All categories

3 years ago

What important example, shown in the picture, is given regarding a real-life

Physics

1 answer:

dolphi86 [110]3 years ago

6 0

Answer:

Explanation:

There is no picture!

You might be interested in

How do you think car makers can design cars to limit cell phone distractions?

Dafna11 [192]

I have two (2) brilliant ideas:

1). Inside the metal that the body of the car is made of, and also between the two sheets of glass that the windows are made of, install a thin layer of material that absorbs RF (radio-wave) energy . . . like the material in the glass window of your microwave oven. Then, no radio waves from the cellular base station can get INTO the car, and no radio waves from your phone can get OUT of the car. The phone can't make a connection to the cellular network, you can't make or receive calls, and you can't connect to Instagram or Brainly, so you might as well just turn it off and save your battery until next time you're outside your car.

2). Somewhere inside the car, like under the dash or in the glove box, install a teeny tiny radio receiver that can recognize the signals coming OUT of your phone. Connect it to the car's electrical system so that when it hears signals from phones inside the car, it it shuts down the car's motor so you can't start or drive. The car only works when phones inside the car are either turned off or in Airplane Mode.

My ideas are so brilliant that I really should patent them, or copyright them, or whatever you do so that other people have to pay you to use your idea. But if you want to use them, that's OK. Just go ahead. I won't mind.

8 0

3 years ago

A student fills a tank of radius r with water to a height of h1 and pokes a small, 1.0 cm diameter hole at a distance h2 from th

Alik [6]

when a hole is made at the bottom of the container then water will flow out of it

The speed of ejected water can be calculated by help of Bernuolli's equation and Equation of continuity.

By Bernoulli's equation we can write

$Po + \frac{1}{2}\rho v_1^2 + \rho g h = Po + \frac{1}{2}\rho v_2^2 + \rho g *0$

Now by equation of continuity

$A_1v_1 = A_2v_2$

$\pi (0.2)^2 v_1 = \pi (0.01)^2 v_2$

from above equation we can say that speed at the top layer is almost negligible.

$v_1 = 0$

now again by equation of continuity

$\rho g h = \frac{1}{2} \rho v^2$

$v = \sqrt{2 g h}$

here we have

$h = h_1 - h_2$

$h = 0.50 - 0.03 = 0.47m$

now speed is given by

$v = \sqrt{2* 9.8 * 0.47}$

$v = 3.03 m/s$

7 0

3 years ago

The Great Sandini is a 60 kg circus performer who is shotfrom a cannon (actually a spring gun). You don't find many men ofhis ca

d1i1m1o1n [39]

Answer:

V=15.3 m/s

Explanation:

To solve this problem, we have to use the energy conservation theorem:

$U_e+K_i+U_{gi}+W_{friction}=K_f+U_{gf}$

the elastic potencial energy is given by:

$U_e=\frac{1}{2}*k*x^2\\U_e=\frac{1}{2}*1100N/m*(4m)^2\\U_e=8800J$

The work is defined as:

$W_{friction}=F_f*d*cos(\theta)\\W_{friction}=40N*2.5m*cos(180)\\W_{friction}=-100J$

this work is negative because is opposite to the movement.

The gravitational potencial energy at 2.5 m aboves is given by:

$U_{gf}=m*g*h\\U_{gf}=60kg*9.8*2.5\\U_{gf}=1470J$

the gravitational potential energy at the ground and the kinetic energy at the begining are 0.

$8800J+0+0+-100J=\frac{1}{2}*62kg*v^2+1470J\\v=\sqrt{\frac{2(8800J-100J-1470J)}{62kg}}\\v=15.3m/s$

3 0

3 years ago

What is #6<br><br> IM GIVING 40 POINTS

frosja888 [35]

Instantaneous velocity, on the other hand, describes the motion of a body at one particular moment in time. Acceleration is a vector which shows the direction and magnitude of changes in velocity. Its standard units are meters per second per second, or meters per second squared. (this is for number 3)

4 0

3 years ago

A loaded 375 kg toboggan is traveling on smooth horizontal snow at 4.50 m/s when it suddenly comes to a rough region. The region

zmey [24]

Answer:

a) The average friction force exerted on the toboggan is 653.125 newtons, b) The rough region reduced the kinetic energy of the toboggan in 92.889 %, c) The speed of the toboggan is reduced in 73.333 %.

Explanation:

a) Given the existence of non-conservative forces (friction between toboggan and ground), the motion must be modelled by means of the Principle of Energy Conservation and the Work-Energy Theorem, since toboggan decrease its speed (associated with due to the action of friction. Changes in gravitational potential energy can be neglected due to the inclination of the ground. Then:

$K_{1} = K_{2} + W_{f}$

Where:

$K_{1}$ , $K_{2}$ are the initial and final translational kinetic energies of the tobbogan, measured in joules.

$W_{f}$ - Dissipated work due to friction, measured in joules.

By applying definitions of translation kinetic energy and work, the expression described above is now expanded and simplified:

$f\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2})$

Where:

$f$ - Friction force, measured in newtons.

$\Delta s$ - Distance travelled by the toboggan in the rough region, measured in meters.

$m$ - Mass of the toboggan, measured in kilograms.

$v_{1}$ , $v_{2}$ - Initial and final speed of the toboggan, measured in meters per second.

The friction force is cleared:

$f = \frac{m\cdot (v_{1}^{2}-v_{2}^{2})}{2\cdot \Delta s}$

If $m = 375\,kg$ , $v_{1} = 4.50\,\frac{m}{s}$ , $v_{2} = 1.20\,\frac{m}{s}$ and $\Delta s = 5.40 \,m$ , then:

$f = \frac{(375\,kg)\cdot \left[\left(4.50\,\frac{m}{s} \right)^{2}-\left(1.20\,\frac{m}{s}\right)^{2}\right]}{2\cdot (5.40\,m)}$

$f = 653.125\,N$

The average friction force exerted on the toboggan is 653.125 newtons.

b) The percentage lost by the kinetic energy of the tobbogan due to friction is given by the following expression, which is expanded and simplified afterwards:

$\% K_{loss} = \frac{K_{1}-K_{2}}{K_{1}}\times 100\,\%$