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3 years ago

How would you graph/sketch 40 lbs at pi/3 south of west

1 answer:

7 0

Well, pi/3 in Radians is equal to 60 degrees, and if your South is negative y and West is negative x, your line segment will be 60 degrees down of left, if you will. Your question is a little confusing, but I'm assuming you're drawing a force diagram, so probably draw the line segment at that angle and label it with the force of gravity- 40 lbs Hope that helps.

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The phrase "hot air rises" describes which type of heat transfer?

77julia77 [94]

Answer:

Option C

Explanation:

C. Convection...

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3 years ago

Read 2 more answers

A car speeds over a hill past point A, as shown in the figure. What is the maximum speed the car can have at point A such that i

Lubov Fominskaja [6]

Answer:

11.8 m/s

Explanation:

At the top of the hill, there are two forces on the car: weight force pulling down (towards the center of the circle), and normal force pushing up (away from the center of the circle).

Sum of forces in the centripetal direction:

∑F = ma

mg − N = m v²/r

At the maximum speed, the normal force is 0.

mg = m v²/r

g = v²/r

v = √(gr)

v = √(9.8 m/s² × 14.2 m)

v = 11.8 m/s

3 0

3 years ago

Please Help Me

JulijaS [17]

1. a=Δv/Δt=(v-vo)/t=(0-25)/5=-25/5=-5 m/s²

The "-" sign shows us that the car has a slow motion

2.a=Δv/Δt=(v-vo)/t=(10-0)/4=10/4=2,5 m/s²

3.they do not have acceleration because they go at constant speed

7 0

4 years ago

assume that the initial speed is 25 m/s and the angle of projection is 53 degree above the hroizontal. the cannon ball leaves th

finlep [7]

Answer:

A. xmax = 131.49 m

B. t = 8.74 s

C. ymax = 220.33 m

Explanation:

A. In order to find the horizontal distance which cannon travels you first calculate the flight time. The flight time can be calculated by using the following formula:

$y=y_o+v_osin\theta-\frac{1}{2}gt^2$ (1)

yo: height from the projectile is fired = 200m

vo: initial velocity of the projectile = 25m/s

g: gravitational acceleration = 9.8 m/s^2

θ: angle between the direction of the initial motion of the ball and the horizontal = 53°

t: time

You need the value of t when the projectile hits the ground. Then, in th equation (1) you make y = 0m.

When you replace the values of all parameters in the equation (1), you obtain the following quadratic formula:

$0=200+(25)sin53\°t-\frac{1}{2}(9.8)t^2\\\\0=200+19.96t-4.9t^2$ (2)

You use the quadratic formula to obtain the value of t:

$t_{1,2}=\frac{-19.96\pm\sqrt{(19.96)^2-4(-4.9)(200)}}{2(-4.9)}\\\\t_{1,2}=\frac{-19.96\pm65.71}{-9.8}\\\\t_1=8.74s\\\\t_2=-4.66s$

You use the positive value because it has physical meaning.

Now, you can calculate the horizontal range of the projectile by using the following formula:

$x_{max}=v_ocos\theta t$

$x_{max}=(25m/s)(cos53\°)(8.74s)=131.49m$

The cannon ball travels a horizontal distance of 131.49 m

B. The cannon ball reaches the canon for t = 8.74s

C. The maximum height is obtained by using the following formula:

$y_{max}=y_o+\frac{v_o^2sin^2\theta}{2g}$ (3)

By replacing in the equation (3) the values of all parameters you obtain:

$y_{max}=200m+\frac{(25m/s)^2(sin53\°)^2}{2(9.8m/s^2)}\\\\y_{mac}=200m+20.33m=220.33m$

The maximum height reached by the cannon ball is 220.33m

3 0

3 years ago

6. Which of the following best describes the situation?

Alla [95]

Answer:

Explanation:

the man is pull the box that is the force and the box rubbing against the ground causes friction

7 0

3 years ago