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2 years ago

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Which of the following quantities are zero for an object traveling in a circle at a constant speed? There may be more than one c

orrect answer. Group of answer choices Velocity Acceleration Net force Radial component of velocity Tangential component of velocity Radial component of acceleration Tangential component of acceleration Radial component of net force Tangential component of net force

1 answer:

ratelena [41]2 years ago

5 0

Answer:

Zero: Tangential component of speed, Tangential component of acceleration,

Explanation:

Let's analyze uniform circular motion, using kinematics and Newton's second law , Tangency component of net force

The speed module is constant, the speed changes since its direction changes and is a vector; If there is a change in speed, there is an acceleration.

With these concepts let's analyze what quantities are zero

Quantities are:

• Acceleration. non-Zero. It is different from zero

• Net force It is different from zero

• Radial speed component. It is different from zero, change the direction

• Tangential component of speed. Zero

• radial acceleration component Non-zero

• Tangential component of acceleration. Zero

• Radial component of net force Non- Zero

• Tangency component of the net force. Zero

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What is the total flux φ that now passes through the cylindrical surface? enter a positive number if the net flux leaves the cyl

trasher [3.6K]

Net flux through the cylindrical surface is given as

$\phi = \frac{q}{epsilon_0}$

here q = enclosed charge in the surface

so here in order to find the value of q

$q = \lambda* L$

so now we have

$\phi = \frac{\lambda * L}{\epsilon_0}$

so this is the total flux

now by Gauss's law we can find the electric field

$\int E.dA = \phi$

$\int E.dA = \frac{\lambda * L}{\epsilon_0}$

$E* 2\pi rL = \frac{\lambda * L}{epsilon_0}$

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

<em>by above expression we can find the electric field at required position</em>

8 0

1 year ago

Two bar magnets are labeled A and B. The ends of each magnet are numbered 1 or 2, but the poles are not labeled. When A1 is brou

vichka [17]

It is definitely letter D. <span>A1 and B1 are like poles, but there is not enough information to tell whether they are north poles or south poles.

A1 and B1 is either both north poles or both south poles. Repulsion of both magnets says it all--like poles always repel while opposite poles always attract. Thus, the best conclusion to this would be choice D.</span>

3 0

2 years ago

Read 2 more answers

Two long, parallel wires carry unequal currents in the same direction. The ratio of the currents is 3 to 1. The magnitude of the

astraxan [27]

Answer:

3A is the larger of the two currents.

Explanation:

Let the currents in the two wires be I₁ and I₂

given:

Magnitude of the electric field, B = 4.0μT = 4.0×10⁻⁶T

Distance, R = 10cm = 0.1m

Ratio of the current = I₁ : I₂ = 3 : 1

Now, the magnitude of a magnetic field at a distance 'R' due to the current 'I' is given as

$B = \frac{\mu_oI}{2\pi R}$

Where $\mu_o$ is the magnitude constant = 4π×10⁻⁷ H/m

Thus, the magnitude of a magnetic field due to I₁ will be

$B_1 = \frac{\mu_oI_1}{2\pi R}$

$B_2 = \frac{\mu_oI_2}{2\pi R}$

given,

B = B₁ - B₂ (since both the currents are in the same direction and parallel)

substituting the values of B, B₁ and B₂

we get

4.0×10⁻⁶T = $\frac{\mu_oI_1}{2\pi R}$ - $\frac{\mu_oI_2}{2\pi R}$

or

4.0×10⁻⁶T = $\frac{\mu_o}{2\pi R}\times (I_1-I_2 )$

also

$\frac{I_1}{I_2} = \frac{3}{1}$

⇒ $I_1 = 3\times I_2$

substituting the values in the above equation we get

4.0×10⁻⁶T = $\frac{4\pi\times 10^{-7}}{2\pi 0.1}\times (3 I_2-I_2)$

⇒ $I_2 = 1A$

also

$I_1 = 3\times I_2$

⇒ $I_1 = 3\times 1A$

⇒ $I_1 = 3A$

Hence, the larger of the two currents is 3A

3 0

2 years ago

A baseball player exerts a force of 100 N on a ball for a distance of 0.5 mas he throws it. If the ball has a mass of 0.15 kg, w

Aloiza [94]

Answer:

25.82 m/s

Explanation:

We are given;

Force exerted by baseball player; F = 100 N

Distance covered by ball; d = 0.5 m

Mass of ball; m = 0.15 kg

Now, to get the velocity at which the ball leaves his hand, we will equate the work done to the kinetic energy.

We should note that work done is a measure of the energy exerted by the baseball player.

Thus;

F × d = ½mv²

100 × 0.5 = ½ × 0.15 × v²

v² = (2 × 100 × 0.5)/0.15

v² = 666.67

v = √666.67

v = 25.82 m/s

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1 year ago

Which formula is used to find fluctuation of the shape of body

Sladkaya [172]

Answer:

varn=n1+1ehvkT–1

Explanation:

This is Einstein's equation.

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1 year ago