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3 years ago

1. What causes some atoms of the same element to have different atomic masses?

Physics

2 answers:

liubo4ka [24]3 years ago

6 0

Answer:

Isotopes are atoms with different atomic masses which have the same atomic number. The atoms of different isotopes are atoms of the same chemical element; they differ in the number of neutrons in the nucleus.

Explanation:

ch4aika [34]3 years ago

5 0

Answer:

isotops sorry if it's wrong

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A circular loop with radius r is rotating with constant angular velocity ω in a uniform electric field with magnitude E. The axi

inn [45]

Answer:

$\Phi_{E} = E\pi r^2 \omega t$

Explanation:

The electric flux is defined as the multiple of electric field and the area that the electric field passes through, such that

$\Phi_{E} = \vec{E}\vec{A}$

When calculating the electric flux, the angle between the directions of electric field and the area becomes important, especially if the angle is changing with time.

The above formula can be rewritten as follows

$\Phi_{E} = EA\cos(\theta)$

where θ is the angle between the electric field and the area of the loop. Note that, the direction of the area of the loop is perpendicular to the plane of the loop.

If the loop is rotating with constant angular velocity ω, then the angle can be written as follows

$\theta = \omega t$

At t = 0, cos(0) = 1 and the electric flux through the loop is at its maximum value.

Therefore the electric flux can be written as a function of time

$\Phi_{E} = E\pi r^2 \omega t$

3 0

3 years ago

What is the force that the gas exerts on each of the six sides of the box when the gas temperature is 20.0∘C?

Taya2010 [7]

Incomplete question as number of moles and length is missing.So I have assumed 3 moles and length of 0.300 m.So the complete question is here:

Three moles of an ideal gas are in a rigid cubical box with sides of length 0.300 m.What is the force that the gas exerts on each of the six sides of the box when the gas temperature is 20.0∘C?

Answer:

The Force act on each side is 2.43×10⁴N

Explanation:

Given data

n=3 mol

L=0.3 m

Temperature=20.0°C=293 K

To find

Force F

Solution

To get force act on each side it would employ by

F=P.A

Where P is pressure

A is Area

First we need to find pressure by applying ideal gas law

$P.V=nRT\\P=\frac{nRT}{V}\\ P=\frac{(3mol)(8.315J/mol)(293K)}{(0.3m*0.3m*0.3m)}\\P=27.069*10^{4}Pa$

So The Force is given as:

$F=P.A\\F=(27.069*10^{4} )(0.3m*0.3m)\\F=2.43*10^{4}N$

The Force act on each side is 2.43×10⁴N

3 0

3 years ago

On average, both arms and hands together account for 13% of a person's mass, while the head is 7.0% and the trunk and legs accou

Feliz [49]

Answer:

Angular velocity, $N_f = 242.36 rpm$

Explanation:

The mass of the skater, M = 74.0 kg

Mass of each arm, $m_{a} = 0.13 * \frac{M}{2}$ ( since it is 13% of the whole body and each arm is considered)

$m_{a} = 0.13 * 37\\m_a = 4.81 kg$

Mass of the trunk, $m_{t} = M - 2m_{a}$

$m_t = 74 - 2(4.81)\\m_{t} = 64.38 kg$

Total moment of Inertia = (Moment of inertia of the arms) + (Moment of inertia of the trunks)

$(I_{T} )_i = 2(\frac{m_{a}L^2 }{12} + m_a(0.5L + R)^2) + 0.5 m_t R^2$

$(I_{T} )_i = 2(\frac{4.81 * 0.7^2 }{12} + 4.81(0.5*0.7 + 0.175)^2) + 0.5 *64.38* 0.175^2\\(I_{T} )_i = 3.052 + 0.986\\(I_{T} )_i = 4.038 kgm^2$

The final moment of inertia of the person:

$(I_{T} )_f = \frac{1}{2} MR^{2} \\(I_{T} )_f = \frac{1}{2} * 74*0.175^{2}\\(I_{T} )_f = 1.133 kg.m^2$

According to the principle of conservation of angular momentum:

$(I_{T} )_i w_{i} = (I_{T} )_f w_{f}\\w_{i} = 68 rpm = (2\pi * 68)/60 = 7.12 rad/s\\4.038 * 7.12 =1.133* w_{f}\\w_{f} = 25.38 rad/s\\w_{f} = \frac{2\pi N_f}{60} \\25.38 = \frac{2\pi N_f}{60}\\N_f = (25.38 * 60)/2\pi \\N_f = 242.36 rpm$

3 0

3 years ago

Two crates, one with mass 5.4 kg and the other with mass 8.2 kg, connected by a light rope. The coefficient of kinetic friction

Gnom [1K]

Answer:

R= 2.5 :ratio of the magnitude of the applied horizontal force to the magnitude of the tension in the rope connecting the blocks

Explanation:

We apply Newton's second law:

∑F=m*a

velocity is constant ,then , a=0

Nomenclature

W: weight

m: mass

N : normal force

Ff: Friction force

μk: coefficient of kinetic friction

T: tension force in the rope

F: applied horizontal force

g: acceleration due to gravity.

Force Calculation

W₁=m₁*g=5.4 kg *9.8m/s²=52.92 N

W₂=m₂*g=8.2 kg *9.8m/s²= 80.36N

∑Fy=0

N₁-W₁=0 , N₁=W₁ = 52.92 N

N₂-W₂=0, N₂=W₂=80.36N

Ff₁= μk* N₁=0.4*52.92 N = 21.16N

Ff₂= μk* N₂=0.4*80.36N = 32.14N

Look at the attached graphic

Free-Body diagram m₁=5.4 kg

∑Fx=0

T- Ff₁=0 , T= Ff₁ , T= 21.16N

Free-Body diagram m₂=8.2 kg

∑Fx=0

F-T- Ff₂=0 , F=T+Ff₂= 21.16N+32.14N=53.3N

Ratio of the magnitude of the applied horizontal force to the magnitude of the tension in the rope connecting the blocks (R)

R= F/T= 53.3N/21.16N = 2.5

3 0

3 years ago

A car stands from rest and attains a speed of 20 m/s in 10 s. What distance will it cover ?

sashaice [31]

Answer:

100m

Explanation:

100m

s=ut+1/2at^2

s= unknown, u=0, a=2, t=10

s=0*10+1/2(2)(10)^2

s=1/2(2)(100)

s=1(100)

displacement = 100 meters

4 0

3 years ago