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serious [3.7K]
3 years ago
8

A car travels 30 kilometers north for 2 hours, then travels 45 kilometers east for 3 hours. The car then drives south for 30 kil

ometers for 1.5 hours, finally travels west for 25 kilometers in 0.5 hours. What is the speed of the car? What is the velocity of the car?
Physics
1 answer:
MrRissso [65]3 years ago
5 0
*Let north and east as positive direction.
1. a = +30 km / 2 hr
a = +15 km/hr

2. a = +45 km / 3 hr
a = +15 km/hr

3. a = -30 km / 1.5 hr
a = -20 km/hr

4. a = -25 km / 0.5 hr
a = -50 km/hr

speed = (15+15+20+50) / 4
= 25 km/hr

velocity:
(horizontal): (+15-50)= -35 km/hr
(vertical): (+15-20)= -5km/hr
v = sqrt(35^2 + 5^2)
v = 35.36 km/hr
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Answer:

d = 1.55 * 10⁻⁶ m

Explanation:

To calculate the distance between the adjacent grooves of the CD, use the formula, d = \frac{m \lambda}{sin(A_{m}) }..........(1)

The fringe number, m = 1 since it is a first order maximum

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Distance between the central maximum and the first order maximum = 1.1 m

Distance between the screen and the CD = 3 m

A_{m} = Angle between the incident light and the diffracted light

From the setup shown in the attachment, it is a right angled triangle in which

sin(A_{m}) = \frac{opposite}{Hypotenuse} \\sin(A_{m}) =\frac{1.1}{\sqrt{1.1^{2}+3^{2}}}

sin(A_{m} ) = 0.344\\A_{m} = sin^{-1} 0.344\\A_{m} = 20.14^{0}

Putting all appropriate values into equation (1)

d = \frac{1* 532*10^{-9} }{0.344 }\\d = 0.00000155 m\\d = 1.55 * 10^{-6} m

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a 10.0 kg sphere is released from rest in an ocean. as it falls, the water applies a resistive force r
dimaraw [331]

The calculated coefficient of kinetic friction is 0.33125.'

The rate of kinetic friction the friction force to normal force ratio experienced by a body moving on a dry, uneven surface is known as k. The friction coefficient is the ratio of the normal force pressing two surfaces together to the frictional force preventing motion between them. Typically, it is represented by the Greek letter mu (). In terms of math, is equal to F/N, where F stands for frictional force and N for normal force.

given mass of the block=10 kg

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now according to principal of conservation of energy we observe,

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10 x 10 x 3= μ(600) +(1125) (0.09)

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The complete question is- A 10.0−kg block is released from rest at point A in Fig The track is frictionless except for the portion between point B and C, which has a length of 6.00m the block travels down the track, hits a spring of force constant 2250N/m, and compresses the spring 0.300m form its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between the block and the rough surface between point Band (C)

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brainly.com/question/13754413

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