charge equal to the elementary charge e, which value is 1.6*10⁻¹⁹ C, we can find the number of protons in excess, doing the following calculation:
$n_{p} =\frac{Q_{n}}{e} = \frac{5.64C}{1.6e-19C} = 3.5 e19 C (2)$

7 0

3 years ago

A block is thrown with an initial velocity of 30.0 m/s at an angle of 25.0o above the horizontal. What is the highest elevation

Serga [27]

The highest elevation reached by the ball in its trajectory is 16.4 m.

To find the answer, we need to know about the maximum height reached in a projectile.

What's the mathematical expression of the maximum height reached in a projectile motion?

The maximum height= U²× sin²(θ)/g
U= initial velocity, θ= angle of projectile with horizontal and g= acceleration due to gravity

What's the maximum height reached by a block that is thrown with an initial velocity of 30.0 m/s at an angle of 25° above the horizontal?

Here, U = 30.0 m/s and θ= 25°
Maximum height= 30²× sin²(25)/9.8

= 16.4m

Thus, we can conclude that the highest elevation reached by the ball in its trajectory is 16.4 m.

Learn more about the projectile motion here:

brainly.com/question/24216590

#SPJ4

3 0

2 years ago

A basketball player is 4.22 m from

max2010maxim [7]

Answer: The height above the release point is 2.96 meters.

Explanation:

The acceleration of the ball is the gravitational acceleration in the y axis.

A = (0, -9.8m/s^)

For the velocity we can integrate over time and get:

V(t) = (9.20m/s*cos(69°), -9.8m/s^2*t + 9.20m/s^2*sin(69°))

for the position we can integrate it again over time, but this time we do not have any integration constant because the initial position of the ball will be (0,0)

P(t) = (9.20*cos(69°)*t, -4.9m/s^2*t^2 + 9.20m/s^2*sin(69°)*t)

now, the time at wich the horizontal displacement is 4.22 m will be:

4.22m = 9.20*cos(69°)*t

t = (4.22/ 9.20*cos(69°)) = 1.28s

Now we evaluate the y-position in this time:

h = -4.9m/s^2*(1.28s)^2 + 9.20m/s^2*sin(69°)*1.28s = 2.96m

The height above the release point is 2.96 meters.

3 0

3 years ago

Read 2 more answers

Can a body have constant speed and still be acclerating ? Give an Example

Allushta [10]

Hi Pupil Here's Your answer :::

➡➡➡➡➡➡➡➡➡➡➡➡➡

An object moving with constant speed can be accelerated if direction of motion changes. For example, an object moving with a constant speed in a circular path has an acceleration because its direction of motion changes continuously.

⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅

Hope this Helps . . . . . . . . .

3 0

4 years ago

The smallest separation resolvable by a microscope is of the order of magnitude of the wavelength used. What energy electrons wo

hjlf

Answer: a) for 150 Angstroms 6.63 *10^-3 eV; b) for 5 Angstroms 6.02 eV

Explanation: To solve this problem we have to use the relationship given by De Broglie as:

λ =p/h where p is the momentum and h the Planck constant

if we consider the energy given by acceleration tube for the electrons given by: E: e ΔV so is equal to kinetic energy of electrons p^2/2m

Finally we have:

eΔV=p^2/2m= h^2/(2*m*λ^2)

replacing we obtained the above values.

6 0

3 years ago