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4 years ago

In each case, you should demonstrate how you worked out your answer, as well as giving the answer.

Physics

2 answers:

Ivanshal [37]4 years ago

7 0

Answer:

First frequency f = 153.846 Hz

Second period T = 1.625 ms

Step by stepplanation:

T = 6.5ms= 6.5*(10^-3)

But T = 1/f

f = 1/T

f = 1(6.5*(10^-3))

f = 153.846 Hz

Now the frequency f is increased by a factor of 3

= 3*153.846

= 461.538

New frequency = 461.538+153.846

= 615.384 Hz

New period T = 1/615.384

Period T = 1.625*10^-3

T= 1.625 ms

irga5000 [103]4 years ago

3 0

Answer:

Explanation:

(a)

Time period, T1 = 6.5 ms

= 6.5 × 10^-3 s

Frequency is the reciprocal of time period.

f1 = 1 / T1 = 1 / (6.5 × 10^-3) = 153.85 Hz

Now the frequency becomes 3 times

f2 = 3 × f1 = 3 × 153.85 = 461.55 Hz

The new period is

T2 = 1 / f2 = 1 / 461.55

= 0.0022 second = 2.22 ms

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A cannon is mounted on a tower above a wide, level field. The barrel of the cannon is 20 m above the ground below. A cannonball

OLga [1]

<u>Answer:</u>

Cannonball will be in flight before it hits the ground for 2.02 seconds

<u>Explanation:</u>

Initial height from ground = 20 meter.

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this the velocity of body in vertical direction = 0 m/s, acceleration = 9.8 $m/s^2$ , we need to calculate time when s = 20 meter.

Substituting

$20=0*t+\frac{1}{2} *9.8*t^2\\ \\ t = 2.02 seconds$

So it will take 2.02 seconds to reach ground.

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3 years ago

Which phrases describe all the outer planets motion? Select two options

11111nata11111 [884]

Answer:

slow revolution and fast rotation

Explanation:

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3 years ago

A 0.03 kg golf ball is hit off the tee at a speed of 34 m/s. The golf club was in contact with the ball for 0.003 s. What is the

Liula [17]

Answer:

The average force on ball by the golf club is 340 N.

Explanation:

Given that,

Mass of the golf ball, m = 0.03 kg

Initial speed of the ball, u = 0

Final speed of the ball, v = 34 m/s

Time of contact, $\Delta t=0.003\ s$

We need to find the average force on ball by the golf club. We know that the rate of change of momentum is equal to the net external force applied such that :

$F=\dfrac{\Delta p}{\Delta t}\\\\F=\dfrac{mv-mu}{\Delta t}\\\\F=\dfrac{mv}{\Delta t}\\\\F=\dfrac{0.03\ kg\times 34\ m/s}{0.003\ s}\\\\F=340\ N$

So, the average force on ball by the golf club is 340 N.

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4 years ago

______ is the total distance traveled divided by the total time of travel.

sweet-ann [11.9K]

Average speed is the answer

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3 years ago

In a rectangular coordinate system, a positive point charge q = 6.50 nC is placed at the point x=0.190 m ,y=0, and an identical

son4ous [18]

Answer:

A) x and y components of the electric field (Ep) at the origin.

Epx = -1620.5 N/C

Epy = -1620.5 N/C

Ep = (1620.5(-i)+1620.5(-j)) N/C

B) Magnitude of the electric field (Ep) at the origin.

Ep= 2291.7 N/C

Explanation:

Conceptual analysis

The electric field at a point P due to a point charge is calculated as follows:

E = k*q/d²

E: Electric field in N/C

q: charge in Newtons (N)

k: electric constant in N*m²/C²

d: distance from charge q to point P in meters (m)

Equivalence

1nC= 10⁻⁹C

Data

K= 9x10⁹N*m²/C²

q₁ = q₂= +6.5nC=+6.5 *10⁻⁹C

d₁=d₂=0.190m

Graphic attached

The attached graph shows the field due to the charges:

Ep₁ : Electric Field at point P (x=0, y=0) due to charge q₁. As the charge q₁ is positive (q₁+) ,the field leaves the charge.

Ep₂: Electric Field at point P (x=0, y=0) due to charge q₂. As the charge q₂ is positive (q₂+) ,the field leaves the charge

Ep: Total field at point P due to charges q₁ and q₂.

Because q₁ = q₂ and d₁ = d₂, then, the magnitude of Ep₁ is equal to the magnitude of Ep₂

Ep₁ = Ep₂ = k*q/d² = 9*10⁹*6.5*10⁻⁹/0.190m² = 1620.5 N/C

Look at the attached graphic :

Epx = Ep₁= -1620.5 N/C

Epy = Ep₂= -1620.5 N/C

A) x and y components of the electric field (Ep) at the origin.

Ep = (1620.5(-i)+1620.5(-j)) N/C

B) Magnitude of the electric field (Ep) at the origin.

$E_{p} =\sqrt{1620.5^{2}+1620.5^{2} } = 2291.7 \frac{N}{C}$

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3 years ago