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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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The force F shown in Figure 4.30 has a moment of 40 Nm about the pivot. Calculate the magnitude

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$\boxed{\sf \tau=rFsin\theta}$

Put values

$\\ \rm\hookrightarrow 40=2Fsin40$

$\\ \rm\hookrightarrow Fsin40=20$

$\\ \rm\hookrightarrow 0.64F=20$

$\\ \rm\hookrightarrow F=31.25N$

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2 years ago

A small child has a wagon with a mass of 10 kilograms. The child pulls on the wagon with a force of 2 Newton’s. What is the acce

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3 years ago

The angle between the two force of magnitude 20N and 15N is 60 degrees (20N force being horizontal) determine the resultant in m

BARSIC [14]

A) The resultant force is 30.4 N at $25.3^{\circ}$

B) The resultant force is 18.7 N at $43.9^{\circ}$

Explanation:

In order to find the resultant of the two forces, we must resolve each force along the x- and y- direction, and then add the components along each direction to find the components of the resultant.

The two forces are:

$F_1 = 20 N$ at $0^{\circ}$ above x-axis

$F_2 = 15 N$ at $60^{\circ}$ above y-axis

Resolving each force:

$F_{1x}=F_1 cos \theta = (20)(cos 0)=20 N\\F_{1y}=F_1 sin \theta =(20)(sin 0)=0 N$

$F_{2x}=F_2 cos \theta = (15)(cos 60)=7.5 N\\F_{2y}=F_2 sin \theta =(15)(sin 60)=13.0 N$

So, the components of the resultant are:

$F_x = F_{1x}+F_{2x}=20+7.5 = 27.5 N\\F_y = F_{1y}+F_{2y}=0+13.0=13.0 N$

And the magnitude of the resultant is:

$F=\sqrt{F_x^2+F_y^2}=\sqrt{27.5^2+13.0^2}=30.4 N$

And the direction is:

$\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{13.0}{27.5})=25.3^{\circ}$

In this case, the 15 N is applied in the opposite direction to the 20 N force. Therefore we need to re-calculate its components, keeping in mind that the angle of the 15 N force this time is

$\theta=180^{\circ}-60^{\circ}=120^{\circ}$