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Sphinxa [80]
2 years ago
12

U-238 has protons and146 neutrons. A particular isotope of plutonium has 94 protons, neutrons, and a mass number of 241. Thorium

- has 90 protons and 137 neutrons.
Physics
2 answers:
Vadim26 [7]2 years ago
6 0

Answer:it's 92+147+227

Explanation:

I just put the answer below and I got it wrong so my answers are right on edgen

enyata [817]2 years ago
5 0

#1

^{238}U

so mass number = 238

mass number = protons + neutrons

given that

neutrons = 146

238 = protons + 146

protons = 92

#2

^{241}Pu

so mass number = 241

mass number = protons + neutrons

given that

Protons = 94

241 = 94 + neutrons

neutrons = 147

#3

^ATh

A = mass number

Protons = 90

Neutrons = 137

A = protons + Neutrons

A = 90 + 137 = 227

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A ladder placed up against a wall is sliding down. The distance between the top of the ladder and the foot of the wall is decrea
Kitty [74]

Answer:

distance changing at rate of 3.94 inches/sec

Explanation:

Given data

wall decreasing at a rate = 9 inches per second

ladder L = 152 inches

distance  h = 61 inches

to find out

how fast is the distance changing

solution

we know that

h² + b² = L²   ..................1

h² + b² = 152²

Apply here derivative w.r.t. time

2h dh/dt + 2b db/dt = 0

h dh/dt + b db/dt = 0

db/dt = - h/b × dh/dt     .............2

and

we know

h = 61

so h² + b² = L²

61² + b² = 152²

b² = 19383

so b = 139.223

and we know dh/dt = -9 inch/sec

so from equation 2

db/dt = -61/139.223  (-9)

so

db/dt = 3.94 inches/sec

distance changing at rate of 3.94 inches/sec

3 0
2 years ago
As you know, loudspeakers are used for communication at sporting events, and in schools or supermarkets. Research loudspeakers o
Ivahew [28]

Sound is invisible, but sometimes we can feel it. when a kettle-drum is thumped with a stick that time the tight drum skin moving up and down very quickly for some time and then it pumped sound waves into the air. Loudspeakers also work in a similar manner.

components of a speaker:

Electromagnet

In order to translate an electrical signal into an audible sound, an electromagnet is used in looudspeakers

 Suspension

suspension is used to  center the voice coil in the gap of the magnet and it exerts a restoring force to keep it there. suspension is used to limits the maximum mechanical excursion of the diaphragm and voice coil.

voice Coil

Voice coil is a coil of wire which is attached to the apex of a loudspeaker cone. It provides the motive force to the cone by the reaction of a magnetic field to the current passing through it.

Cone

In a  loudspeaker, a  thin, semi-rigid membrane attached to the voice coil, which moves in a magnetic gap due to which it vibrates  and produced sound. It is known as diaphragm and It can also be called a cone, though not all speaker diaphragms are cone-shaped.

Surround

surround is also known as front suspension. surround is used to join the cone to the chassis. Together with the suspension it controls the cone excursion, and it is also used to  determine how energy which is  travelling through the cone is absorbed, and how the speaker limits when it reaches the ends of its travel.

Dustcap

dustcap is used to protect the voice coil from dust and dirt. It is a part of the cone.

How speaker work

The outer part of the cone is fastened to the outer part of the loudspeaker's circular metal rim. The inner part is fixed to  voice coil that sits just in front of a permanent magnet . When loudspeaker is hooked up to a stereo, electrical signals feed through the speaker cables into the coil. It will convert the coil into an electromagnet magnet . when electricity flows back and forth in the cables, the electromagnet either repels or attract the permanent magnet. This moves the coil back and forth, and it pull and push the loudspeaker cone. Like a drum skin vibrating back and forth, the moving cone pumps sounds out into the air.In this way loudspeakers works.

4 0
2 years ago
Greg walks on a straight road from his home to a convenience store 3.0 km away with a speed of 6.0 km/h. On reaching the store h
VladimirAG [237]

This question was apprently selected from the "Sneaky Questions" category.

The store is 3 km from his home, and he walks there with a speed of 6 km/hr.  So it takes him (3 km) / (6 km/hr)  =  1/2 hour to get to the store.

That's 30 minutes.  So the whole part-(a.) of the question refers to only that part of the trip, and we don't care what happens once he reaches the store.  

a). Over the first 30 minutes of his travel, Greg walks 3.0 km on a straight road, and he ends up 3.0 km away from where he started.

Average speed = (distance/time) = (3.0 km) / (1/2 hour) = <em>6.0 km/hr</em>

Average velocity = (displacement/time) = (3.0 km) / (1/2 hour) = <em>6.0 km/hr</em>

There's probably some more questions in part-(b.) where you'd need to use Greg's return trip to find the answers, but johnaddy210 is only asking us for part-(a.).

8 0
2 years ago
Read 2 more answers
A 38 kg crate rests on a floor. A horizontal pulling force of 170 N is needed to start the crate
Mandarinka [93]

Answer:

0.456033049

Explanation:

F=\mu N where N=mg hence

F=\mu mg where m is mass of object, g is acceleration due to gravity whose value is taken as 9.81 m/s^{2}, \mu is the coefficient of static friction and F is the applied force.

Making \mu the subject we obtain

\mu=\frac {mg}{N} and substituting m for 38 Kg, g for 9.81 m/s^{2} and 170 N for  F we obtain

\mu=\frac{170} {38*9.81}=0.456033049

Therefore, the coefficient of static friction is 0.456033049

5 0
2 years ago
A professor's office door is 0.89 m wide, 2.0 m high, and 4.0 cm thick; has a mass of 25 kg ; and pivots on frictionless hinges.
taurus [48]
In order to answer this question ... strange as it may seem ...
we only need one of those measurements that you gave us
that describe the door.

The door is hanging on frictionless hinges, and there's a torque
being applied to it that's trying to close it.  All we need to do is apply
an equal torque in the opposite direction, and the door doesn't move.

Obviously, in order for our force to have the most effect, we want
to hold the door at the outer edge, farthest from the hinges.  That
distance from the hinges is the width of the door ... 0.89 m.

We need to come up with 4.9 N-m of torque,
applied against the mechanical door-closer.

Torque is (force) x (distance from the hinge).

                                    4.9 N-m  =  (force) x (0.89 m) 

Divide each side by 0.89m:    Force = (4.9 N-m) / (0.89 m)

                                                             =  5.506 N .
7 0
2 years ago
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