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WARRIOR [948]
2 years ago
11

If AB has a bearing of following 234° 51' 48" and a anti-clockwise angle from AB to C is measured as 80° Calculate the bearing A

C. (Enter as numeric value of ddd.mmss e.g. 100° 20' 30" would be entered as 100.2030. Marked out of 5.00 P
Physics
1 answer:
Musya8 [376]2 years ago
4 0

Answer:

the bearing of the angle AC will be equal to 154°51' 48"

Explanation:

given,

bearing of  line AB = 234° 51' 48"

C is  anti clockwise angle from 80° from AB      

bearing of line AC = ?                  

To calculate the bearing of line AB  80° anticlockwise movement

bearing of the AC =  234° 51' 48"  - 80°                

                              = 154°51' 48"                

The bearing can be represented as 154.5148

hence, the bearing of the angle AC will be equal to 154°51' 48"

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A box slides down a 30.0° ramp with an acceleration of 1.20 m/s^2. Determine the coefficient of kinetic friction between the box
Zielflug [23.3K]

m = mass of the box

N = normal force on the box

f = kinetic frictional force on the box

a = acceleration of the box

μ = coefficient of kinetic friction

perpendicular to incline , force equation is given as

N = mg Cos30                                         eq-1

kinetic frictional force is given as

f = μ N

using eq-1

f = μ mg Cos30    


parallel to incline , force equation is given as

mg Sin30 - f = ma

mg Sin30 - μ mg Cos30  = ma

"m" cancel out

a = g Sin30 - μ g Cos30

inserting the values

1.20 = (9.8) Sin30 - (9.8) Cos30 μ

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3 years ago
Highlight two factors which shows that heat from the sun does reach the earth's surface by convection.
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radiation

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2 years ago
A 120 kg box is on the verge of slipping down an inclined plane with an angle of inclination of 47º. What is the coefficient of
Alex_Xolod [135]

Given :

A 120 kg box is on the verge of slipping down an inclined plane with an angle of inclination of 47º.

To Find :

The coefficient of static friction between the box and the plane.

Solution :

Vertical component of force :

mg\ sin\ \theta =  120\times 10 \times sin\ 47^\circ{}=877.62 \ N

Horizontal component of force(Normal reaction) :

mg\ cos\ \theta =  120\times 10 \times cos\ 47^\circ{}=818.40 \ N

Since, box is on the verge of slipping :

mg\ sin\ \theta= \mu(mg \ cos\ \theta)\\\\\mu = tan \ \theta\\\\\mu = tan\ 47^o\\\\\mu = 1.07

Therefore, the coefficient of static friction between the box and the plane is 1.07.

Hence, this is the required solution.

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kondor19780726 [428]

Answer:

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r is the distance between the center of the two masses

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