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3 years ago

An 80 kg object has a µk = 0.35 and a µs = 0.60. Assuming it is on a flat surface

Physics

1 answer:

Over [174]3 years ago

6 0

Answer:

What is the normal force on the object (draw a diagram if needed)

784 N

How much force is required to get the object to start to move from rest (max static friction)?

470.4 N

When the object starts moving, what is the force of kinetic friction?

274.4 N

If the moving object has a tension force of 300N to the right pulling it, what is the net force on the object in the horizontal direction? 784 N

What is the acceleration (with direction) of the object based on your answer for part d? Remember that a = net force/m.

Explanation:

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Light of wavelength 480 nm illuminates a double slit, and the interference pattern is observed on a screen. At the position of t

vova2212 [387]

Answer:

BC =980nm

Explanation:

Given wavelength =490nm

m=2

Required to find (OB - OC) = BC (more distant slit - nearest slit ) = dsin theta

We have dsin theta = m*wavelength

or BC = m* wavelength = 2*490nm

= 980nm = 98.000A

4 0

3 years ago

Why is the moon's umbra much smaller during a solar eclipse?

Stolb23 [73]

Answer:

the sun is larger than the moon so it creates a shadow

7 0

2 years ago

Microscopes are often equipped with blue filters. give a reason for using blue light in a microscope

timama [110]

The blue light removes red parts of the color spectrum giving it a cooler color

8 0

3 years ago

I drove from Oklahoma to California (1452.9 miles) in 23 hours? What was my average speed?

storchak [24]

Answer:

63 miles per hour

Explanation:

you just divide miles and hours to get your average speed

5 0

2 years ago

A horizontal circular curve on a highway is designed for traffic moving at 60 km/h. If the radius of the curve is 150 m, and if

KengaRu [80]

Answer:

The value of the correct angle of banking for the road is $\theta = 67.76$ °

Explanation:

Given data

Velocity (v) = 60 $\frac{m}{s}$

Radius = 150 m

The velocity of the car in this case is given by

$v = \sqrt{r g \tan \theta}$

$v^{2} = r g \tan \theta$

$\tan \theta = \frac{v^{2} }{rg}$

Put all the values in above formula we get

$\tan \theta = \frac{60^{2} }{(150)(9.81)}$

$\tan \theta =$ 2.446

$\theta = 67.76$ °

Therefore the value of the correct angle of banking for the road is $\theta = 67.76$ °

4 0

3 years ago