The orbital period of Jupiter is 11.86 years. What is its distance from the sun?

an archer stands 40.0m from the target. if the arrow is shot horizontally with a velocity of 90.0 m/s, how far above the bull's

pshichka [43]

Ok, first we need to have in mind that to findthe distance we need to find also the time then we use the kinematic equations in the x-direction

delta x = Vox*t + 1/2*a*t^2 
since there is no horizontal acceleration, a = 0 m/s/s 
delta x = Vox*t 
40.0 m = 90 m/s * t 
t = 4/9 s 

Now using this time you can calculate the vertical displacement due to gravity. 
delta y = Voy*t + 1/2*g*t^2 
since there is no initial vertical velocity, Voy = 0m/s 
delta y = 1/2*g*t^2 
delta y = 1/2*(-9.81 m/s/s)*(4/9 s)^2 
solve for delta y and you will have the answer

3 0

3 years ago

Three forces act on an object. a 3 n force acts due east and a 8 n force acts due north. if the net force on the object is zero,

Bingel [31]

The third force C is in the opposite direction of the sum of the first two forces, A+B, but the magnitude is the same as A+B

Since A & B are at right angles we can just use Pythagoras and trig:

A+B magnitude = sqrt(6^2 + 3^2) = 6.71

A+B angle = arctan(6/3) deg north of east = 63.4 deg (first quadrant)

So C magnitude = 6.71 (same as A+B)

[C angle = 63.4 deg south of west (third quadrant) = -180+63.4 deg polar = -116.6 deg

6 0

3 years ago

A dart is thrown from 1.50 m high at 10.0 m/s toward a target 1.73 m from the ground. At what angle was the dart thrown?

Triss [41]

Answer:

The angle of projection is 12.26⁰.

Explanation:

Given;

initial position of the dart, h₀ = 1.50 m

height above the ground reached by the dart, h₁ = 1.73 m

maximum height reached by the dart, Hm = h₁ - h₀ = 1.73 m - 1.50 m= 0.23 m

velocity of the dart, u = 10 m/s

The maximum height reached by the projectile is calculated as;

$H_m = \frac{u^2sin^2 \theta}{2g}$

where;

θ is angle of projection

g is acceleration due to gravity = 9.8 m/s²

$H_m = \frac{u^2sin^2 \theta}{2g}\\\\sin^2 \theta = \frac{H_m \ \times \ 2g}{u^2} \\\\sin^2 \theta = \frac{0.23 \ \times \ 2(9.8)}{10^2} \\\\sin ^2\theta =0.04508\\\\sin \theta = \sqrt{0.04508} \\\\sin \theta = 0.2123\\\\\theta = sin^{-1}(0.2123)\\\\\theta = 12.26^0$

Therefore, the angle of projection is 12.26⁰.

6 0

2 years ago

When you exert 75 N on a jack to lift a 6000 N car, what is the jack’s actual mechanical advantage? Show your work.

professor190 [17]

Answer:

80

Explanation:

the mechanical advantage is the ratio of the load to the effort so it doesn't have units.to calculate it you use the formula

mechanical advantage=load/effort

in this case the load is 6000N and the effort is 75N

Ma=6000/75

=80

I hope this helps

3 0

3 years ago

Which feature of the sun is a section that is cooler than its surroundings?

Snowcat [4.5K]

Sunspot. Sunspots are temporary cooler areas on the Sun's surface caused by changes in its magnetic field.

7 0

4 years ago

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