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A water-skier is being pulled by a tow rope attached to a boat. As the driver pushes the throttle forward, the skier accelerates. A 76.2-kg water-skier has an initial speed of 5.0 m/s. Later, the speed increases to 10.4 m/s. Determine the work done by the net external force acting on the skier.

Answers

Answer 1

Answer:

Answer:

Work done will be equal to 3186.396 J

Explanation:

We have mass m = 76.2 kg

Initial velocity u = 5 m/sec

Final velocity v = 10.4 m/sec

We have to find the work done

From work energy theorem work done is equal to change in kinetic energy

$w=(1)/(2)mv^2-(1)/(2)mu^2$

$w=(1)/(2)* 76.2* 10.4^2-(1)/(2)* 76.2* 5^2$

w = 3168.396 J

So work done will be equal to 3186.396 J

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Lightning can sometimes occur on hot and humid summer evenings when there are no thunderstorms.(A) True(B) False

4 A wheel starts from rest and has an angular acceleration of 4.0 rad/s2. When it has made 10 rev determine its angular velocity.]

Answers

The rate of change of angulardisplacement is defined as angular velocity. The angular velocity will be 22.41rad/s.

What is angular velocity?

The rate of change of angular displacement is defined as angular velocity. Its unit is rad/sec.

ω = θ t

Where,

θ is the angle of rotation,

tis the time

ω is the angular velocity

The given data in the problem is;

u is the initialvelocity=0

α is the angularacceleration = 4.0 rad/s²

t is the time period=

n is the number of revolution = 10 rev

From Newton's second equation of motion in terms of angular velocity;

$\rm \omega_f^2 - \omega_i^2 = 2as \n\n \rm \omega_f^2 - 0 = 2* 4 * 62.83 \n\n \rm \omega_f= 22.41 \ rad/sec$

Hence the angular velocity will be 22.41 rad/s.

To learn more about angularvelocity refer to the link

brainly.com/question/1980605

Answer:

$w_f$ = 22.41rad/s

Explanation:

First, we know that:

a = 4 rad/s^2

S = 10 rev = 62.83 rad

Now we know that:

$w_f^2-w_i^2=2aS$

where $w_f$ is the final angular velocity, $w_i$ the initial angular velocity, a is the angular aceleration and S the radians.

Replacing, we get:

$w_f^2-(0)^2=2(4)(62.83)$

Finally, solving for $w_f$ :

$w_f$ = 22.41rad/s

The volume control on a stereo is designed so that three clicks of the dial increase the output by 10 dB. How many clicks are required to increase the power output of the loudspeakers by a factor of 100?

Answers

Answer:

300 clicks...

Explanation:

Output on 3 clicks = 10 dB

Increasing 10 by a factor of 100 equals 1000 dB so,

Its simple math, clicks will also increase in the same ratio and it shall take 300 clicks to increase the volume by a factor of 100.

Which of the following is not a risk associated with using legal drugs without medical supervision

Answers

Answer:

paying too much on the black market instead of getting a prescription

Explanation:

i just took the quiz

Answer:

Paying too much on the black market instead of getting a prescription

Explanation:

The rest of the options are risks associated with using legal drugs without medical supervision.

TsAn electric light bulb mixer is used for 19.2 seconds. In that time, it transfers 1
536 J of energy.
Calculate the power output of the cake mixer.
A
I DONT KNOW

Answers

And electric light bulb mix it is used for nine. Two seconds in that temperature transfers one 536 J of energy calculate shin is 2+2 is 4-1 that’s three quick maths exclamation is calculate the power work output of the kitchen

Two bullets of the same size, mass and horizontal velocity are fired at identical blocks, only one is made of steel and the other is made of rubber. The steel bullet has a perfectly inelastic collision with the block, while the rubber bullet has an elastic collision. Which bullet is more likely to knock over the block, or are both equally likely to do so? Justify your choice based on physics principles.

Answers

Answer and Explanation:

Since we're discussing shots, the significant thing is the way the energy is changed over as there is deceleration of the bullet to a halt when it hits something.

Kinetic Energy is relative to mass times speed squared, so in reality, the 2 cases given have practically indistinguishable Kinetic energy. The measure of energy is authoritative, so the two cases will do generally a similar harm given, obviously we look at situations when all the kinetic energy is spent.

One contrast that will be effectively obvious is that the weapon in the case of heavy bullet will recoil more.
One can consider energy assimilation as force times separation distance, and energy ingestion as a product of force and time.
Henceforth, the heavier yet more slow bullet with a similar energy will venture to every part of a similar separation in the engrossing material, but since of bigger force, will take a more drawn out time doing it.
It will along these lines, additionally, give a more noteworthy "kick" to the object that absorbs.

The atmosphere on Venus consists mostly of CO2. The density of the atmosphere is 65.0 kg/m3 and the bulk modulus is 1.09 x 107 N/m2. A pipe on a lander is 75.0 cm long and closed at one end. When the wind blows across the open end, standing waves are caused in the pipe (like blowing across the top of a bottle). a) What is the speed of sound on Venus? b) What are the first three frequencies of standing waves in the pipe?

Answers

Answer:

a. 409.5 m/s b. f₁ = 136.5 Hz, f₂ = 409.5 Hz, f₃ = 682.5 Hz

Explanation:

a. The speed of sound v in a gas is v = √(B/ρ) where B = bulk modulus and ρ = density. Given that on Venus, B = 1.09 × 10⁷ N/m² and ρ = 65.0 kg/m³

So, v = √(B/ρ)

= √(1.09 × 10⁷ N/m²/65.0 kg/m³)

= √(0.01677 × 10⁷ Nm/kg)

= √(0.1677 × 10⁶ Nm/kg)

= 0.4095 × 10³ m/s

= 409.5 m/s

b. For a pipe open at one end, the frequency f = nv/4L where n = mode of wave = 1,3,5,.., v = speed of wave = 409.5 m/s and L = length of pipe = 75.0 cm = 0.75 m

Now, for the first mode or frequency, n = 1

f₁ = v/4L

= 409.5 m/s ÷ (4 × 0.75 m)

= 409,5 m/s ÷ 3 m

= 136.5 Hz

Now, for the second mode or frequency, n = 2

f₂ = 3v/4L

= 3 ×409.5 m/s ÷ (4 × 0.75 m)

= 3 × 409,5 m/s ÷ 3 m

= 3 × 136.5 Hz

= 409.5 Hz

Now, for the third mode or frequency, n = 5