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Licemer1 [7]
3 years ago
13

Question 2 of 10 Which illustration represents low accuracy and low precision?

Physics
1 answer:
Zina [86]3 years ago
6 0

Answer:

D is the correct answer

Explanation:

A) shows low accuracy and high precision since they missed the bull's-eye but are all grouped together.      

B) Shows high accuracy and high precision

C) Shows high precision

D  is the only one that shows both low accuracy and low precision

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A 5 N force pushes on the right side of a box. At the same time, a 10 N force pushes on the left side of the box. What happens t
olya-2409 [2.1K]

The answer is C as there is more force on the left side ( excess of 5 N) which therefore pushed it to the right with a force of 5 N!


8 0
3 years ago
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A rock weighing 98 newtons is pushed off the edge of a bridge 50 meters above the ground. What was the potential energy of the r
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Ep = 4900 because Ep = wh
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Answer the following in 1-2 complete sentences:
mojhsa [17]

Answer:

The graph with a more steeper slope

Explanation:

If the x axis represented time and the y axis represented speed the steeper the slope the more faster something was moving in a shorter amount of time. On the other hand a less steeper slope means an object is gradually picking up speed in a longer amount of time thus makeing it slower

HOPE THIS HELPED:)

8 0
3 years ago
The stopping sight distance for a car traveling at 50 mph is 461 ft (including both perception-reaction distance and braking dis
Ivan

Answer:

2.08 s

Explanation:

We are given that

Speed,v=50mph=73.3ft/s

1 mile=5280 feet

1 hour=3600 s

Distance,d=461 ft

t=2.5 s

v'=60 mph=88 ft/s

We have to find the perception reaction time.

Perception reaction distance=d_p=vt=73.3\ti es 2.5=183.25 feet

d_p=d'_p

d'_p=v't

t'=\frac{183.25}{88}=2.08 s

4 0
3 years ago
Sarah throws a ball directly upward at the edge of a cliff with a starting velocity of 3.0 m/s
strojnjashka [21]

Answer:

The correct answer is t = 0.92s

Explanation:

Initial velocity v​0 = 3.0 m/s

Displacement Δy = ?

Acceleration a = -9.8m/s2

Final velocity v = -6.0m/s

Time t=? Target unknown

We can use the kinematic formula missing Δy to solve for the target unknown t:

V=v0+at

We can rearrange the equation to solve to t:

V-v0=at

t= v-v0/a

Substituting the known value into the kinematic formula gives:

t= (-6.0m/s)-(3.0m/s)

————————————

-9.8m/s2

= -9m/s

—————-

-9.8m/s2

=0.92s

6 0
3 years ago
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