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Gekata [30.6K]
3 years ago
15

Which element's atoms have the greatest average number of neutrons ?

Physics
2 answers:
Mkey [24]3 years ago
6 0
I am pretty sure this is uranium. it has 140 neutrons. 
elena55 [62]3 years ago
3 0
That would be 180 neutrons hope i helped.
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You are designing a generator with a maximum emf 8.0 V. If the generator coil has 200 turns and a cross-sectional area of 0.030
shutvik [7]

Answer:

7.1 Hz

Explanation:

In a generator, the maximum induced emf is given by

\epsilon= 2\pi NAB f

where

N is the number of turns in the coil

A is the area of the coil

B is the magnetic field strength

f is the frequency

In this problem, we have

N = 200

A=0.030 m^2

\epsilon=8.0 V

B = 0.030 T

So we can re-arrange the equation to find the frequency of the generator:

f=\frac{\epsilon}{2\pi NAB}=\frac{8.0 V}{2\pi (200)(0.030 m^2)(0.030 T)}=7.1 Hz

4 0
3 years ago
Read 2 more answers
An object is thrown upwards with the speed of 40.0 km/hr in 5 second, calculate the distance in millimeters (mm).
sertanlavr [38]
V=(40km/hr)(hr/3600s)(1000000mm/km)
v=11111.1mm/s

v=d/t
d=vt
d=(11111.1mm/s)(5s)
d=55555mm
d=5.56x10^4mm
4 0
3 years ago
Think of the last five things you read. Now think of a type of reading you havent done recenty
GarryVolchara [31]

Answer:

1) Half Girlfriend

2) I am Malala

3) Diamond fire

4)I too had a love story

5)Your Dream are Now Mine

Explanation:

I don't have a same pattern to read but mostly of romantic and fiction. I read less autobiographies but when I read I am Malala , it was an inspiring one.My favorite type of reading is romantic types and least favorite is non fiction especially History

6 0
3 years ago
Marcia flew her ultralight plane to a nearby town against a head wind of 15 km/h in 2h 20 min. the return trip under the same wi
insens350 [35]

Let the distance between the towns be d and the speed of the air be s.

distance = speed * time

convert the minutes time into hours.

When flying into the wind, ground speed will be air speed MINUS wind speed, hence the against the wind trip is described by:

d

s−15

=

7

3

return trip is then :

d

s+15

=

7

5

Cross-multiplying both we get the two-variable system:

3d=7∗(s−15)5d=7∗(s+15)

3d=7s−1055d=7s+105

subtract first equation from second equation we get

2d=210d=105km

Substitute the value of d in the above equations for s.

5∗105=7s+1057s=420s=60km/hr

8 0
3 years ago
Read 2 more answers
A 70.0-kg person throws a 0.0480-kg snowball forward with a ground speed of 33.5 m/s. A second person, with a mass of 55.0 kg, c
saw5 [17]

Answer:

The final velocity of the thrower is \bf{3.88~m/s} and the final velocity of the catcher is \bf{0.029~m/s}.

Explanation:

Given:

The mass of the thrower, m_{t} = 70~Kg.

The mass of the catcher, m_{c} = 55~Kg.

The mass of the ball, m_{b} = 0.0480~Kg.

Initial velocity of the thrower, v_{it} = 3.90~m/s

Final velocity of the ball, v_{fb} = 33.5~m/s

Initial velocity of the catcher, v_{ic} = 0~m/s

Consider that the final velocity of the thrower is v_{ft}. From the conservation of momentum,

&& m_{t}v_{ft} + m_{b}v_{fb} = (m_{t} + m_{b})v_{it}\\&or,& v_{ft} = \dfrac{(m_{t} + m_{b})v_{it} - m_{b}v_{fb}}{m_{t}}\\&or,& v_{ft} = \dfrac{(70 + 0.0480)(3.90) - (0.0480)(33.5)}{70}\\&or,& v_{ft} = 3.88~m/s

Consider that the final velocity of the catcher is v_{fc}. From the conservation of momentum,

&& (m_{c} + m_{b})v_{fc} = m_{b}v_{it}\\&or,& v_{fc} = \dfrac{m_{b}v_{it}}{(m_{c} + m_{b})}\\&or,& v_{fc} = \dfrac{(0.048)(33.5)}{(55.0 + 0.0480)}\\&or,& v_{fc} = 0.029~m/s

Thus, the final velocity of thrower is 3.88~m/s and that for the catcher is 0.029~m/s.

8 0
3 years ago
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