All categories

3 years ago

9 10

Physics

1 answer:

Alchen [17]3 years ago

8 0

The type of circuit the situation models is: C) a parallel circuit because there is more than one path

You might be interested in

Use dimensional analysis to determine how the linear acceleration a in m/s2 of a particle traveling in a circle depends on some,

astraxan [27]

I have to say, i love this kind of problems.

So, we got the linear acceleration, as we all know, linear the acceleration its, in dimensional units:

$[a]=[\frac{distance}{time^2}]$ .

Now, we got the radius

$[r] = [distance]$

the angular frequency

$[\omega] = \frac{1}{s}$

and the mass

$[m]=[mass]$ .

Now, the acceleration doesn't have units of mass, so it can't depend on the mass of the particle.

The distance in the acceleration has exponent 1, and so does in the radius. As the radius its the only parameter that has units of distance, this means that the radius must appear with exponent 1. Lets write

$a \propto r$ .

The time in the acceleration has exponent -2 As the angular frequency its the only parameter that has units of time, this means that the angular frequency must appear, but, the angular frequency has an exponent of -1, this means it must be squared

$a \propto r \omega^2$ .

We are almost there. If this were any other problem, we would write:

$a = A r \omega^2$

where A its an dimensionless constant. Its common for this constants to appears if we need an conversion factor. If we wanted the acceleration in cm/s^2, for example. Luckily for us, the problem states that there is no dimensionless constant involved, so:

$a = r \omega^2$

5 0

3 years ago

A load of 250 kg is hung by a crane's cable. The load is pulled by a horizontal force such that the cable makes a 30 angle to th

Anna [14]

Answer:

1250 $\sqrt{3}$

Explanation:

Let's look at this in a simple manner, because it is.

The crane weights 250Kg. Okay.

Since it is hung, there is the acceleration of gravity being applied on it (10m/s²)

Since F = m * a

F = 250 * 10

F = 2500

Now we know that the downward Force is 2500N.

To find the force that is being applied on that 30° angle, we can multiply our 2500N by cos30°, which happens to be $\frac{\sqrt{3}}{2}$ .

Therefore, the force pulling the box in the cable's direction is:

$\frac{2500\sqrt{3} }{2} = 1250\sqrt{3}$

If you have any questions feel free to comment.

Have a great one and mark brainliest if it helped, please

8 0

2 years ago

a boy throws a small stone into a pond. waves spread out from where the stone hits the water and travel to the side of the pond.

stich3 [128]

Answer: 1.6Hz

foe[vqefmvkeqmvkevkefmvqelkfveklveqv

4 0

3 years ago

A ball rolls off the edge of a horizontal roof at 10 m/s. 2 seconds later, the speed of the ball will be? a. 10 m/s b. 19.6m/s c

disa [49]

The ball rolls horizontally thus it is not affected by the gravitational acceleration. We assume that velocity is constant since we are not given acceleration. Therefore, acceleration is zero. The speed of the ball after 2 seconds will still be the same given we have a constant velocity system. Option A is the answer.

6 0

4 years ago

Read 2 more answers

A spring with a pointer attached is hanging next to a scale marked in millimeters. Three different packages are hung from the sp

SOVA2 [1]

solution;

$the expression for force applied on the spring due to the load is\\f=k\Delta x\\here,\Delta x is the extension in the spring due to appling force\\given three case as following\\110N=k(40-x_{o})----------1\\240N=k(60-x_{o})----------2\\w=k(30-x_{o})-------------3\\$ $To calculate the accrual length of the spring,solve th equation 1 and 2\\\frac{110N}{240N}=\frac{k(40-x_{o})}{k(60-x_{o})}\\0.458=\frac{k(40-x_{o})}{k(60-x_{o})}\\0.458(60mm-x_{o})=(40mm-x_{o})\\x_{o}(1-0.458)=(40-60(0.458))mm\\x_{o}\frac{12.52}{0.542}\\=23.1mm\\to calculate the force on the spring in case,\\solve the equation 1 and 2\\\frac{110}{w}=\frac{k(40-x_{o})}{k(60-x_{o})}\\\frac{110}{w}=\frac{(40mm-23.1mm)}{30mm-23.1mm}\\w=\frac{110}{2.45}=44.9N$

8 0

3 years ago