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Sergeeva-Olga [200]
1 year ago
6

A long, thin rod parallel to the y-axis is located at x = -1.0 cm and carries a uniform linear charge density of +1.0 nC/m. A se

cond long, thin rod parallel to the z-axis is located at x = +1.0 cm and carries a uniform linear charge density of -1.0 nC/m.

Physics
1 answer:
Bad White [126]1 year ago
8 0

Answer:

Hhere is the other part of the question ; What is the net electric field due to these rods at the origin? (epsilon not = 8.85*10^-12)

net electric field = 3600N/C

Explanation:

The detailed calculation is as shwon below

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A child is riding a merry-go-round that has an instantaneous angular speed of 1.25 rad/s and an angular acceleration of 0.745 ra
skelet666 [1.2K]

Answer:

So the acceleration of the child will be 8.05m/sec^2

Explanation:

We have given angular speed of the child \omega =1.25rad/sec

Radius r = 4.65 m

Angular acceleration \alpha =0.745rad/sec^2

We know that linear velocity is given by v=\omega r=1.25\times 4.65=5.815m/sec

We know that radial acceleration is given by a=\frac{v^2}{r}=\frac{5.815^2}{4.65}=7.2718m/sec^2

Tangential acceleration is given by

a_t=\alpha r=0.745\times 4.65=3.464m/sec^

So total acceleration will be a=\sqrt{7.2718^2+3.464^2}=8.05m/sec^2

7 0
1 year ago
Why does carpet tend to produce differences in static electricity more that hardwood or tile floors
Makovka662 [10]

Answer:

This is because the rubbing releases negative charges, called electrons, which can build up on one object to produce a static charge. For example, when you shuffle your feet across a carpet, electrons can transfer onto you, building up a static charge on your skin.

Explanation:

This is because the rubbing releases negative charges

4 0
2 years ago
In the winter sport of bobsledding, athletes push their sled along a horizontal ice surface and then hop on the sled as it start
Hitman42 [59]

Answer:

v_f = 16.6 m/s

Explanation:

As we know by force equation that force along the inclined planed due to gravity is given as

F_g = mg sin\theta

so the acceleration due to gravity along the plane is given as

a = \frac{F_g}{m}

now we have

a = g sin\theta

a = (9.81 sin4.0)

a = 0.68 m/s^2

now we know that

v_f^2 - v_i^2 = 2 a d

v_f^2 - 9.2^2 = 2(0.68)(140)

v_f = 16.6 m/s

4 0
2 years ago
WallyGPX accelerates from 0 m/s to 8 m/s in 3 seconds. What is his acceleration? Is this acceleration higher than that of a car
olga nikolaevna [1]

My Phone is +2348181686682

4 0
2 years ago
An insurance company hired your group to help investigate an insurance claim following a car accident. In the accident, two cars
Romashka-Z-Leto [24]

Answer:

v=8m/s

Explanation:

To solve this problem we have to take into account, that the work done by the friction force, after the collision must equal the kinetic energy of both two cars just after the collision. Hence we have

W_{f}=E_{k}\\W_{f}=\mu N=\mu(m_1+m_1)g\\E_{k}=\frac{1}{2}[m_1+m_2]v^2

where

mu: coefficient of kinetic friction

g: gravitational acceleration

We can calculate the speed of the cars after the collision by using

W_f=(0.7)(1650kg+1900kg)(9.8\frac{m}{s^2})=24353J\\24353J=\frac{1}{2}(1650kg+1900kg)v^2\\v=\sqrt{\frac{24353J}{1775kg}}=3.70\frac{m}{s}

Now , we can compute the speed of the second car by taking into account the conservation of the momentum

P_b=P_a\\m_1v_1+m_2v_2=(m_1+m_2)v\\\\v_2=\frac{(m_1+m_2)v-m_1v_1}{m_1}\\\\v_2=\frac{(1650kg+1900kg)(3.7\frac{m}{s})-(1650kg)(16\frac{m}{s})}{(1650kg)}\\\\v_2=8\frac{m}{s}

the car did not exceed the speed limit

Hope this helps!!

7 0
1 year ago
Read 2 more answers
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