25 POINTS In outerspace, do Newton's laws still apply? A) yes B) No

To simulate deafness, put ear plugs or cotton balls into your ears and put your hands over your ears to block out all sound. Des

yulyashka [42]

Just so you know, total deafness is quite different than partial deafness.

Partial deafness can be any number of things. It can be an inability to make out the contents of high notes (notes with high frequencies and few harmonics).

Partial deafness can be the inability to hear low notes. I only know one person who suffers from that.

Your response to speech is quite different. Usually you have to have people look right at you. Lip reading is an art. Most deaf people between what they partially hear and lip reading can make out about 1/2 of what is said. Guessing usually takes care of the rest. Course you can get everything all muddled by guessing.

It does change your way of life. The TV is often turned up high enough that people living 4 houses down can list to the 6 o'clock news when you do.

Music is hard to make out. The overtones are lost.

Earphones have very tinny sound.

People are not as enjoyable as they once were.

6 0

3 years ago

An ideal transformer has 60 turns on its primary coil and 300 turns on its secondary coil. If 120 V at 2.0 A is applied to the p

Taya2010 [7]

Explanation:

It is given that,

Number of turns in primary coil, $N_p=60$

Number of turns in secondary coil, $N_s=300$

Voltage in primary coil, $V_p=120\ V$

Current in primary coil, $I_p=2\ A$

We have to find the voltage and current in the secondary coli. Firstly calculating the voltage in secondary coil as :

$\dfrac{N_p}{N_s}=\dfrac{V_p}{V_s}$

$\dfrac{60}{300}=\dfrac{120}{V_s}$

$V_s=720\ Volts$

Now, calculating the current present in the secondary coil as :

$\dfrac{N_p}{N_s}=\dfrac{I_s}{I_p}$

$\dfrac{60}{300}=\dfrac{I_s}{2\ A}$

$I_s=0.4\ A$

Hence, this is the required solution.

3 0

3 years ago

Cylindrical beaker of height 0.100 mm and negligible weight is filled to the brim with a fluid of density rhorhorho = 890 kg/m3k

Nesterboy [21]

Incomplete part of the question

A ball of density ρb = 5000 kg/m3 and volume V = 60.0 cm3 is then submerged in the fluid, so that some of the fluid spills over the side of the beaker. The ball is held in place by a stiff rod of negligible volume and weight. Throughout the problem, assume the acceleration due to gravity is g = 9.81 m/s2 . What is the weight Wb of the ball? Express your answer numerically in newtons.

What is the reading W2 of the scale when the ball is held in this submerged position? Assume that none of the water that spills over stays on the scale. Calculate your answer from the quantities given in the problem and express it numerically in newtons.

What is the force Fr applied to the ball by the rod? Take upward forces to be positive (e.g., if the force on the ball is downward, your answer should be negative). Express your answer numerically in newtons.

The rod is now shortened and attached to the bottom of the beaker. The beaker is again filled with fluid, the ball is submerged and attached to the rod, and the beaker with fluid and submerged ball is placed on the scale.

What weight W3 does the scale now show?

Answer:

(a) 2.94 N

(b) 1 N

(c) 2.42 N

(d) 3.42 N

Explanation:

(a)

From the definition of density, it's mass per unit volume hence mass is a product of density and volume. To get weight, we multiply mass by acceleration due to gravity

The weight of the ball is $W=\rho g V$

Where $\rho$ is the density, V is volume and g is acceleration due to gravity

Substituting density for 5000 Kg/m3 and g for 9.8 m/s2 and v for 0.00006 m3 then

$W= 5000 kg/m^{3} * 9.8 m/s^{2} * 0.00006 m^{3}=2.94 N$

(b)

Because the ball is being held up mostly by the rod, the fluid pressure on the bottom of the cylinder is just the same as before.

The scale does not "know" the ball is there at all.

That's why it still reads 1 N.

Therefore, the reading is 1 N

(c)

The buoyant force of the fluid on the ball is equal to the weight of the displaced fluid, namely,

$890 kg/m^{3} * 9.8 m/s^{2} * 0.00006 m^{3} = 0.52 N$

so the force needed for the rod to hold up the ball is 2.94 N - 0.52 N = 2.42 N.

(d)

Now the scale "feels" the weight of the ball,

so the scale reads the weight of the ball

PLUS the weight of the original fluid

MINUS the weight of the fluid that was displaced

= 2.94 N + 1.00 N - 0.52 N = 3.42 N

6 0

3 years ago

PLEASE HELP, THANK YOU!.. :)

Firlakuza [10]

Answer:

First Reaction:

$^{234}U$ => $^{230} Th + ^{4}He$

Second Reaction:

$^{230} Th$ => $^{226} Ra + ^{4}He$

Combined Reaction:

$^{234} U$ => $^{226}Ra + 2( ^{4} He)$

8 0

3 years ago

Someone plzzz help it’s my test it’s 50 summative points I’ll mean a lot of you can help thank youuu!!

DaniilM [7]

Answer:if you go to g○○gle there's a camera icon next to the search bar, click on that and take a photo of your test, then they'll find the answers that other people answered or needed, hope that helps

Explanation:

8 0

3 years ago

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