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3 years ago

If the period of a wave is 0.05 s, then what is its frequency?

Physics

2 answers:

yKpoI14uk [10]3 years ago

8 0

Frequency = 1/time period = 1/0.05 = 20s^-1.

statuscvo [17]3 years ago

6 0

Answer:

20 $s^{-1}$

Solution:

We Know that frequenency is defined as the number of times something happens within a specific period of time.

Frequency is related with time by following formula

Frequency= $\frac{1}{T}$

Here T is time period=.05

Therefore,

Frequency= $\frac{1}{.05}$

Frequency=20 $s^{-1}$

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The current supplied by a battery slowly decreases as the battery runs down. Suppose that the current as a function of time is:

ludmilkaskok [199]

Answer: 8.1 x 10^24

Explanation:

I(t) = (0.6 A) e^(-t/6 hr)

I'll leave out units for neatness: I(t) = 0.6e^(-t/6)

If t is in seconds then since 1hr = 3600s: I(t) = 0.6e^(-t/(6 x 3600) ).

For neatness let k = 1/(6x3600) = 4.63x10^-5, then:

I(t) = 0.6e^(-kt)

Providing t is in seconds, total charge Q in coulombs is

Q= ∫ I(t).dt evaluated from t=0 to t=∞.

Q = ∫(0.6e^(-kt)

= (0.6/-k)e^(-kt) evaluated from t=0 to t=∞.

= -(0.6/k)[e^-∞ - e^-0]

= -0.6/k[0 - 1]

= 0.6/k

= 0.6/(4.63x10^-5)

= 12958 C

Since the magnitude of the charge on an electron = 1.6x10⁻¹⁹ C, the number of electrons is 12958/(1.6x10^-19) = 8.1x10^24 to two significant figures.

5 0

4 years ago

A small object with a 5.0-mC charge is accelerating horizontally on a friction-free surface at 0.0050 m/s2 due only to an electr

kolbaska11 [484]

Answer:

$0.002 N/C$

Explanation:

Parameters given:

Charge of object, q = 5 mC = $5 * 10^{-3} C$

Acceleration of object, a = $0.005 m/s^2$

Mass of object, m = 2.0 g

The Electric field exerts a particular force on the object, causing it to accelerate (Electrostatic force).

We know that Electrostatic force, F, is given in terms of Electric field, E, as:

F = qE

This means that the object exerts a force of -qE on the Electric force (Action with equal and opposite reaction).

The object also has a force, F, due to its acceleration a. This force is the product of its mass and acceleration. Mathematically:

F = ma

Equating the two forces of the object, we get:

-qE = ma

=> $E = \frac{-ma}{q}$

Solving for E, we have:

$E = \frac{-2 * 10^{-3} * 0.005}{5 * 10^{-3}} \\\\\\E = -0.002 N/C$

The magnitude will be:

$|E| = |-0.002| N/C = 0.002 N/C$

The electric field has a magnitude of 0.002 N/C.

4 0

4 years ago

You throw a ball upward from ground level with initial upward speed v0. What is the max height of the trajectory?

Inga [223]

Answer:

The max height of the ball is y = -1/2 (v0²/g).

It takes the ball t = -2 · v0/g to hit the ground.

The speed of the ball when it hits the ground is v = -v0.

Explanation:

The height and velocity of the ball is given by the following equations:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height of the ball at time t

y0 = initial height

v0 = initial velocity

t = time

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity at time t

When the ball is at max height, the velocity is 0. So, let´s find the time at which the velocity of the ball is 0.

v = v0 + g · t

0 = v0 + g · t

t = -v0/g

Now, replacing t = -v0/g in the equation of height, we will obtain the maximum height:

y = y0 + v0 · t + 1/2 · g · t² (y0 = 0 because the origin of the frame of reference is located on the ground)

y = v0 · t + 1/2 · g · t²

Replacing t:

y = v0 · (-v0/g) + 1/2 · g · (-v0/g)²

y = -(v0²/g) + 1/2 · (v0²/g)

y = -1/2 (v0²/g)

The max height of the ball is y = -1/2 (v0²/g). Remember that g is negative.

Since the acceleration of the ball is always the same, the time it takes the ball to impact the ground will be twice the time it takes to reach its max height, t = -2 v0/g.

However, let´s calculate that time knowing that at that time the height is 0:

y = y0 + v0 · t + 1/2 · g · t²

0 = v0 · t + 1/2 · g · t²

0 = t · ( v0 + 1/2 · g · t)

0 = v0 + 1/2 · g · t

-2 · v0/g = t

It takes the ball t = -2 · v0/g to hit the ground.

Let´s use the equation of velocity at final time (t = -2 · v0/g):

v = v0 + g · t

v = v0 + g · ( -2 · v0/g)

v = v0 - 2· v0

v = -v0

The speed of the ball when it hits the ground is v = -v0.

7 0

4 years ago

A helium balloon ride lifts up passengers in a basket. Assume the density of air is 1.28 kg/m3 and the density of helium in the

marysya [2.9K]

Answer:

Explanation:

1 ) the volume of helium in the balloon when fully inflated

= 4/3 π R³

= 4/3 x 3.14 x 5.3³

= 623.3 m³

2 ) force of gravity on the entire system ,but with no people

= (mass of the empty balloon and basket + mass of helium )x g

=( 129 + 623.3 x.18 )x 9.8

= 2363.7 N

3 ) magnitude of the buoyant force on the entire system (but with no people )

( Volume of empty balloon + basket + volume of helium ) density of air x g

( 0.057 + 623.3 ) x 1.23 x 9.8

= 7513.94 N

4 ) magnitude of the force of gravity on each person

Mass of each person x g

= 71 x 9.8

= 695.8 N

6 0

3 years ago

Explain why incremental development is the most effective approach for developing business software systems. Why is this model l

Ksenya-84 [330]

Explanation:

Typically, business software technologies are complex, and software strenuous. Business software applications are also often upgraded for changes in business goals or procedures. Real-time systems usually require a lot of hardware components that are quite difficult to change and cannot be upgraded Usually, actual-time safety critical systems that required to be built based on well-planned processes.

3 0

3 years ago