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2 years ago

This circuit has five bulbs (A, B, C, D, and E). Select which bulbs would turn off if bulb C was unscrewed.

Physics

1 answer:

LiRa [457]2 years ago

4 0

Answer:

C, A and E

Explanation:

In theory you always assume that the prostive is going to the negtive and since C, A and E are in series they all will turn off will turn off

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A book on a 2-meter high shelf has a mass of 0.4 kg. What is its potential energy?

poizon [28]

Answer:

$\boxed {\boxed {\sf 7.84 \ Joules}}$

Explanation:

The formula for potential energy is:

$PE=m*g*h$

where m is the mass, g is the gravitational acceleration, and h is the height.

The mass of the book is 0.4 kilograms. The gravitational acceleration on Earth is 9.8 m/s². The height of the book is 2 meters.

$m=0.4 \ kg \\g=9.8 \ m/s^2 \\h=2\ m$

Substitute the values into the formula.

$PE=(0.4 \ kg)(9.8 \ m/s^2)(2 \ m)$

Multiply the first two numbers.

0.4 kg*9.8 m/s²= 3.92 kg*m/s²
If we convert the units now, the problem will be much easier later on.
1 kg*m/s² is equal to 1 Newton. So, our answer of 3.92 kg*m/s² is equal to 3.92 N

$PE=(3.92 \ N )(2 \ m)$

Multiply.

3.92 N* 2 m=7.84 N*m
1 Newton meter is equal to 1 Joule (this is why we converted the units).
Our answer is equal to 7.84 Joules.

$PE=7.84 \ J$

6 0

2 years ago

Read 2 more answers

Which state of matter has the MOST energy?

Nonamiya [84]

Answer:

Gas

Explanation:

The gas state of matter has the most energy because of how freely the molecules move

4 0

2 years ago

Lakes

Nataliya [291]

I think is between A&B
I think I would answer B

6 0

2 years ago

Force → F = ( − 8.0 N ) ˆ i + ( 6.0 N ) ˆ j acts on a particle with position vector → r = ( 3.0 m ) ˆ i + ( 4.0 m ) ˆ j . What a

andrey2020 [161]

Explanation:

It is given that,

Force, $F=(-8\ N)i+(6\ N)j$

Position vector, $r=(3i+4j)\ m$

(a) The torque on the particle about the origin is given by :

$\tau=F\times r\\\\\tau=(-8i+6j)\times (3i+4j)\\\\\tau=(-50k)\ N-m$

(b) To find the angle between r and F use dot product formula as :

$F{\cdot} r=|F||r|\ \cos\theta\\\\\cos\theta=\dfrac{F{\cdot} r}{|F| |r|}\\\\\cos\theta=\dfrac{(-8i+6j){\cdot} (3i+4j)}{\sqrt{(-8)^2+6^2} \times \sqrt{3^2+4^2} }\\\\\cos\theta=\dfrac{-24+24}{\sqrt{(-8)^2+6^2} \times \sqrt{3^2+4^2} }\\\\\cos\theta=0\\\\\theta=90^{\circ}$

Hence, this is the required solution.

8 0

2 years ago

Read 2 more answers

A box rests on top of a flat bed truck. The box has a mass of m = 16.0 kg. The coefficient of static friction between the box an

3241004551 [841]

Answer:

1) 1.31 m/s2

2) 20.92 N

3) 8.53 m/s2

4) 1.76 m/s2

5) -8.53 m/s2

Explanation:

1) As the box does not slide, the acceleration of the box (relative to ground) is the same as acceleration of the truck, which goes from 0 to 17m/s in 13 s

$a = \frac{\Delta v}{\Delta t} = \frac{17 - 0}{13} = 1.31 m/s2$

2)According to Newton 2nd law, the static frictional force that acting on the box (so it goes along with the truck), is the product of its mass and acceleration

$F_s = am = 1.31*16 = 20.92 N$

3) Let g = 9.81 m/s2. The maximum static friction that can hold the box is the product of its static coefficient and the normal force.

$F_{\mu_s} = \mu_sN = mg\mu_s = 16*9.81*0.87 = 136.6N$

So the maximum acceleration on the block is

$a_{max} = F_{\mu_s} / m = 136.6 / 16 = 8.53 m/s^2$

4)As the box slides, it is now subjected to kinetic friction, which is

$F_{\mu_s} = mg\mu_k = 16*9.81*0.69 = 108.3 N$

So if the acceleration of the truck it at the point where the box starts to slide, the force that acting on it must be at 136.6 N too. So the horizontal net force would be 136.6 - 108.3 = 28.25N. And the acceleration is

28.25 / 16 = 1.76 m/s2

5) Same as number 3), the maximum deceleration the truck can have without the box sliding is -8.53 m/s2

3 0

3 years ago