All categories

4 years ago

The time that is required for a vibrating object to complete one full cycle is called : Group of answer choices

Physics

1 answer:

disa [49]4 years ago

4 0

Answer:

Correct option : (c) period.

Explanation:

The time that is required for a vibrating object to complete one full cycle is called the time period. If f is the frequency of a wave, then the relation between the frequency and the time period is given by :

$T=\dfrac{1}{f}$

These are the characteristics of a wave. Some other characteristics are wavelength, amplitude, intensity etc. So, the correct option is (c) "period".

You might be interested in

Viewers of Star Trek hear of an antimatter drive on the Starship Enterprise. One possibility for such a futuristic energy

Dominik [7]

a) 0.261 T

b) this field strength is obtainable with today's technology

Explanation:

The force experienced by a charged particle moving perpendicular to a magnetic field is given by

$F=qvB$

where

q is the charge

v is the velocity

B is the strength of the field

This force is perpendicular to the motion of the particle, which therefore moves in a circular path; and so, this force acts as centripetal force, so we can write:

$qvB=m\frac{v^2}{r}$

where

m is the mass of the particle

r is the radius of the circle

In this problem, we have:

$q=1.6\cdot 10^{-19}C$ (magnitude of the charge of antiprotons)

$v=5.00 \cdot 10^7 m/s$ (velocity)

$m=1.67\cdot 10^{-27}kg$ (mass of antiprotons)

r = 2.00 m (radius)

Therefore, we can re-arrange the equation and solve to find B, the magnetic field strength:

$B=\frac{mv}{qr}=\frac{(1.67\cdot 10^{-27})(5.00\cdot 10^7)}{(1.6\cdot 10^{-19})(2.00)}=0.261 T$

The strength of the magnetic field calculated in part A) is

$B=0.261 T$

This is indeed a very strong magnetic field. In fact, by comparison, the Earth's magnetic field has a strength of about

$B_{earth}=5\cdot 10^{-5} T$

However, there are current technologies available that are able to produce such strong fields. For instance, the superconducting magnets in the LHC (Large Hadron Collider) are able to produce magnetic fields of strength up to 8 Tesla (8 T).

Therefore, we can say that this field strength is obtainable with today's technology.

5 0

3 years ago

all of the following methods are ways to correct a run-on sentence except ,placing a semicolon between the compete sentences.put

olchik [2.2K]

All of the following methods are ways to correct a run-on sentence except: Putting a comma in between the complete sentences.

One of the uses of semi-colon is to separate two complete or independent sentences. A conjunction can also be used such as: and, however and but, to connect two complete sentences and transform them into a compound sentence.

6 0

3 years ago

Two very large parallel sheets are 5.00 cm apart. sheet a carries a uniform surface charge density of -9.70 μc/m2 , and sheet b,

MAVERICK [17]

Question is missing. Found on internet the complete text of the problem:

"<span>Two very large parallel sheets are 5.00 cm apart. Sheet A carries a uniform surface charge density of −9.70µC/m2, and sheet B, which is to the right or A, carries a uniform charge density of −11.5 µC/m2. Assume the sheets are large enough to be treated as infinite. Find the magnitude and direction of the net electric field these sheets produce at a point (a) 4.00 cm to the right of sheet A; (b) 4.00 cm to the left of sheet A; (c) 4.00 cm to the right of sheet B."

Solution:

(a) The electric field produced by a uniformly charged sheet at any distance is given by
</span> $E= \frac{\sigma}{2 \epsilon _0}$
where $\sigma$ is the charge density and $\epsilon _0 = 8.85\cdot 10^{-12} F/m$ is the vacuum permittivity.

First of all, let's compute the fields generated by the two sheets separately. The two densities of charge are $\sigma_A = -9.70 \mu C/m^2=-9.70\cdot 10^{-6}C/m^2$ and $\sigma_B = -11.5 \mu C/m^2 = -11.5\cdot 10^{-6} C/m^2$ .

