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3 years ago

What type of circuit is illustrated

Physics

1 answer:

Varvara68 [4.7K]3 years ago

4 0

The bulbs are in parallel.

The picture doesn't show enough of the circuit to tell
whether it's also open or shorted. It may be either, or
it may be neither.

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A radio has a 1.3 A current. If it has a resistance of 35 Ω, what is the potential difference?

DedPeter [7]

Answer:

Explanation:

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3 years ago

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Where is the magnetic south pole compared to the geographical north pole?

Alex777 [14]

Currently, the magnetic south pole lies about ten degrees distant from the geographic north pole, and sits in the Arctic Ocean north of Alaska. The north end on a compass therefore currently points roughly towards Alaska and not exactly towards geographic north.

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3 years ago

suppose an electrically charged ruler transfers some of its charge by contact to a tiny plastic sphere. will the ruler and the s

poizon [28]

Answer:

Here ball and rod will repel each other as they are of similar charges

Explanation:

As we know that the two charges attract or repel each other by electrostatic force

This force is given as

$F = \frac{kq_1q_2}{r^2}$

so we know if two charges are similar in nature then they will repel each other and if the two charges are opposite in nature then they will attract each other

So here when rod touch the ball then it transfer its charge to the ball and due to similar charges in ball and rod they both repel each other

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3 years ago

What is the momentum of a 950 kg car moving at 10.0 m/s?

pentagon [3]

Answer:

<h3>The answer is 9500 kgm/s</h3>

Explanation:

The momentum of an object can be found by using the formula

<h3>momentum = mass × velocity</h3>

From the question

mass = 950 kg

velocity = 10.0 m/s

We have

momentum = 950 × 10

We have the final answer as

Hope this helps you

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3 years ago

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You pull straight up on the string of a yo-yo with a force 0.35 N, and while your hand is moving up a distance 0.16 m, the yo-yo

jarptica [38.1K]

Answer:

a) 0.138J

b) 3.58m/S

c) (1.52J)(I)

Explanation:

a) to find the increase in the translational kinetic energy you can use the relation

$\Delta E_k=W=W_g-W_p$

where Wp is the work done by the person and Wg is the work done by the gravitational force

By replacing Wp=Fh1 and Wg=mgh2, being h1 the distance of the motion of the hand and h2 the distance of the yo-yo, m is the mass of the yo-yo, then you obtain:

$Wp=(0.35N)(0.16m)=0.056J\\\\Wg=(0.062kg)(9.8\frac{m}{s^2})(0.32m)=0.19J\\\\\Delta E_k=W=0.19J-0.056J=0.138J$

the change in the translational kinetic energy is 0.138J

b) the new speed of the yo-yo is obtained by using the previous result and the formula for the kinetic energy of an object:

$\Delta E_k=\frac{1}{2}mv_f^2-\frac{1}{2}mv_o^2$

where vf is the final speed, vo is the initial speed. By doing vf the subject of the formula and replacing you get:

$v_f=\sqrt{\frac{2}{m}}\sqrt{\Delta E_k+(1/2)mv_o^2}\\\\v_f=\sqrt{\frac{2}{0.062kg}}\sqrt{0.138J+1/2(0.062kg)(2.9m/s)^2}=3.58\frac{m}{s}$

the new speed is 3.58m/s

c) in this case what you can compute is the quotient between the initial rotational energy and the final rotational energy

$\frac{E_{fr}}{E_{fr}}=\frac{1/2I\omega_f^2}{1/2I\omega_o^2}=\frac{\omega_f^2}{\omega_o^2}\\\\\omega_f=\frac{v_f}{r}\\\\\omega_o=\frac{v_o}{r}\\\\\frac{E_{fr}}{E_{fr}}=\frac{v_f^2}{v_o^2}=\frac{(3.58m/s)}{(2.9m/s)^2}=1.52J$

hence, the change in Er is about 1.52J times the initial rotational energy

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3 years ago

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