Can I please get help on this one.

Samuel is preventing a boulder from rolling down a hill with a slope of 35°. If the boulder

Rashid [163]

Answer:

1,809.98 lb*m/s^2

Step-by-step explanation:

First, we want to know how much weight of the boulder is projected along the path in which the boulder can move.

The weight of the boulder is:

W = 322lb*9.8 m/s^2 = (3,155.6 lb*m/s^2)

This weight has a direction that is vertical, pointing downwards.

Now, we know that the angle of the hill is 35°

The angle that makes the direction of the weight and this angle, is:

(90° - 35°)

(A rough sketch of this situation can be seen in the image below)

Then we need to project the weight over this direction, and that will be given by:

P = W*cos(90° - 35°) = (3,155.6 lb*m/s^2)*cos(55°) = 1,809.98 lb*m/s^2

This is the force that Samuel needs to exert on the boulder if he wants the boulder to not roll down.

3 0

3 years ago

Find the system of inequalities which represents the shaded region on the coordinate plane.

dybincka [34]

Answer:

given, y≥2x+1 and

y>

2

1

x−1

first, draw the graph for equations y=2x+1 and y=

2

1

x−1

for y=2x+1

substitute y=0 we get, 2x+1=0⟹x=−0.5

substitute x=0 we get, y=1

therefore, y=2x+1 line passes through (0.5,0) and (0,1) as shown in fig.

Hence, y≥2x+1 includes the region above the line.

for y=

2

1

x−1

substitute y=0 we get,

2

1

x−1=0⟹x=2

substitute x=0 we get, y=−1

therefore, y=2x+1 line passes through (2,0) and (0,-1) as shown in fig.

Hence, y>

2

1

x−1 includes the region above the line.

the intersection region is the shaded region as shown in above figure which includes I, II and III quadrants.

Therefore, quadrant IV has no solution please make as brainlelist

7 0

3 years ago

An observer in a hot air balloon sights a building that is 50 m from the balloon's launch point. The balloon has risen 165 m. Wh

Kaylis [27]

Notice the picture,
recall your SOH, CAH, TOA
$\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad \qquad 
% cosine
cos(\theta)=\cfrac{adjacent}{hypotenuse}

\\ \quad \\
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}$

you have,
opposite side, 165
adjacent side, 50
and the angle

that means, we'll need Mrs. tangent
thus
$\bf tan(\theta)=\cfrac{opposite}{adjacent}\implies tan(\theta)=\cfrac{165}{50}
\\ \quad \\
 tan^{-1}\left[ tan(\theta) \right]=tan^{-1}\left[ \cfrac{165}{50} \right]
\\ \quad \\
\theta=tan^{-1}\left[ \cfrac{165}{50}\right]
\\ \uparrow \\
 \textit{angle of elevation}\iff\textit{angle of depression}$

4 0

3 years ago

If a1= 1 and an = -5an-1 then find the value of a6.

anyanavicka [17]

The answer is 6 (:::::)

4 0

3 years ago

Slope, Y intercepts, linear equations, ect

Rudiy27

To graph an equation with one x variable you first go on the y (vertical axis) and put a point at the number that doesnt have the x (12). Sometimes this number could be zero. The +12 is where it crosses the y axis. next to make the line you look at the number infront of x (1). that is the slope. It is basically 1 over 1. this means you go up one and right one, up one and right one, or down one and left one etc. this will make a line. If the number infront of x was -2 then it is seen as -2 over one. this slope is drawn by going down one and right one or up two and left one

5 0

4 years ago