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Naily
3 months ago
10

A small glass bead charged to 5.0 nCnC is in the plane that bisects a thin, uniformly charged, 10-cmcm-long glass rod and is 4.0

cmcm from the rod's center. The bead is repelled from the rod with a force of 840 μNμN.What is the total charge on the rod?
Physics
1 answer:
Sav [3.1K]3 months ago
6 0

Answer:

The glass rod has a total charge of 47.8 nC.

Explanation:

Details provided:

Charge = 5.0 nC

Length of glass rod = 10 cm

Force = 840 μN

Distance = 4.0 cm

To evaluate the electric field intensity emitted by a uniformly charged rod of length L at a perpendicular distance from its midpoint

We will calculate the electric field

Using the formula for electric field intensity

E=\dfrac{kQ}{x\sqrt{(\dfrac{L}{2})^2+x^2}}

Where Q represents the charge on the rod

The force acting on the charged bead placed in the electric field of strength E

Using the force formula

F=qE

Insert the values into the formula

F=q\times\dfrac{kQ}{x\sqrt{(\dfrac{L}{2})^2+x^2}}

Now we calculate the total charge on the rod

Q=\dfrac{Fx\sqrt{(\dfrac{L}{2})^2+x^2}}{kq}

Substituting values into the equation

Q=\dfrac{840\times10^{-6}\times4.0\times10^{-2}\sqrt{(\dfrac{10.0\times10^{-2}}{2})^2+(4.0\times10^{-2})^2}}{9\times10^{9}\times5.0\times10^{-9}}

Q=47.8\times10^{-9}\ C

Q=47.8\ nC

Thus, the total charge on the rod is 47.8 nC.

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In pottery class, you throw a pot from a lump of wet clay. your pot's mass is 5.5 kg. after the pot is fired, it's mass is 4.9 k
ValentinkaMS [3465]

To find the volume, we can utilize the ratio of mass to density, as shown by:

volume = mass / density

 

A. when mass = 5.5 kg = 5500 g; density = 1.60 g/cm^3

volume = 5500 g / (1.60 g/cm^3)

resulting in volume = 3,437.5 cm^3

By rounding according to significant digits:

volume = 3,400 cm^3 = 3.4 L

 

B. when mass = 4.9 kg = 4900 g; density = 1.36 g/cm^3

volume = 4900 g / (1.36 g/cm^3)

calculating gives volume = 3,602.94 cm^3

Considering significant digits:

<span>volume = 3,600 cm^3 = 3.6 L</span>

8 0
2 months ago
A champion athlete can produce one horsepower (746 W) for a short period of time. The number of 16-cm-high steps a 70-kg athlete
Sav [3153]

Answer:

407 steps

Explanation:

Based on the question,

P = mgh/t........... Equation 1

Where P stands for power, m is mass, g denotes gravity, h is height, and t represents time.

Rearranging the equation to solve for h, we have:

h = Pt/mg............. Equation 2

Providing values: P = 746 W, t = 1 minute = 60 seconds, m = 70 kg.

Given constant: g = 9.8 m/s²

By substituting into equation 2

h = 746(60)/(70×9.8)

h = 44760/686

h = 65.25 m

h = 6525 cm

Calculating number of steps: 6525/16

The resulting number of steps = 407 steps

6 0
3 months ago
The acceleration of segment D is m/s2. Rank segments A, B , C from least accelerations to greatest acceleration. Least
Ostrovityanka [3204]

Answer:

D, C, B, A

Explanation:

The derivative from a velocity-time graph provides the acceleration value.

Segment A

\frac{dy}{dx} = \frac{15m/s}{1s} = 15m/s^2

Segment B

\frac{dy}{dx} = \frac{5m/s}{1s} = 5m/s^2

Segment C

\frac{dy}{dx} = \frac{0m/s}{2s} = 0m/s^2

Segment D

\frac{dy}{dx} = \frac{-20m/s}{1s} = -20m/s^2

Sorted from the lowest to the highest acceleration:

D, C, B, A

8 0
2 months ago
Read 2 more answers
A suspension system is being designed for a 1300 kg vehicle. When the vehicle is empty, its static deflection is measured as 2.5
Keith_Richards [3271]

Answer: damping coefficient = 1.5×10^5Ns/m

Explanation:

Refer to the attached file for the solution

3 0
2 months ago
THE RELATIVE ANGLE AT THE KNEE CHANGES FROM 0O TO 85O DURING THE KNEE FLEXION PHASE OF A SQUAT EXERCISE. IF 10 COMPLETE SQUATS A
ValentinkaMS [3465]

Answer: total angular distance = 1700° and 29.7 rad

   the total angular displacement = 0

Explanation:

This is a breakdown of the calculations needed.

The task is to determine both the total angular distance and displacement experienced by the knee.

To calculate the distance traveled by the knee, consider that when squatting, the knee bends 85° to lower, and then another 85° to return to standing (upright). Thus, the cumulative angular movement during the squat totals 170°.

For 10 squats, the knee must undergo 170° motion multiplied by 10, resulting in:

10 * 170° = 1700°

As such, the total angular distance reached is 1700°.

Now converting this to radians since both degree and radian outputs are required:

Since 2π equals 360°, it follows that one degree equates to about 57.3°;

∅ (rad) = ∅ (deg) * 2π/360°

∅ (rad) = 1700° * 2π/360° = 29.7 rad

∅ (rad) = 29.7 rad

For the second part, remember that angular displacement is determined by the angular distance divided by time, leading to a displacement of zero because the knee's ending position is the same as its starting position.

I hope this is helpful!!!!

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5 0
3 months ago
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