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3 years ago

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A dog sitting at the edge of the pool notices that 5 waves pass in fromt of his paws over a span of 1.72 seconds. He watches one

wave in particular and notices that it goes from one end to other other 13 m in 13.54 seconds. What is the length of each wave?

2 answers:

gayaneshka [121]3 years ago

8 0

I think it si 20.3 second hope it helps

Fantom [35]3 years ago

6 0

20.3 sec so ye good luck I hope I helped you

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A point charge of 4.0 µC is placed at a distance of 0.10 m from a hard rubber rod with an electric field of 1.0 × 103 . What is

yuradex [85]

Since

Electric potential energy = qV

Where V = Ed

Hence
Electric potential energy = q(Ed) --- (1)

Since E = 1.0 * 10^3 N/C
d = 0.10 m
q = 4 * 10^-6 C

Plug in the values in (1)
(1) => Electric potential energy = 4 * 10^-6(1.0 * 10^3 * 0.10)
Electric potential energy = 400 μJ

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3 years ago

A 1100 kg car rounds a curve of radius 68 m banked at an angle of 16 degrees. If the car is traveling at 95 km/h, will a frictio

Mariulka [41]

Answer:

Yes. Towards the center. 8210 N.

Explanation:

Let's first investigate the free-body diagram of the car. The weight of the car has two components: x-direction: towards the center of the curve and y-direction: towards the ground. Note that the ground is not perpendicular to the surface of the Earth is inclined 16 degrees.

In order to find whether the car slides off the road, we should use Newton's Second Law in the direction of x: F = ma.

The net force is equal to $F = \frac{mv^2}{R} = \frac{1100\times (26.3)^2}{68} = 1.1\times 10^4~N$

Note that 95 km/h is equal to 26.3 m/s.

This is the centripetal force and equal to the x-component of the applied force.

$F = mg\sin(16) = 1100(9.8)\sin(16) = 2.97\times10^3$

As can be seen from above, the two forces are not equal to each other. This means that a friction force is needed towards the center of the curve.

The amount of the friction force should be $8.21\times 10^3~N$

Qualitatively, on a banked curve, a car is thrown off the road if it is moving fast. However, if the road has enough friction, then the car stays on the road and move safely. Since the car intends to slide off the road, then the static friction between the tires and the road must be towards the center in order to keep the car in the road.

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3 years ago

A particle with charge 3.20×10−19 c is placed on the x axis in a region where the electric potential due to other charges increa

lys-0071 [83]

Answer:

-5 V

Explanation:

The charged particle (which is positively charged) moves from point A to B, and its kinetic energy increases: it means that the particle is following the direction of the field, so its potential energy is decreasing (because it's been converted into potential energy), therefore it is moving from a point at higher potential (A) to a point at lower potential (B). This means that the value

vb−va

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We can calculate the potential difference between the two points by using the law of conservation of energy:

$\Delta K+ \Delta U=0\\\Delta K + q\Delta V=0$

where:

$\Delta K=+1.6\cdot 10^{-18} J$ is the change in kinetic energy of the particle

$q=3.2\cdot 10^{-19} C$ is the charge of the particle

$\Delta V =V_b-V_a$ is the potential difference

Re-arranging the equation, we can find the value of the potential difference:

$\Delta V=V_b-V_a = -\frac{\Delta K}{q}=-\frac{1.6\cdot 10^{-18} J}{3.2\cdot 10^{-19} C}=-5 V$

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3 years ago

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Neko [114]

Answer:

I think its 2. It would decrease

Explanation:

Hope this helps

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3 years ago

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