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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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The parietal lobe is the portion of the cerebral cortex located below the temporal lobe.
jasenka [17]

Answer:

False

Explanation:

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3 years ago
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A 3.5 x 10-6 C charge is located 0.28 m from a 2.8 x 10-6 C charge. What is the magnitude of the force being exerted on the smal
Margarita [4]
The electrostatic force between two charges is given by Coulomb's law:
F=k_e  \frac{q_1 q_2}{r^2}
where
ke is the Coulomb's constant
q1 is the first charge
q2 is the second charge
r is the separation between the two charges

By substituting the data of the problem into the equation, we can find the magnitude of the force between the two charges:
F=(8.99 \cdot 10^{9} Nm^2C^{-2} ) \frac{(3.5 \cdot 10^{-6} N/C)(2.8 \cdot 10^{-6}N/C)}{(0.28m)^2}=1.2 N
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3 years ago
Agility is the
olga_2 [115]

Answer:

combination of strength and speed

Explanation:

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8 0
3 years ago
The driver of a 1750 kg car traveling on a horizontal road at 110 km/h suddenly applies the brakes. Due to a slippery pavement,
Alexeev081 [22]

Answer: a=-2.4525 m/s^2

d=s=190.3 m

Explanation:The only force that is stopping the car and causing deceleration is the frictional force Fr

Fr = 25% of weight

W=mg

W=1750*9.81

W=17167.5

Hence

Fr=\frac{25}{100} * -17167.5\\\\Fr=-4291.875 N

Frictional force is negative as it acts in opposite direction

According to newton second law of motion

F=ma

hence

a=Fr/m

a=-4291.875/1750\\a=-2.4525

given

u= 110 km/h

u=110*1000/3600

u=30.55 m/s

to get t we know that final velocity v=0

v^2=u^2+2as\\0=30.55^2-2*2.4525*s\\s=190.34m

3 0
2 years ago
A 20.0-N weight slides down a rough inclined plane which makes an angle of 30 degree with the horizontal. The weight starts from
Ulleksa [173]

Answer:

1270.64\ \text{J}

Explanation:

m = Mass of object = \dfrac{mg}{g}

mg = Weight of object = 20 N

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

v = Final velocity = 15 m/s

u = Initial velocity = 0

d = Distance moved by the object = 150 m

\theta = Angle of slope = 30^{\circ}

f = Force of friction

fd = Work done against friction

The force balance of the system is

\dfrac{1}{2}m(v^2-u^2)=(mg\sin\theta-f)d\\\Rightarrow \dfrac{1}{2}mv^2=mg\sin\theta d-fd\\\Rightarrow fd=mg\sin\theta d-\dfrac{1}{2}mv^2\\\Rightarrow fd=20\times \sin 30^{\circ}\times 150-\dfrac{1}{2}\times \dfrac{20}{9.81}\times 15^2\\\Rightarrow fd=1270.64\ \text{J}

The work done against friction is 1270.64\ \text{J}.

8 0
3 years ago
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