All categories

3 years ago

Please help me with this question. Every help is appreciated.

Physics

1 answer:

Brilliant_brown [7]3 years ago

5 0

Answer:

Change in KE = +1.96×10^4 J while the change in ME = 0 J

You might be interested in

How far can a person run in 15 minutes if he runs at an average speed of 16 km/hr?

anygoal [31]

Answer:

4km

Explanation:

15 minutes is 1/4 of an hour.

1/4 of 16 is 4.

3 0

4 years ago

Calculate the (absolute) pressure at the bottom of a neighborhood swimming pool 30.0 m by 8.0 m whose uniform depth is 2.0 m. Th

Alika [10]

Answer:

Total pressure= 120945[Pa]

Force exerted = 29026800 [N] or 29.02*10^6 [N]

Explanation:

We know that the total pressure is the result of the sum of the atmospheric pressure plus the manometric pressure. The equation is:

$Ptotal=Patm + Pman$

In this problem we know the atmospheric pressure 101.325x10^3 [Pa], therefore we need to find the manometric pressure.

The manometric pressure in the bottom of the swimming pool depends only on the water column of water generated (depth of the swimming pool)

$Pman = density*g*h$

where:

density = density of the water 1000 [kg/m^3]

g= gravity [m/s^2]

h= column of water (meters)

replacing the values:

$Pman= 1000 *9.81* 2 = 19620 [Pa]\\\\$

The total pressure will be:

$Ptotal= 101325+19620 = 120945 [Pa]\\\\$

The force exerte on the bottom is defined by the following expression:

$Pressure=Force/area\\\\Force= Pressure*Area\\\\Area = 30m*8m= 240 m^2Force= 120945*240\\Force= 29026800N or 2958 Ton$

4 0

4 years ago

As a result of friction, the angular speed of a wheel changes with time according to dθ/dt = ω0e^−σt where ω0 and σ are constant

NNADVOKAT [17]

Hi there!

We can use the initial conditions to solve for w₀.

It is given that:

$\frac{d\theta}{dt} = w_0e^{-\sigma t}$

We are given that at t = 0, ω = 3.7 rad/sec. We can plug this into the equation:

$\omega(0)= \omega_0e^{-\sigma (0)}\\\\3.7 = \omega_0 (1)\\\\\omega_0 = \boxed{3.7 rad/sec}$

Now, we can solve for sigma using the other given condition:

$2 = 3.7e^{-\sigma (8.6)}\\\\.541 = e^{-\sigma (8.6)}\\\\ln(.541) = -\sigma (8.6)\\\\\sigma = \frac{ln(.541)}{-8.6} = \boxed{0.0714s^{-1}}$

The angular acceleration is the DERIVATIVE of the angular velocity function, so:

$\alpha(t) = \frac{d\omega}{dt} = -\sigma\omega_0e^{-\sigma t}\\\\\alpha(t) = -(0.0714)(3.7)e^{-(0.0714) (3)}\\\\\alpha(t) = \boxed{-0.213 rad\sec^2}$

The angular displacement is the INTEGRAL of the angular velocity function.

$\theta (t) = \int\limits^{t_2}_{t_1} {\omega(t)} \, dt\\\\\theta(t) = \int\limits^{2.5}_{0} {\omega_0e^{-\sigma t}dt\\\\$

$\theta(t) = -\frac{\omega_0}{\sigma}e^{-\sigma t}\left \| {{t_2=2.5} \atop {t_1=0}} \right.$

$\theta = -\frac{3.7}{0.0714}e^{-0.0714 t}\left \| {{t_2=2.5} \atop {t_1=0}} \right. \\\\\theta= -\frac{3.7}{0.0714}e^{-0.0714 (2.5)} + \frac{3.7}{0.0714}e^{-0.0714 (0)}$

$\theta = 8.471 rad$

Convert this to rev:

$8.471 rad * \frac{1 rev}{2\pi rad} = \boxed{1.348 rev}$

We can begin by solving for the time necessary for the angular speed to reach 0 rad/sec.

$0 = 3.7e^{-0.0714t}\\\\t = \infty$

Evaluate the improper integral:

$\theta = \int\limits^{\infty}_{0} {\omega_0e^{-\sigma t}dt\\\\$

$\lim_{a \to \infty} \theta = -\frac{\omega_0}{\sigma}e^{-\sigma t}\left \| {{t_2=a} \atop {t_1=0}} \right.$

$\lim_{a \to \infty} \theta = -\frac{3.7}{0.0714}e^{-0.0714a} + \frac{3.7}{0.0714}e^{-0.0714(0)}\\\\ \lim_{a \to \infty} \theta = \frac{3.7}{0.0714}(1) = 51.82 rad$

Convert to rev:

$51.82 rad * \frac{1rev}{2\pi rad} = \boxed{8.25 rev}$

8 0

3 years ago

The electrical energy that moves through a surge protector is 1.8 kilowatt-hours. How many watts of power will the surge protect

11Alexandr11 [23.1K]

The surge protector doesn't use energy. The current that carries Energy may pass through it but a good surge protector won't consume any of that energy. It'll just pass through the surge protector. The current that carrys the energy passes through it to be used by devices further on down the line, such as a toaster or a welder. If those devices are using 1.8 kilowatts of power, then that means that 6,480,000 Joules of energy pass through the surge protector in one hour. They all come out the other end and keep going.

4 0

4 years ago

Two ice skaters stand together. They push off and travel directly away from each other, the boy with a velocity of v = +0.35 m/s

DIA [1.3K]

Answer:

-0.55m/s

Explanation:

Given that: For the boy

Weight = 745N

Velocity = +0.35 m/s

Mass of the boy = ?

g = 9.81m/s^2

W = mg

745 = m×9.81

m = 75.94kg

For the girl

Given that:

Weight = 477 N

g = 9.81m/s^2

m = ?

W = mg

477 = m×9.81m/s^2

m = 48.62kg

To solve for the v of the girl, the two has to add up

48.62kg×v + 75.94kg×+0.35 m/s = 0

48.62v + 26.579 = 0

48.62v = - 26.579

v = -26.579/48.62

v = -0.5466

v = -0.55m/s

Hence, the velocity of the girl is -0.55m/s.

The negative sign is as a result of the two of them moving is opposite direction.

5 0

3 years ago