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2 years ago

5

A wave has an amplitude of 0.2 cm.

1 answer:

Anettt [7]2 years ago

3 0

The distance between the resting point and maximum height of the wave is 0.2 cm.

The amplitude is measured from the resting point up to the highest point of the wave.

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A long straight rod experiences several forces, each acting at a different location on the rod. All forces are perpendicular to

enot [183]

Answer:

1. C

2.C

Explanation:

1. The rod is perpendicular to every axis and forces are acting on every location. If the torque on the left side is zero, this indicates that forces with respect to their distance on the left side is zero and doesn't account for the net force at a point.

2. If the net torque about every point on every axis is zero, the rod will be rotational because each axis will yield a magnitude of zero which obeys the principle of rotation at a point.

6 0

3 years ago

Suppose you take a trip that covers 180 km and takes 3 hours to make. Your average speed is A. 30 km/h. B. 60 km/h. C. 180 km/h.

Ivenika [448]

B. 60 kilometers per hour

5 0

3 years ago

Two identical small metal spheres with q1 > 0 and |q1| > |q2| attract each other with a force of magnitude 72.1 mN when se

Brrunno [24]

1) $+2.19\mu C$

The electrostatic force between two charges is given by

$F=k\frac{q_1 q_2}{r^2}$ (1)

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

When the two spheres are brought in contact with each other, the charge equally redistribute among the two spheres, such that each sphere will have a charge of

$\frac{Q}{2}$

where Q is the total charge between the two spheres.

So we can actually rewrite the force as

$F=k\frac{(\frac{Q}{2})^2}{r^2}$

And since we know that

r = 1.41 m (distance between the spheres)

$F= 21.63 mN = 0.02163 N$

(the sign is positive since the charges repel each other)

We can solve the equation for Q:

$Q=2\sqrt{\frac{Fr^2}{k}}=2\sqrt{\frac{(0.02163)(1.41)^2}{8.98755\cdot 10^9}}}=4.37\cdot 10^{-6} C$

So, the final charge on the sphere on the right is

$\frac{Q}{2}=\frac{4.37\cdot 10^{-6} C}{2}=2.19\cdot 10^{-6}C=+2.19\mu C$

2) $q_1 = +6.70 \mu C$

Now we know the total charge initially on the two spheres. Moreover, at the beginning we know that

$F = -72.1 mN = -0.0721 N$ (we put a negative sign since the force is attractive, which means that the charges have opposite signs)

r = 1.41 m is the separation between the charges

And also,

$q_2 = Q-q_1$

So we can rewrite eq.(1) as

$F=k \frac{q_1 (Q-q_1)}{r^2}$

Solving for q1,

$Fr^2=k (q_1 Q-q_1^2})\\kq_1^2 -kQ q_1 +Fr^2 = 0$

Since $Q=4.37\cdot 10^{-6} C$ , we can substituting all numbers into the equation:

$8.98755\cdot 10^9 q_1^2 -3.93\cdot 10^4 q_1 -0.141 = 0$

which gives two solutions:

$q_1 = 6.70\cdot 10^{-6} C\\q_2 = -2.34\cdot 10^{-6} C$

Which correspond to the values of the two charges. Therefore, the initial charge q1 on the first sphere is

$q_1 = +6.70 \mu C$

8 0

3 years ago

..................,,,,,,,,,

Fiesta28 [93]

Answer:

.....,.,.,.,.,.,.,.,.,.,.,.,..,.,.,.,.,.,.,,.,.┌(・。・)┘♪

7 0

2 years ago

Read 2 more answers

An automobile traveling on a straight, level road has an initial speed v when the brakes are applied. In coming to rest with a c

Molodets [167]

Answer:

4x

Explanation:

Use $v^{2} = u^{2} +2as$ to do the question.

For first instance,

$0 = v^{2} +2ax$ -------------------( 1 )

for second instance,

$0 = (2v)^{2} +2as$ -----------------( 2 )

So by (1) and (2),

s = 4x

3 0

3 years ago