Answer:
Acceleration, 
Explanation:
Given that,
Mass of Suzie, m = 76 kg
Initial speed of Suzie, u = 50 mph = 22.35 m/s
Finally, it stops, v = 0
Time, t = 1.2 s
Let a is the Suzie’s average acceleration. It is equal to the rate of change of velocity. Its formula is given by :
So, the average acceleration of Suzie’s is 
and she is decelerating. Hence, this is the required solution.
The Iroquois depended on the natural resources around them to meet all of their basic needs. Because they lived in the Eastern Woodlands of North America, their food, clothing, and shelter, as well as the materials for making their tools and weapons.Hunting tools were a bow and arrow, tomahawks and clubs. They ate corn, moose, beans and deer and many other foods. Beaver skin was used for capes.
They used reeds to make baskets and they used animal hides as clothing.
The Iroquois made weapons out of bones, wood, stones, and feathers. The Iroquois used wood, stone and feathers to make arrows.Hunting tools were a bow and arrow, tomahawks and clubs.
Answer:
That is true.
Explanation:
When the input force is applied to the wheel, as it is with a doorknob, the axle turns over a shorter distance but with greater force, so the mechanical advantage is greater than 1.
This question is solved using an available similar problem as data provided for the forces was not given.
Repeat the same steps outlined for your problem.
Regards.
Answer:
F = 1.598 KN , Q = 90 degree (+ y-axis)
Explanation:
Sum of Forces in x-direction to the left (+)
2 cos (30) + 3cos (60) + F*cos (Q) = F_a ..... 1
Sum of Forces in y-direction to the up (+)
2 sin (30) + F*sin (Q) - 3 sin (60) ...... 2
Using Eq 2 and solve:
F*sin (Q) = 1.598 KN
F_min when sin (Q) is max, max possible value of sin(Q) = 1 @ Q = 90 degrees.
Hence,
F_min = 1.598 KN
Using Eq 1 @ Q = 90 degrees and F = 1.598 KN:
F_a = 2 cos (30) + 3cos (60) = 3.2 KN
Answer:
42.3 MV
Explanation:
d = diameter of the metal sphere = 2.15 m
r = radius of the metal sphere
diameter of the metal sphere is given as
d = 2r
2.15 = 2 r
r = 1.075 m
Q = charge on sphere = 5.05 mC = 5.05 x 10⁻³ C
Potential near the surface is given as
V = 4.23 x 10⁷ volts
V = 42.3 MV
