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PtichkaEL [24]
3 years ago
14

Carlos is playing darts. He throws a dart at the bullseye, but it hits the outer ring of the dartboard instead. He aims for the

bullseye again and hits it with a second dart. Carlos threw three more darts which landed close to his first shot. What does the second throw indicate about his accuracy and precision? The second throw shows poor accuracy but high precision. The second throw shows accuracy but poor precision. The second throw shows both accuracy and precision. The second throw shows poor accuracy and poor precision.
Physics
1 answer:
xz_007 [3.2K]3 years ago
6 0

Answer:

The second throw shows both accuracy and precision.

Explanation:

The second throw shows both accuracy and precision because he knew in which degree and the amount of force that can be applied to the dart so that it hits the center of the bullseye or close to it. Due to knowing of these two things the Carlos is able to hit the dart close to the target at the second time so in this way he got both accuracy and precision at the second turn of throwing darts.

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Consider a large truck carrying a heavy load, such as steel beams. A significant hazard for the driver is that the load may slid
nataly862011 [7]

Answer:

A)

the minimum stopping distance for which the load will not slide forward relative to the truck is 14 m

B)

data that were not necessary to the solution are;

a) mass of truck and b) mass of load

Explanation:

Given that;

mass of load m_{LS} = 10000 kg

mass of flat bed m_{FB} = 20000 kg

initial speed of truck v_{0} = 12 m/s

coefficient of friction between the load sits and flat bed μs = 0.5

A) the minimum stopping distance for which the load will not slide forward relative to the truck.

Now, using the expression

Fs,max = μs F_{N}     -------------let this be equation 1

where F_{N} = normal force = mg

so

Fs,max = μs mg

ma_{max} = μs mg

divide through by mass

a_{max} = μs g    ---------- let this be equation 2

in equation 2, we substitute in our values

a_{max} = 0.5 × 9.8 m/s²

a_{max} = 4.9 m/s²

now, from the third equation of motion

v² = u² + 2as

v_{f}² = v_{0}² + 2aΔx

where v_{f} is final velocity ( 0 m/s )

a is acceleration( - 4.9 m/s² )

so we substitute

(0)² = (12 m/s)² + 2(- 4.9 m/s² )Δx

0 = 144 m²/s² - 9.8 m/s²Δx

9.8 m/s²Δx = 144 m²/s²

Δx = 144 m²/s² /  9.8 m/s²

Δx = 14 m

Therefore, the minimum stopping distance for which the load will not slide forward relative to the truck is 14 m

B) data that were not necessary to the solution are;

a) mass of truck and b) mass of load

3 0
3 years ago
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