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Viefleur [7K]
3 years ago
5

Where is most of the mass of an atom located? in the nucleus in the orbits in the electrons it is split between the nucleus and

the orbits
Physics
2 answers:
yulyashka [42]3 years ago
7 0
Most of the mass is located in the nucleus as suggested by Rutherford's gold foil experiment.
Lostsunrise [7]3 years ago
7 0

Answer:

in the nucleus

Explanation:

The majority of mass of an atom consists in the nucleus which consists of protons and neutrons. The mass of a proton is 1.6726219 × 10⁻²⁷ kilograms and the mass of neutron is 1.674929 x 10⁻²⁷ kg while the mass of an electron is 9.109390 x 10⁻³¹ kg.

As it can be seen that the masses of both the protons and neutrons are higher than that of an electron the most mass lies in the nucleus.

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Can someone check my answer thanks ! 2.The ability to do work or cause change is called A.velocity. B.energy.(I PICK B.) C.trans
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2 is B.  3 is D.   4 is C.   I think 5 is A.   6 is A.   7 is D.   I think you are all correct. Good Luck!
7 0
3 years ago
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A 6.0-kilogram cart initially traveling at 4.0 meters per second east accelerates uniformly at 0.50 meter per second squared eas
VMariaS [17]

Answer: 5.5m/s

Explanation:

vf=vi+at

vf= 4.0m/s + (0.50m/s^2)(3.0s)

3 0
3 years ago
4. Lead has a density of 11.5g/cmº. A rectangular block of lead measures 7cm x5cmx2cm.
lisov135 [29]

Answer:

(a) 70cm³

(b) 805 grams

Explanation:

(a) V = L×B×H

= 7cm×5cm×2cm

= 35cm×2cm

= 70cm³

(b) Mass = Volume × Density

= 70cm³ × 11.5g/cm³

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3 0
3 years ago
A car going at v = 29.7 m/s (67 mph) rounds a curve of radius R = 50.0 m, where the road is banked at an angle of θ = 30.0°. Wha
Nikolay [14]

Answer:

μ = 0.6

Explanation:

given,

speed of car = 29.7 m/s

Radius of curve = 50 m

θ = 30.0°

minimum static friction = ?

now,

writing all the forces acting along y-direction

N cos θ - f sinθ = mg

N cos θ -μN sinθ = mg

N = \dfrac{m g}{cos\theta-\mu sin \theta}

now, writing the forces acting along x- direction

N sin θ + f cos θ = F_{net}

N cos θ + μN sinθ = F_{net}

\dfrac{m g}{cos\theta-\mu sin \theta}(cos \theta + \mu sin\theta)=F_{net}

taking cos θ from nominator and denominator

F_{net} =\dfrac{tan\theta + \mu}{1-\mutan\theta}. mg

\dfrac{mv^2}{r}=\dfrac{tan\theta + \mu}{1-\mutan\theta}. mg

\dfrac{v^2}{r}=\dfrac{tan\theta + \mu}{1-\mutan\theta}g

\mu=\dfrac{v^2 -r g tan\theta}{v^2tan\theta + r g}

now, inserting all the given values

\mu=\dfrac{29.7^2 -50 \times 9.8tan 30^0}{29.7^2\times tan 30^0 +50 \times 9.8}

μ = 0.6

7 0
3 years ago
At a stoplight, a truck traveling at 15 m/s passes a car as it starts from rest. The truck travels at constant velocity and the
Art [367]
The time that the car take to catch up to truck would be :

1/2 ar^2  = vt

1/2 (3) t^2 = 15t

1.5  t^2  =  15t

t ^2    =   10s

Hope this helps
4 0
3 years ago
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