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3 years ago

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Two aluminum soda cans are charged and repel each other, hanging motionless at an angle. Which of the forces on the left can has

the greatest magnitude? T FE they are all the same impossible to tell mg Changed: Your submitted answer was incorrect. Your current answer has not been submitted.

1 answer:

swat323 years ago

7 0

Answer:

the tension is the greatest force the system

Explanation:

For this exercise we use Newton's second law, in the equilibrium condition, this means that the acceleration is zero a = 0

X-axis (horizontal)

FE - Tₓ = 0

Fe = Tₓ

Y axis (vertical)

$T_{y}$ - W = 0

T_{y} = W

let's use trigonometry for the stress components

sin θ = Tₓ / T

Tₓ = T sin θ

cos θ = T_{y} / T

T_{y} = T cos θ

we substitute

FE = T sin θ

W = T cos θ

Since the sine and cosine function have values between 0 and 1, the voltage must be greater than the electrical force and greater than the weight of the body.

Therefore the tension is the greatest force the system

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An electron that has an instantaneous velocity of ???? = 2.0 × 106 m ???? ???? + 3.0 × 106 m ???? ???? is moving through the uni

Setler79 [48]

Explanation:

It is given that,

Velocity of the electron, $v=(2\times 10^6i+3\times 10^6j)\ m/s$

Magnetic field, $B=(0.030i-0.15j)\ T$

Charge of electron, $q_e=-1.6\times 10^{-19}\ C$

(a) Let $F_e$ is the force on the electron due to the magnetic field. The magnetic force acting on it is given by :

$F_e=q_e(v\times B)$

$F_e=1.6\times 10^{-19}\times [(2\times 10^6i+3\times 10^6j)\times (0.030i-0.15j)]$

$F_e=-1.6\times 10^{-19}\times (-390000)(k)$

$F_e=6.24\times 10^{-14}k\ N$

(b) The charge of electron, $q_p=1.6\times 10^{-19}\ C$

The force acting on the proton is same as force on electron but in opposite direction i.e (-k). Hence, this is the required solution.

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3 years ago

zysi [14]

Answer:

yes it doesn't matter

Explanation:

it doesn't matter because troughs and crests are the same and either can be used

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2 years ago

A physics student swings a pail of water in a vertical circle 1.0 m in radius at a constant speed. If the water is NOT to spill

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Answer:

(A) 3.1 m/s

(B) 2.0 s

Explanation:

At the minimum speed, the force of gravity equals the centripetal force.

mg = m v² / r

v = √(gr)

v = √(9.8 m/s² × 1.0 m)

v = 3.1 m/s

The time is the circumference divided by the speed.

t = (2π × 1.0 m) / (3.1 m/s)

t = 2.0 s

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3 years ago

A mass of 0.5 kg hangs motionless from a vertical spring whose length is 1.10 m and whose unstretched length is 0.50 m. Next the

ser-zykov [4K]

Answer:

The maximum length during the motion is $L_{max} = 1.45m$

Explanation:

From the question we are told that

The mass is $m =0.5 kg$

The vertical spring length is $L = 1.10m$

The unstretched length is $L_{un} = 1.30m$

The initial speed is $v_i = 1.3m/s$

The new length of the spring $L_{new} = 1.30 m$

The spring constant k is mathematically represented as

$k = -\frac{F}{y}$

Where F is the force applied $= m * g = 0.5 * 9.8=4.9N$

y is the difference in weight which is $=1.10-0.50=0.6m$

The negative sign is because the displacement of the spring (i.e its extension occurs against the force F)

Now substituting values accordingly

$k = \frac{4.9}{0.6}$

$= 8.17 N/m$

The elastic potential energy is given as $E_{PE} = \frac{1}{2} k D^2$

where D is this the is the displacement

Since Energy is conserved the total elastic potential energy would be

$E_T = initial \ elastic\ potential \ energy + kinetic \ energy$

$E_T = \frac{1}{2} k D_{max}^2 = \frac{1}{2} k D^2 + \frac{1}{2} mv^2$

Substituting value accordingly

$\frac{1}{2} *8.17 *D_{max}^2 =\frac{1}{2} * 8.17*(1.30 - 0.50)^2 + \frac{1}{2} * 0.5 *1.30^2$

$4.085 * D_{max}^2 = 3.69$

$D^2_{max} = 0.9033$

$D_{max} = 0.950m$

So to obtain total length we would add the unstretched length

So we have

$L_{max} = 0.950 + 0.5 = 1.45m$

5 0

3 years ago

Read 2 more answers

A 269-turn solenoid is 102 cm long and has a radius of 2.3 cm. It carries a current of 3.9 A. What is the magnetic field inside

RUDIKE [14]

Answer:

Magnitude of the magnetic field inside the solenoid near its centre is 1.293 x 10⁻³ T

Explanation:

Given;

number of turns of solenoid, N = 269 turn

length of the solenoid, L = 102 cm = 1.02 m

radius of the solenoid, r = 2.3 cm = 0.023 m

current in the solenoid, I = 3.9 A

Magnitude of the magnetic field inside the solenoid near its centre is calculated as;

$B = \frac{\mu_o NI}{l} \\\\$

Where;

μ₀ is permeability of free space = 4π x 10⁻⁷ m/A

$B = \frac{4\pi*10^{-7} *269*3.9}{1.02} \\\\B = 1.293 *10^{-3} \ T$

Therefore, magnitude of the magnetic field inside the solenoid near its centre is 1.293 x 10⁻³ T

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3 years ago