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1 year ago

Which part or parts of the light bulb must be touched for the light bulb to light?

1 answer:

3 0

<u>The tip of the bulb touches the </u>negative end<u> of the battery</u> parts of the light bulb must be touched for the light bulb to light.

What is negative end?

The grounded connection of a car battery is called a negative terminal, and it is typically identified by a minus (-) sign. Typically, it is black and color-coded. The other terminal is a positive terminal that is typically color-coded red or orange and denoted by a plus (+) sign. The grounded terminal attaches to an object that is grounded on the car, usually a bolt in the chassis. The starter motor, as well as other switches, relays, and accessories, are connected to the positive terminal. The car won't start if the battery isn't properly grounded. These days, the majority of car batteries are packed with a charged acid solution that is very explosive and gaseous. In order to keep the battery's inner plates immersed, distilled water may occasionally need to be supplied to the fill-caps.

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A scuba diver measures an increase in pressure of around 10^5 Pa upon descending by 10 m, what is the change in force per square

IceJOKER [234]

Answer:

Explanation:

Given that, .

Pressure around scuba is

P = 10^5 Pa

1 Pa = 1 N/m²

Then

P = 10^5 N/m²

Descending height

h = 10m

Change in force per unit square centimetre

We know that,

Pressure = Force / Area

Then,

Force / Area is the required question we are finding

Then,

Force / Area = 10^5 N / m²

So, let convert the m² to cm²

100cm = 1m

(100cm)² = (1m)²

10⁴cm² = 1m²

Then,

Force / Area = 10^5 N/m² × 1m² / 10⁴cm²

Force / Area = 10 N/cm²

So, the force per unit square centimeters is 10.

7 0

2 years ago

Help me with the question in the image I provided

inna [77]

Answer: 10 s, 30 m/s , 150 m

Explanation:

Given

The speed of motorcyclist is $u_1=15\ m/s$

The initial speed of a police motorcycle is $u_2=0\ m/s$

acceleration of police motorcycle is $a=3\ m/s^2$

Police will catch the motorcyclist when they traveled equal distances

distance traveled by motorcyclist in time t is

$\Rightarrow s_1=15\times t$

Distance traveled by Police in time t is

$\Rightarrow s_2=u_2t+\frac{1}{2}at^2\\\Rightarrow s_2=0+0.5\times 3\times t^2$

put $s_1=s_2$

$\Rightarrow 15t=0.5\times 3\times t^2\\\\\Rightarrow t(1.5t-15)=0\\\\\Rightarrow t=\dfrac{15}{1.5}=10\ s$

Police officer's speed at that time is

$\Rightarrow v_2=u_2+at\\\Rightarrow v_2=0+3\times 10=30\ m/s$

Distance traveled by each vehicle is

$\Rightarrow s_1=s_2=15\times t=15\times 10=150\ m$

8 0

2 years ago

You pick up a 3.4-kg can of paint from the ground and lift it to a height of 1.8 m. (a) how much work do you do on the can of pa

MariettaO [177]

(a) For the work-energy theorem, the work done to lift the can of paint is equal to the gravitational potential energy gained by it, therefore it is equal to

$W=mg\Delta h$

where m=3.4 kg is the mass of the can, g=9.81 m/s^2 is the gravitational acceleration and $\Delta h=1.8 m$ is the variation of height. Substituting the numbers into the formula, we find

$W=(3.4 kg)(9.81 m/s^2)(1.8 m)=60.0 J$

(b) In this case, the work done is zero. In fact, we know from its definition that the work done on an object is equal to the product between the force applied F and the displacement:

$W=Fd$

However, in this case there is no displacement, so d=0 and W=0, therefore the work done to hold the can stationary is zero.

(c) In this case, the work done is negative, because the work to lower the can back to the ground is done by the force of gravity, which pushes downward. Its value is given by the same formula used in part (a):

$W=mg \Delta h=(3.4 kg)(9.81 m/s^2)(-1.8 m)=-60.0 J$