Sheet a gives an electric field of
$E_A= \frac{\sigma_A}{2\epsilon _0}= \frac{-9.7\cdot 10^{-6} C/m^2}{2\cdot 8.85\cdot 10^{-12} F/m} = -5.48\cdot 10^5 V/m$
where the negative sign means the field points towards sheet A, in any point of the space.

The electric field produced by sheet B is given by:
$E_B = \frac{\sigma_A}{2\epsilon _0}= \frac{-11.5\cdot 10^{-6} C/m^2} {2\cdot 8.85\cdot 10^{-12} F/m} =-6.50\cdot 10^{5} V/m$
and again, the negative sign means that the field at any point of the space points towards sheet B.

The point at which we have to compute the total field is at 4.00 cm right of sheet A. Since the two sheets are 5.00cm far apart, it means that this point is between the two sheets. Therefore, in this point the two fields point into opposite directions. Therefore, the total field is
$E=E_1-E_2= -5.48\cdot 10^5 V/m - (-6.50\cdot 10^{5} V/m)=1.02\cdot 10^5 V/m$
And the direction is towards sheet B, since it has a field with stronger intensity.

(b) Field at 4.00 cm to the left of sheet A: in this point of the space, the two fields point towards same direction (on the right, towards both sheet A and sheet B). So, the total field is simply the sum of the two fields:
$E=E_1+E_2=-11.98\cdot 10^5 V/m$
towards right.

(c) Field at 4.00 cm to the right of sheet B. As before, the two fields in this point have same direction (both towards left, pointing towards both sheet A and sheet B). And so, the total field is simply the sum of the two fields:
$E=E_1+E_2=11.98 \cdot 10^5 V/m$
towards left.

7 0

4 years ago

We would like to use the relation V(t)=I(t)RV(t)=I(t)R to find the voltage and current in the circuit as functions of time. To d

drek231 [11]

Answer:

V = -RC (dV/dt)

Solving the differential equation,

V(t) = V₀ e⁻ᵏᵗ

where k = RC

Explanation:

V(t) = I(t) × R

The Current through the capacitor is given as the time rate of change of charge on the capacitor.

I(t) = -dQ/dt

But, the charge on a capacitor is given as

Q = CV

(dQ/dt) = (d/dt) (CV)

Since C is constant,

(dQ/dt) = (CdV/dt)

V(t) = I(t) × R

V(t) = -(CdV/dt) × R

V = -RC (dV/dt)

(dV/dt) = -(RC/V)

(dV/V) = -RC dt

∫ (dV/V) = ∫ -RC dt

Let k = RC

∫ (dV/V) = ∫ -k dt

Integrating the the left hand side from V₀ (the initial voltage of the capacitor) to V (the voltage of the resistor at any time) and the right hand side from 0 to t.

In V - In V₀ = -kt

In(V/V₀) = - kt

(V/V₀) = e⁻ᵏᵗ

V = V₀ e⁻ᵏᵗ

V(t) = V₀ e⁻ᵏᵗ

Hope this Helps!!!

5 0

4 years ago

A competitive go-cart driver is traveling 32 m/s. He sees a caution flag go up, so he slows at a rate of -1.5 m/s^2 in 10.8 s. W

Svetllana [295]

Answer:

15.8 m/s

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 32 m/s.

Acceleration (a) = – 1.5 m/s²

Time (t) = 10.8 s.

Final velocity (v) =?

Acceleration is simply defined as the rate of change of velocity with time. Mathematically, it is expressed as:

Acceleration (a) = [final velocity (v) – initial velocity (u)] / time (t)

a = (v – u) /t

With the above formula, we can obtain the final velocity of go-cart driver as follow:

Initial velocity (u) = 32 m/s.

Acceleration (a) = – 1.5 m/s²

Time (t) = 10.8 s.

Final velocity (v) =?

a = (v – u) /t

– 1.5 = (v – 32) / 10.8

Cross multiply

(v – 32) = –1.5 × 10.8

v – 32 = – 16.2

Collect like terms

v = – 16.2 + 32

v = 15.8 m/s

Therefore, the final velocity of go-cart driver is 15.8 m/s.

5 0

3 years ago