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2 years ago

Write one use of magnetic force and frictional force

Physics

1 answer:

grandymaker [24]2 years ago

5 0

Answer:

<em>1.Magnetic Force - Used in cranes to separate metal junk from junkyards.</em>

<em>2.Frictional Force- Helps us to walk.</em>

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Calculate the Potential Energy of an object that has a mass of 14-kg and is located at a height of 24-m?

inessss [21]

Potential Energy= 24m * 14kg * 9.8N/kg = 3292.8J

4 0

3 years ago

A car initially traveling at 24 m/s slams on the brakes and moves forward 196 m before coming to a complete halt. What was the m

Marrrta [24]

Answer:

$-1.47 m/s^2$

Explanation:

We can use the following SUVAT equation to solve the problem:

$v^2 - u^2 = 2ad$

where

v = 0 is the final velocity of the car

u = 24 m/s is the initial velocity

a is the acceleration

d = 196 m is the displacement of the car before coming to a stop

Solving the equation for a, we find the acceleration:

$a=\frac{v^2-u^2}{2d}=\frac{0-(24)^2}{2(196)}=-1.47 m/s^2$

4 0

3 years ago

A particle with a mass of 0.500 kg is attached to a horizontal spring with a force constant of 50.0 N/m. At the moment t = 0, th

svp [43]

a) $x(t)=2.0 sin (10 t) [m]$

The equation which gives the position of a simple harmonic oscillator is:

$x(t)= A sin (\omega t)$

where

A is the amplitude

$\omega=\sqrt{\frac{k}{m}}$ is the angular frequency, with k being the spring constant and m the mass

t is the time

Let's start by calculating the angular frequency:

$\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{50.0 N/m}{0.500 kg}}=10 rad/s$

The amplitude, A, can be found from the maximum velocity of the spring:

$v_{max}=\omega A\\A=\frac{v_{max}}{\omega}=\frac{20.0 m/s}{10 rad/s}=2 m$

So, the equation of motion is

$x(t)= 2.0 sin (10 t) [m]$

b) t=0.10 s, t=0.52 s

The potential energy is given by:

$U(x)=\frac{1}{2}kx^2$

While the kinetic energy is given by:

$K=\frac{1}{2}mv^2$

The velocity as a function of time t is:

$v(t)=v_{max} cos(\omega t)$

The problem asks as the time t at which U=3K, so we have:

$\frac{1}{2}kx^2 = \frac{3}{2}mv^2\\kx^2 = 3mv^2\\k (A sin (\omega t))^2 = 3m (\omega A cos(\omega t))^2\\(tan(\omega t))^2=\frac{3m\omega^2}{k}$

However, $\frac{m}{k}=\frac{1}{\omega^2}$ , so we have

$(tan(\omega t))^2=\frac{3\omega^2}{\omega^2}=3\\tan(\omega t)=\pm \sqrt{3}\\$

with two solutions:

$\omega t= \frac{\pi}{3}\\t=\frac{\pi}{3\omega}=\frac{\pi}{3(10 rad/s)}=0.10 s$

$\omega t= \frac{5\pi}{3}\\t=\frac{5\pi}{3\omega}=\frac{5\pi}{3(10 rad/s)}=0.52 s$

c) 3 seconds.

When x=0, the equation of motion is:

$0=A sin (\omega t)$

so, t=0.

When x=1.00 m, the equation of motion is:

$1=A sin(\omega t)\\sin(\omega t)=\frac{1}{A}=\frac{1}{2}\\\omega t= 30\\t=\frac{30}{\omega}=\frac{30}{10 rad/s}=3 s$

So, the time needed is 3 seconds.

d) 0.097 m

The period of the oscillator in this problem is:

$T=\frac{2\pi}{\omega}=\frac{2\pi}{10 rad/s}=0.628 s$

The period of a pendulum is:

$T=2 \pi \sqrt{\frac{L}{g}}$

where L is the length of the pendulum. By using T=0.628 s, we find

$L=\frac{T^2g}{(2\pi)^2}=\frac{(0.628 s)^2(9.8 m/s^2)}{(2\pi)^2}=0.097 m$

5 0

3 years ago

Nick is so excited to Trick-or-Treat, he RUNS down the street to the house that has the BIG candy bars! He ran 203 meters. It to

sergiy2304 [10]

Answer:

it would take him 1 minute to run 304.5 meters and 1 second to run 5.075 meters

Explanation:

8 0

2 years ago

Read 2 more answers

oil is flowing through a tube that has two different cross-sectional areas. at position a where the radius of the tube is 7.0 cm

AleksAgata [21]

According to the continuity equation, the rate at which mass enters the system equals the rate at which mass exits in any steady state process.

An equation that explains the movement of a particular quantity is a continuity equation, also known as a transport equation. Although it can be applied generally to any significant quantity, it is extremely simple and useful when used with preserved quantities.

The radius is seven centimeters, and the mass flow rate is 0 to 5 kg/s. Find the mass flow rate at a point with a 3.5 cm radius. We can consequently deduce that based on the equation. As we all know, the mass flow rate is constant.

If the rate of mass entering and leaving the system is equal, the rate of mass leaving the system should be processed.

The mass flow rate air section A and the mass flow rated section B are equivalent, according to the continuity equation. Mass flow rate in section B is therefore 0.02, or five kilograms per second.

To know more about continuity equation, click on the link below:

brainly.com/question/19566865

#SPJ4

5 0

1 year ago