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3 years ago

Please help!

Physics

2 answers:

mariarad [96]3 years ago

4 0

It's hard to tell how the choices are laid out, and they have no labels.

The parallel circuits are the two circuits in your second picture.

One of them is the one with 3 switches, 2 batteries, and 2 light bulbs.

The other one is in the same picture. It's the one where all the circuit components ... the battery and 3 resistors ... are in the 4 vertical lines.

netineya [11]3 years ago

3 0

Answer:

The two circuits in the second picture.

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Three strings, attached to the sides of a rectangular frame, are tied together by a knot as shown in the figure. The magnitude o

Anna [14]

The magnitude of the tension in the string marked A is 52.5N

Generally, the equation for is mathematically given as

Let's take θ be an angle at A

So, tanθ = 3/8

Let's take α be an angle at B (Below X)

tanα = 5/4

Let's take β be an angle at C (Below x)

tanβ = 1/6

First we take the Horizontal Components

74.9cos(9.46°) = Acos(20.6°) + Bcos(51.3°)

By solving the equation, we get

A = 78.9 - 0.668B … (1)

Now, we take the vertical components

74.9sin(9.46°) + Asin(20.6°) = Bsin(51.3°)

By solving the equation, we get

40.07 = 1.015B

B = 39.5N

By substituting the value of B in equation (1)

A = 78.9 - 0.6668× 39.5

A = 52.5N

Hence, the magnitude of the tension in the string marked A is 52.5N

Learn more about Tension here brainly.com/question/2287912

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2 years ago

Friction provides the force needed for a car to travel around a flat, circular race track. What is the maximum speed at which a

blagie [28]

Answer:

The maximum speed at which the car can safety travel around the track is 18.6m/s.

Explanation:

Since the car is in circular motion, there has to be a centripetal force $F_c$ . In this case, the only force that applies for that is the static frictional force $f_s$ between the tires and the track. Then, we can write that:

$f_s=F_c$

And since $f_s\leq \mu N$ and $F_c=\frac{mv^{2}}{r}$ , we have:

$\mu N\geq \frac{mv^{2}}{r}$

Now, if we write the vertical equation of motion of the car (in which there are only the weight and the normal force), we obtain:

$N-mg=0\\\\\\implies N=mg$

Substituting this expression for $N$ and solving for $v$ , we get:

$\mu mg\geq \frac{mv^{2}}{r}\\\\v\leq \sqrt{\mu gr}$

Finally, plugging in the given values for the coefficient of friction and the radius of the track, we have:

$v\leq \sqrt{(0.42)(9.81m/s^{2})(84.0m)}\\\\v\leq 18.6m/s$

It means that in its maximum value, the speed of the car is equal to 18.6m/s.

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