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chubhunter [2.5K]
3 years ago
10

A force of 10 N making an angle 30 with horizontal .its horizontal component wll be​

Physics
1 answer:
Norma-Jean [14]3 years ago
7 0

Answer:

10 cos (30)N

Explanation:

So, if we were to visualize it for a sec

we would see 10 as the hypotenuse, 30 as the angle between and would be solving for the adjacent side

So, using basic trig rules,

using the fact that cos(30)=x/10

x=10 cos (30)N

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Find the resultant of the following forces: 3.0N east, 4.0N west, and 5.0N in a direction north 60° west.
zzz [600]

Answer:

5.5 N at 50.8° north of west.

Explanation:

To find the resultant of these forces, we have to resolve each force along the x- and y-direction, then find the components of the resultant force, and then calculate the resultant force.

The three forces are:

F_1=3.0 N (east)

F_2=4.0 N (west)

F_3=5.0 N (at 60° north of west)

Taking east as positive x-direction and north as positive y-direction, the components of the forces along the 2 directions are:

F_{1x}=3.0 N\\F_{1y}=0

F_{2x}=-4.0 N\\F_{2y}=0

F_{3x}=-(5.0)(cos 60^{\circ})=-2.5 N\\F_{3y}=(5.0)(sin 60^{\circ})=4.3 N

Threfore, the components of the resultant force are:

F_x=F_{1x}+F_{2x}+F_{3x}=3.0+(-4.0)+(-2.5)=-3.5 N\\F_y=F_{1y}+F_{2y}+F_{3y}=0+0+4.3=4.3 N

Therefore, the magnitude of the resultant force is

F=\sqrt{F_x^2+F_y^2}=\sqrt{(-3.5)^2+(4.3)^2}=5.5 N

And the direction is:

\theta=tan^{-1}(\frac{F_y}{|F_x|})=tan^{-1}(\frac{4.3}{3.5})=50.8^{\circ}

And since the x-component is negative, it means that this angle is measured as north of west.

7 0
3 years ago
What is the area of space around a magnet called?
erma4kov [3.2K]

Answer:

These acts of attraction and repulsion are called “magnetism”, and the magnetic space around a magnet is called the “magnetic field”

Explanation:

8 0
3 years ago
2. Fracture mechanics. A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plan
ikadub [295]

Answer:

the critical flaw is subject to detection since this value of ac (16.8 mm) is greater than the 3.0 mm resolution limit.  

Explanation:

This problem asks that we determine whether or not a critical flaw in a wide plate is subject to detection  given the limit of the flaw detection apparatus (3.0 mm), the value of KIc (98.9 MPa m), the design stress (sy/2 in which s y = 860 MPa), and Y = 1.0.

ac=1/\pi (\frac{Klc}{Ys} )^{2}\\ ac=1/\pi(\frac{98.9}{(1)(860/2)} )^{2}\\  ac=0.0168m\\ac=16.8mm

Therefore, the critical flaw is subject to detection since this value of ac (16.8 mm) is greater than the 3.0 mm resolution limit.  

5 0
4 years ago
Power is the product of_____.
tia_tia [17]
Heya user☺☺

All options are wrong here.

The correct answer is..

Work/Time.

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3 0
3 years ago
The Steamboat Geyser in Yellowstone National Park shoots water into the air at 48.0 m/s. How
qwelly [4]

Answer:

The maximum height reached by the water is 117.55 m.

Explanation:

Given;

initial velocity of the water, u = 48 m/s

at maximum height the final velocity will be zero, v = 0

the water is going upwards, i.e in the negative direction of gravity, g = -9.8 m/s².

The maximum height reached by the water is calculated as follows;

v² = u² + 2gh

where;

h is the maximum height reached by the water

0 = u² + 2gh

0 = (48)² + ( 2 x -9.8 x h)

0 = 2304 - 19.6h

19.6h = 2304

h = 2304 / 19.6

h = 117.55 m

Therefore, the maximum height reached by the water is 117.55 m.

7 0
3 years ago
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